35 This entry had its birth in my fooling around with my 1966 attempt to determine the length of 1/4 of a circle's circumference without Pi. tpo = term position sqrt = square root sq = square nnss = natural number summation sequence tpo0 0/4 = 0 (0/4 = 0) + 1 = 1 - tpo1 4/4= 1 8/4 = 2 8 + 1 = 9 2 - 1 = 1 2 x 2 3 x 3 - tpo2 16/4 = 4 24/4 = 6 24 + 1 = 25 6 - 4 = 2 4 x 4 5 x 5 - tpo3 36/4 = 9 48/4 = 12 48 + 1 = 49 12 - 9 = 3 6 x 6 7 x 7 - tpo4 64/4 = 16 80/4 = 20 80 + 1 = 81 20 - 16 = 4 8 x 8 9 x 9 - tpo5 100/4 = 25 120/4 = 30 120 + 1 = 121 30 - 25 = 5 10 x 10 11 x 11 - tpo6 144/4 = 36 168/4 = 42 168 + 1 = 169 42 - 36 = 6 12 x 12 13 x 13 - tpo7 196/4 = 49 224/4 = 56 224 + 1 = 225 56 - 49 = 7 14 x 14 15 x 15 - tpo8 256/4 = 64 288/4 = 72 288 + 1 = 289 72 - 64 = 8 16 x 16 17 x 17 - tpo9 324/4 = 81 360/4 = 90 360 + 1 = 361 90 - 81 = 9 18 x 18 19 x 19 Am letting tpo0 be compressed. Thereafter, each tpo visibly consists of two terms displaying division by "4". As can be seen, an sq results in the first term of each tpo. In an odd tpo, an odd sq; in an even tpo, an even sq. Am defining each set by the results in it: odd-even set and even-even set. General: - 1) Each set's tpo = the difference between the division-by-4 results for that set. 2) The sum of each set's results divided by its tpo = the sqrt of the square for that set's 2nd term. 3) The square for a set's 2nd term is its intial integer + 1. 4) Excluding 0, the sum of each set's results = every 2nd term in a paired natural number summation sequence. 5) Including 0, summing the division-by-4 result of the 2nd term of a set with the division-by-4 result of the following set's 1st term = every 1st term in a paired natural number summation sequence. 6) tpo times (tpo + 1) = division-by-4 result for the 2nd term in that tpo's set. 7) In each set/ the first term's division-by-4 result = the square of that set's tpo. 8) Deriving from its intial integer, the 1st sqrt in each set is always an even integer. Therefore, if dividing its square by "4" equals an even integer, it is the first square root/square term in an even- even tpo set. Therefore the difference between the (2nd sq - 1) and the first square/ will be an even integer, and that integer will be that set's tpo. 9) "2" times a tpo = the square root of that tpo's initial integer. Still, how can one know when an integer is a square. Using the above, one knows immediately that when the initial integer in the first term of a tpo is divided by "4"/ the result is also a square. One knows too that the sqrt of that sq = that set's tpo. Further, one knows that an odd sq - 1 is evenly divisible by eight, and that the resulting integer is a term in the natural number summation sequence. Here is an example I think might be helpful. I got it from the 3.14159265 expansion of Pi; so I already know the answer. 8464/4 = 2116 2116/4 = 529 528/8 = 66 Depending on how the natural number summation sequence is treated, "66" is either the 11th or 12th term in that sequence. Admittedly, it is easier to ignore 0 and see "66" as the 11th term, but what one is trying to do might not make that the best choice. 1 3 6 10 15 21 28 36 45 55 66 78 So, the squares are proved, but what is the square root of 529 and what is the square root of 2116 and what is the square root of 8464? The square root of 2116 equals the tpo where the initial integer is 8464. 2 times the sqrt of 529 = the sqrt of 2116, and 2 times the sqrt of 2116 = the sqrt of 8464. I'm taking a sabbatical. Okay, PM 8 is drawing near, and here is the setup: 1 2 3 wherein each integer is a square root in the first group. Am approaching the problem as if I do not know anything. The even square root rules, but first this: the infinite set of odd integers runs between the squares, connecting one to the next. And, yes, I am leaving 0 behind. About those odd integers: 1) the 1st one is "1" less than 2 x the even square root - 2) the 2nd one is "1" more than 2 x the even square root - 3) if an odd integer is the starting point, then that integer + 1 = a sum which when divided by "2" = the square root of the square it is connecting to - - - Back to 1 2 3: new look: 1 x 1 = 1 2 x 2 = 4 3 x 3 = 9 because 1 + 3 = 4 and 4 + 5 = 9. Note also that 3 + 5 = 8 as does 4 + 4; and 8 into 8 = 1. [nnss] Next group: 3 4 5 - the connecting odd integers here are 7 and 9 wherein (7 + 1)/2 = 4 and (9 + 1)/2 = 5. 3 x 3 = 9 4 x 4 = 16 5 x 5 = 25 because 9 + 7 = 16 and 16 + 9 = 25. Remember also (because I forgot) that (2 x the even sqrt) - 1 = the 1st connecting odd integer and (2 x the even sqrt) + 1 = the 2nd connecting odd integer. That's the method, but it still does not reveal the answer I am seeking. The odd integer connecting 9 x 9 to 100 is 19. (19 + 1)/2 = 10 and 2 x 10 = 20. Another way to look at 10 x 10 is 5 x 20, and 20 is important here in that (9 x 9) - 1 = 80 and (11 x 11) - 1 = 120. 20 + 20 = 40 = 19 + 21. [ A further point of interest which applies generally is: (2 x the lesser odd sqrt) + 1 = the connecting integer related to it, and (2 x the greater odd sqrt) - 1 = the integer related to it. ] This might be useful if only a square root is known. Back to 8464 and the paired nnss terms and 66 (term 11). That 66/6 = 11 is interesting, but that 11 x 2 = 22 may be more than just interesting: 22 + 22 = 44 = 21 + 23 44 x 2 = 88 529 - 88 = 441 440/8 = 55 (term 10 in the nnss). This says that 23 x 23 = 529, and that 46 x 46 = 2116, and that 46 is the tpo, and that 92 x 92 = 8464. By the way, 22 x 22 = 484, and 45 is the integer that connects 484 to 529, and 45 + 1 = 46, and 46/2 = 23. May 26, 2008 addenda: 10) Regarding 2nd term in each two-term set: (odd sq - 1)/tpo = 4 x (tpo + 1) and (odd sq - 1)/4 = tpo x (tpo + 1) and (odd sq - 1)/4 divided by (tpo + 1) = tpo. 11) Regarding 1st term in each two-term set: sq/4tpo = tpo. # Brian A. J. Salchert
Thinking Lizard
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- brian (baj) salchert
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Saturday, May 24, 2008
sw00903math--squares-sets-from-4-divisor
Thursday, May 22, 2008
sw00902v-OdysseusSongs10
Venturings "Interlude" The leaf of the sea with its serrated edge and its surface roughened and cracking and its shadowy veins and veins of white locked in a dance rocks on the earth like the tormented spirit of someone dying or dead The skin of the sea with its curved horizon trembles like the head of a drum or a shook note or a frightened finger whose quaking cannot so much be seen as felt, the curious blessings of distance in our eyes. I see, my friends, though I see not; my ears challenge the ears of birds; the good and evil that ride in the winds I know as well as a teller of truth; my eager tastebuds imagine the food of gods. - Brian A. J. Salchert
sw00901v-OdysseusSongs09
Venturings "Under the Cyclops" Giant's eye: a shot in the light to put in the dark. Polyphemus, he's no genius, ho, ho, ho. O, Polyphemus, O, O, Polyphemus, ate of my men, two, four, six, ten, ho, ho, ho. Now Polyphemus cannot find us in the wool. So long, monster, we know you're glad to see us go. - Brian A. J. Salchert
sw00900v-OdysseusSongs08
Venturings "Trials" One-eyed nightmare devourer of men Island witch wander of men into swine Voices of illusion entrancers Envious men of Odysseus opening the sack from Aeolus in sight of your homeland's stubble fires my head lowers loosers of evil winds Rock whirlpool batterer anemone of water Ghosts in the land of constant dark revealers of sorrows Fair Calypso's Ogygian strander of a man - Brian A. J. Salchert
sw00899v-OdysseusSongs07
Venturings "Praying by the Sea" Swimming in the gray dawn, Telemachus, his fingers combing the waves, shifts his shaking body, crouched on a rock, and calls to Pallas Athena; (music depicting call for courage) she answers him. (music depicting the giving of courage) - Brian A. J. Salchert
sw00898v-OdysseusSongs06
Venturings "Preparing to Find Odysseus" Overhead, two eagles, fiercely fighting, Halitherses, reader of omens, warns: Though twenty years absent, Odysseus, to the sorrow of some, will return. And Telemachus, by Minerva urged and aided, makes ready to find his father, and his faithful mother, Penelope, to spite her brash suitors, leave behind. Let the sun be your food, the wind your clothes, the earth your shelter, and the water, let the water answer your heart. But the suitors, led by Eurymachus, having so long vied for this prize, this Penelope, this woman rare, are not about to go home. So Eurymachus chides Halitherses and tells him they fear no man and will continue to live off Telemachus until Penelope quits her game. Boom boom boom Eee-tool eee-ew-tool eee-tool Telemachus, then, addressing them and asking for a ship and twenty men to look here and there for his father: Sparta, Pylos, wherever his spirit leads, says that for another twelve months he will suffer their wasting if he hears his father's alive; but if not, will quickly come back. Let the sun be your food, the wind your clothes, the earth your shelter, and the water, let the water answer your heart. And will ritualize Odysseus' death as is proper and as he desires, and build to his memory a barrow, and command his mother to marry again. But the suitors, through Leiocritus, do not believe he will leave; and his Mentor, having railed the Ithacans for fearing to care, put down, as a moon that would brighten a day. Boom boom boom Eee-tool eee-ew-tool eee-tool - Brian A. J. Salchert
sw00897v-OdysseusSongs05
Venturings "His Presence" Now the mightiest gods watch and allow Odysseus trials while Telemachus, his one and glorious son, by Minerva made most comely, strides forth with his sword and dogs, and marvels all. - Brian A. J. Salchert
sw00896v-OdysseusSongs04
Venturings "Passages" Aeneas gone the dream of someone else I image how Odysseus and his foolish men foolish because they ate of the cattle Of Hyperion god of the sun and so not allowed to get back home across the proud water The Trojans gone their dreams their bloodied lives I image how Old Neptune shakes Odysseus' prows to revenge for his son Polyphemus the loss of his one eye to crack the Achaeans' hearts crack them mad across the proud water The long war gone dreams answered dreams denied I image how the gods and men twisted by fates by whims attempt to straighten to begin anew vainly strive to become better than today better than they are across the proud water dance as the moon dance as the sun dance as the fish-hunting gulls across the proud water sream in the storms weep in the winds die in the deserts of hopelessness across the proud water - Brian A. J. Salchert
sw00895v-OdysseusSongs03
Venturings "Singing the Man" Odysseus Odysseus over the war and its wounds you are the man who with this seventh rising my heart and the gods have drawn to the center of my myth you are the man protected in battle while so many died around you who bolstered every Achaean by your firm stance and the deep green of your words and whose cunning kept you well and will keep you through journeys not even I yet know who with this voice chant to any who will hear me across our Aegean and all the waters of angry Neptune as Dawn now in her saffron robes disappears from us and you Odysseus - Brian A. J. Salchert
sw00894v-OdysseusSongs02
Venturings "Remembering" On this jutting of rock, my hair weaving, I watch for this seventh day in a row how the rosy-fingered Dawn floats on our cold Aegean in my dark eyes O the exploits of that Trojan honorable and proud Hector the light of Priam And the flashing of spears before me hold my aged face and the clever and patient Achaeans move my aged voice O the glories of Achilles blessed and unblessed shining in his dream Birds of prey - Brian A. J. Salchert
sw00893v-OdysseusSongs01
Re-found these ten circa 1983 pieces yesterday. Am including them as a section in Venturings. Odysseus Songs Entrance Part I: Odysseus Remembering Singing the Man Passages Part II: Telemachus His Presence Preparing to Find Odysseus Praying by the Sea Part III: Odysseus at Sea Trials Under the Cyclops Interlude ~ ~ ~ "Entrance" Leaving his woman, his servants, his son, and his carefully acquired possessions, Odysseus, to help defend Achaean honor by rescuing Helen from Priam's Paris, gathering the warriors of his realm, sails the unpredictable waves of times not come of water and gods and men: sails to Troy. But a strong spirit can live through any loved one's absence, feel the winds of sadness in his bowels, and still look up; and clever and wise, still make his dreams do, still move out and on and in. - Brian A. J. Salchert
Wednesday, May 21, 2008
sw00892math--tpo-double-view-paired-6nET
34 ET = Elimination Table Explanation of entry title: tpo means term position - this is the first time I am using two tpo streams together - the reason is/ I will be showing both the usual integer results of a 6n Elimination Table (all integers "3" does not divide into evenly) and the integers required using the formula 1 + 6n to get the squares of the usual integer results; BUT this will be done in both cases through the use of magic numbers. Therefore, instead of dealing with each usual integer result separately, the minor result and its related major result will be dealt with as a pair, and so also the integers which when multiplied by "6" will get those respective products needed to get the respective squares (once + 1 has been added) for the pair's members. The intial minor member is "5" and the initial major member is "7". Their respective integers to be multiplied by "6" are "4" and "8". A single tpo (in this case tpo1) will suffice for both magic number streams. Showing this, however, might be a bit tricky. Was working on this last night, and went through several changes, but this is how it began (without showing "1"): The times 6 squares row: 4 8 20 28 48 60 88 104 140 160 The 6n minor and major square root row: 5 7 11 13 17 19 23 25 29 31 Multiples of 6 apply throughout, but only in the first set is that multiple the same integer: "1". So, each tpo serves two functions: one for the square roots stream and another for the squares stream. Have concluded that envisioning 1 x (1 x 6) = 6, 2 x (2 x 6) = 24, 3 x (3 x 6) = 54, 4 x (4 x 6) = 96, 5 x (5 x 6) = 150, and so on, is the best way to see each tpo. These are my reasons: 6 -1 = 5 and 6 + 1 = 7, and 1 + 1 = 2 so that 6 - 2 = 4 and 6 + 2 = 8. Just don't forget that the tpo# represents the n for "6" in the square roots stream. This then is how it works: tpo4 nets 24-1 (23) and 24 + 1 (25) for the square roots, wherein 24 is the magic number in that stream; and 24 x 4 (96) from which 4 + 4 must be subtracted (88) and to which 4 + 4 must be added (104) for the squares, wherein 96 is the magic number in that stream, and once + 1 is added to 6 x 88 the resulting integer will be the square of 23, and once + 1 is added to 6 x 104 the result will be the square of 25. The magic numbers for the first five pairs are: 6 and 6, 12 and 24, 18 and 54, 24 and 96, and 30 and 150. Knowing the tpo# isall that is necessary as long as the handling of it uses the approaches shown above. Each term position represents a set of four integers. # Brian A. J. Salchert
Monday, May 19, 2008
sw00891sl14--like-1-2-3
Like 1 2 3 Like 1 2 3 like 1 2 3 it's over the fence like 1 2 3 Like 1 2 3 like 1 2 3 it's over the fence like 1 2 3 (change melody) Like 1 2 3-e like 1 2 3-e it's over the fence like 1 2 3 Like 1 2 3-e like 1 2 3-e it's over the fence like 1 2 3 (Repeat entire lyrics as often as you like.) Brian A. J. Salchert 2008-05-19
Sunday, May 18, 2008
sw00890math--Goldbach-conjecture-fiddlings
33 Every even number is the sum of two primes. Since "2" is a prime number, "4" is the first even number, and since 2 + 2 = 4 and 3 + 3 = 6 is allowed, that is how my fiddlings will be. PM 9:01 - Have already changed my mind on this. 3 + 5 = 8 and 5 + 7 = 12 are the only solutions for 8 and 12. 3 + 7 = 10 5 + 7 = 12 3 + 11 = 14 3 + 13 = 16 7 + 11 = 18 7 + 13 = 20 3 + 19 = 22 11 + 13 = 24 3 + 23 = 26 11 + 17 = 28 13 + 17 = 30 13 + 19 = 32 11 + 23 = 34 17 + 19 = 36 19 + 19 = 38 17 + 23 = 40 19 + 23 = 42 13 + 31 = 44 17 + 29 = 46 19 + 29 = 48 13 + 37 = 50 23 + 29 = 52 23 + 31 = 54 13 + 43 = 56 29 + 29 = 58 17 + 43 = 60 19 + 43 = 62 23 + 41 = 64 13 + 53 = 66 31 + 37 = 68 23 + 47 = 70 29 + 43 = 72 31 + 43 = 74 17 + 59 = 76 31 + 47 = 78 37 + 43 = 80 23 + 59 = 82 41 + 43 = 84 7 + 79 = 86 41 + 47 = 88 43 + 47 = 90 19 + 73 = 92 41 + 53 = 94 13 + 83 = 96 37 + 61 = 98 - 0 pattern 10 through 100: 3 7 7 3 3 7 7 3 3 7 7 3 3 7 7 3 3 7 7 3 2, 4, 6, 8 patterns not yet defined, but there are final-digit limitations. At 112/ 41 + 71 or 29 + 83, the point being that both combinations work with f-d-2 sums. At 114/ 53 + 61 and 7 + 107 work. At 116/ 3 + 113 and 7 + 109 work. At 118/ 17 + 101 and 29 + 89 work. 118 is also 59 + 59. As to 0, the pattern I have shown is not the only possible one since f-d-1 and f-d-9 sum to 0, but it is the only possible one--if 5 + 5 is excluded--before 11 + 19. - 47 + 53 = 100 43 + 59 = 102 43 + 61 = 104 23 + 73 = 106 47 + 61 = 108 43 + 67 = 110 AM 9:27 5-19-2008 - The above merely begins to prove Goldbach's conjecture, and I actually don't think that is why I am doing this. Curiosity is the real reason. However, this fiddling might be a backdoor way of uncovering a prime number fact or two through final digits. Consider these: 1 and 9 can work for 0 here and there, but 3 and 7 might work for 0 infinitely. 5 and 7 work for 2 once only and where no other primes can, but otherwise 1 and 1 works for 2, and 3 and 9 might work for 2 consistently. 2 and 2 works for 4 once only and where no other primes can, but otherwise 7 and 7 works for 4, and 1 and 3 work for 4. 3 and 3 works for 6 where it alone can work, but also beyond, and 7 and 9 also work for 6. 3 and 5 work for 8 where they alone can work, but beyond "8" 1 and 7 work along with 9 and 9. - I think these limits are meaningful. # Brian A. J. Salchert
Friday, May 16, 2008
sw00889a--math-project
On April 24th I began a math project which, with the entry initiated yesterday and completed today, has expanded to 13 entries. I have substantially done what I set out to do, and so do not expect any new entries. My core goal was to provide a paired twin primes proof, not for the infinity of them, but for where they are and how to find them. The basics for this are simple, but the details/ are somewhat complex. Did not count them, but I combined a number of approaches. I do not do math the way professional mathematicians do, partly because I am not comfortable with how they do math. I invent my own terminologies and methods, yet I strive to be rigorous, and use examples and define my terms. A GRE I diligently prepared for/ and took at the University of Florida in 1984 was the goad behind all my math investigations since. Poetry and mathematics have become my central interests. In both disciplines I am not your usual venturer. from Mantis 5 Stanford University Bob Perelman A Discussion: Poetry and Discipline ~ # Brian A. J. Salchert 2008-05-16
Thursday, May 15, 2008
sw00888math--DFnote-and-squares-da-patterns
32 DF = Descension Family da = digit addition nnss = natural number summation sequence DF note: As God would have it, when I began unstacking the high pile of manilla folders holding writings of mine from my computer stool, the first group did not separate as normal, revealing a sheet of lined paper on which I had penned some math thoughts on 7-31-06. I let it be and moved what I held/ over to its usual location on my bed. Later, I read what I'd noted, and found it of interest in one particular way: an nDF descends to a value which is n - 1. If n = 10 (5 x 5), then (5 + 5) - 1 = the final resultant for 10DF. This formula is not in the note, but the note's examples imply it. My project today is a continuation of the da patterns for the decades pertaining to odd squares, and possibly to even squares. Want to be more explicit, and want to investigate if and where intersectings occur. - First, a brief review: The 2 5 8 pattern for the only group in which paired twin primes can be found looks like this: [ Insert: Decade # equals (decade's f-d-5 integer plus 5) divided by 10. Example: (15 + 5)/10 = 2 ] 2 5 8 - 15/3 = 5 (5) 45/3 = 15 (6) 75/3 = 25 (7) 2 5 8 - 105/3 = 35 (8) 135/3 = 45 (9) 165/3 = 55 (1) [ Insert: The 2nd row decades are 11 (2), 14 (5), 17 (8). ] 2 5 8 - 195/3 = 65 (2) 225/3 = 75 (3) 255/3 = 85 (4) 2 5 8 - 285/3 = 95 (5) 315/3 = 105 (6) 345/3 = 115 (7) 2 5 8 - 375/3 = 125 (8) 405/3 = 135 (9) 435/3 = 145 (1) 2 5 8 - 465/3 = 155 (2) 495/3 = 165 (3) 525/3 = 175 (4) : [ Insert: Have decided to use the f-d-5 integers for the f-d-1 and f-d-9 groups. ] 1 9 4 - [1] 5/3 = 1.6 (1) [81] 85/3 = 28.3 (9) [121] 125/3 = 41.6 (4) 1 9 4 - [361] 121.6 (1) [441] 148.3 (9) [841] 281.6 (4) 7 9 7 - [961] 321.6 (7) [1521] 508.3 (9) [1681] 561.6 (7) 7 9 7 - [2401] 801.6 (7) [2601] 868.3 (9) [3481] 1161.6 (7) 4 9 1 - [3721] 1241.6 (4) [4761] 1588.3 (9) [5041] 1681.6 (1) 4 9 1 - [6241] 2081.6 (4) [6561] 2188.3 (9) [7921] 2641.6 (1) : 1 5 8 - [9] 1.6 (9) [49] 15 (4) [169] 55 (7) 2 8 1 - [289] 95 (1) [529] 175 (7) [729] 241.6 (9) 1 2 5 - [1089] 361.6 (9) [1369]455 (1) [1849] 615 (4) 5 2 1 - [2209] 735 (4) [2809] 935 (1) [3249] 1081.6 (9) 1 8 2 - [3969] 1231.6 (9) [4489] 1495 (7) [5329] 1775 (1) 8 5 1 - [5929] 1975 (7) [6889] 2295 (4) [7569] 2521.6 (9) In the first group's first column is a 95. In the third group's first column is a 95. That says the same decade is being referenced in both. That decade is 29. That says 289 is a square and therefore decade 29 is eliminated. This is what I mean by intersectings. More to come. PM 2:24 - I don't do math the way professional mathematicians do; yet I strive to be rigorous. Ideally, I would like to be able to ascertain all the essential qualities of an integer just by closely observing it. I am seeking methods which will apply into infinity, but I prefer methods which do not require intricate formulas. Of the 18 terms in the first cycle of the f-d-9 group, 12 intersect with terms in group 1. Of the 18 terms in the first cycle of the f-d-1 group, 0 instersect with terms in group 1. Of those that interact, four are of the da2 kind, four of the da5, and four of the da8. The square in the first da5 is 49, and the square in the first da8 is 169. In groups 2 and 3 the order is irregular within each cycle, but does have a pattern. In group 1 the order is regular, but group 1's order does not depend on where square's occur but rather on where "3" divides evenly into an f-d-5 integer. In group 2 "3" does not divide evenly into any f-d-5 integer. In group 3 "3" divides evenly into an f-d-5 integer 12 times in each 18-term cycle. In group 3 the first da8 is in R1C3, but in the first group it is in R2C3. What does this tell about what is in R1C3 in group 1? That it does not have an f-d-9 square in it, but that some other eliminator of a form not here being tested might be lurking there remains a possibility. I believe it's time for a different chart. These are decade # / da charts: - Group 1 2 / 2 5 / 5 8 / 8 11 / 2 14 / 5 17 / 8 20 / 2 23 / 5 26 / 8 29 / 2 32 / 5 35 / 8 38 / 2 41 / 2 44 / 8 47 / 2 50 / 5 53 / 8 : Group 3 1 / 1 5 / 5 17 / 8 29 / 2 53 / 8 73 / 1 109 / 1 137 / 2 185 / 5 221 /5281 / 2 325 / 1 397 / 1 449 / 8 533 / 2 593 / 8 689 / 5 757 / 1 After some figuring, figured out 689 / 5 in group 3 is in R5C2 of cycle 13 in group 1, which cycle's 18th term is 701, there being a 54 spread in each cycle after cycle 1. 12 x 54 = 648; 648 + 53 = 701. 701, 698, 695, 692, 689. (689 x 10) - 1 = 6889. More to come. 05/16/08 AM 7:32 - A little something that provides both an nnss magic number and the number to minus from and plus to it: 2 x 0 x 0 = 0 2 x 1 x 1 = 2 2 x 2 x 2 = 8 2 x 3 x 3 = 18 2 x 4 x 4 = 32 2 x 5 x 5 = 50 2 x 6 x 6 = 72 for generating the natural number summation sequence: 0 1 3 6 10 15 21 28 36 45 55 66 78 and dividing one of these magic numbers by "2" begets a square as with 50/2 - a different view of these nnss magic numbers: 0 2 8 18 32 50 72 2 6 10 14 18 22 so (4 x 0) + 2 (4 x 1) + 2 (4 x 2) + 2 (4 x 3) + 2 (4 x 4) + 2 (4 x 5) + 2 - Also, as can be seen, I have separated the nnss into sets. Adding the integers in a set gets an even square, and subtracting a set's lesser # from its greater # gets the even square's square root. Adding the integers between sets gets an odd square, and subtracting the lesser # from the greater # gets the odd square's square root. - a way to allow the term position (tpo) of an nnss magic number to equal the second and third square root associated with that number: 0's tpo is 0 2's tpo is 1 8's tpo is 2 18's tpo is 3 32's tpo is 4 Thus, for this sequence, tpo 3 = 2 x 3 x 3. 18 - 3 = 15 18 + 3 = 21 - Another view of the squares: 0 1 2 3 4 5 8 9 10 15 16 17 24 25 26 35 36 37 48 49 50 During lunch break I uncovered digit addition relationships between square roots and their squares. Examples follow: 1 1 = 1 1 = 1 19 = 1 361 = 1 28 = 1 784 = 1 73 = 1 5329 = 1 2 2 = 2 4 = 4 11 = 2 121 = 4 38 = 2 1444 = 4 47 = 2 2209 = 4 3 3 = 3 9 = 9 | 6 = 6 36 = 9 | 9 = 9 81 = 9 4 4 = 4 16 = 7 13 = 4 169 = 7 22 = 4 484 = 7 103 = 4 10609 = 7 5 5 = 5 25 = 7 14 = 5 196 = 7 23 = 5 529 = 7 41 = 5 1681 = 7 7 7 = 7 49 = 4 16 = 7 256 = 4 25 = 7 625 = 4 61 = 7 3721 = 4 8 8 = 8 64 = 1 17 = 8 289 = 4 26 = 8 676 = 4 71 = 8 5041 = 4 Note the 1 4 9 7 7 9 4 1 9 routine. To start: If a square root is of the 3, 6, or 9 variety, its square will be of the 9 variety as in 21 = 3 441 = 9. Sqrts of the 4 and 5 variety net squares of the 7 variety. Sqrts of the 2 and 7 variety net squares of the 4 variety. Sqrts of the 1 and 8 variety net squares of the 1 variety. All this helps in that knowing the da of a square limits its possible square roots. The square "6889" = 31 = 4 and so signals that its sqrt must be a da2 or a da7. One is still left guessing, or trying other methods. 6888/4 = 1722 1722/3 = 574 574/7 = 82 3 x 4 x 7 = 84 82 x 84 = 6888 83 x 83 = 6889 83 = 11 = 2 May 17, 2008 note - Stating the implicit: Every nine integers. # Brian A. J. Salchert
Wednesday, May 14, 2008
sw00887math--my-paired-twin-primes-proof
31 is nothing fancy, but a bit complex. The goal is to discern how/where/why paired twin primes can be found. Here is a simple example using decade two. (The first decade is: 0 1 2 3 4 5 6 7 8 9.) (In a decade pattern scheme, decade 10 is actually a 1 because digit addition is used to determine decade patterns.) 10 | 11 12 13 14 15 16 17 18 19 Note: Only an integer whose final digit is "5" can be at the center of a paired twin primes set because an integer whose final digit is "5" will be encountered if the center is at an integer whose final digit is "1" or "3" or "7" or "9" (11 + 4 = 15 13 + 2 = 15 17 - 2 = 15 19 - 4 = 15). Step 1. That integer whose final digit is "5" must be evenly divisible by "3". Step 2. In the decade being investigated/ the integers whose final digits are "2" and "8" must be evenly divisible by "6". Step 3. Also, one of the integers noted in Step 2 must be evenly divisible by "4". In the example/ that integer is 12. Interestingly, in every even decade "4" divides evenly into the final-digit-2 integer; and in every odd decade "4" divides evenly into the final-digit-8 integer. [ Formula for finding Decade #: (decade's final-digit-5 integer + 5) divided by 10 ] (285 + 5)/10 Note 1 --- This is where complexity arrives. --- Example: 280 281 282 283 284 285 286 287 288 289 Odd-integer squares follow this final-digit pattern: 1 9 5 9 1 as in 1 x 1 = 1, 3 x 3 = 9, 5 x 5 = 25, 7 x 7 = 49, 9 x 9 = 81. Three ways to find them are: 1. (a natural number summation sequence term times 8) + 1 2. (an integer times an adjacent integer times 4) + 1 note: the sum of the two adjacent integers chosen equals the square root 3. (an even integer times an adjacent even integer) + 1 note: the odd integer between the adjacent even integers is the square root I'm going to use the latter here. 288/4 = 72 72/2 = 36 36/2 = 18 288/18 = 16 18 - 1 or 16 + 1 17 Also:
18/2 = 9 72/9 = 8 8 x 9 = 72 8 + 9 = 17 17 is 289's square root. 17 x 17 = 289. [ sub-note: 5 35 65 95 (x 3) These are 90-apart locations, but 30-apart locations also sometimes contain paired twin primes. Both the 30-apart and the 90-apart decades begin at decade 2. 5 (15/3) is in decade 2 and 95 (285/3) is in decade 29. 5 x 1 = 5 5 x 7 = 35 5 x 13 = 65 5 x 19 = 95 90-apart 5 x 1 = 5 5 x 3 = 15 5 x 5 = 25 5 x 7 = 35 30-apart To get the actual integers/ multiply the products in these intentionally-collapsed sequences by "3". ] Note 2 -- 0 4 8 and 2 6 are the two available division-by-4 patterns in each decade Note 3 -- Regarding the odd squares: Since the odd squares pattern is 1 9 5 9 1 | 1 9 5 9 1, beginning at the fifth odd square (81), adjacent occurrences of squares having "1" as a final digit/ happen every 5 squares after 81 and after 121. Squares having "1" as a final digit/ also, starting at 1/ occur every ten odd squares (1 x 1 and 19 x 19). They interlock, however, as 19 x 19 is adjacent to 21 x 21. Integers: 1, 81, 121, 361, 441, 841, 961, 1521, 1681 Decades: 1 9 13 37 45 85 97 153 169 The final-digit-9 squares run: Integers: 9, 49, 169, 289, 529, 729, 1089, 1369, 1849 Decades: 1 5 17 29 53 73 109 137 185 Note 4 -- final-digit-5 integers evenly divisible by "3" 15 45 75 105 135 165 195 225 255 285 - 315 345 375 405 435 465 495 525 555 585 - 615 645 675 705 735 765 795 825 855 885 [ sub-note: The decade pattern for those decades in which "3" divides evenly into a final-digit-5 integer is: 2 5 8. Therefore, only those decades can contain, but may not contain, paired twin primes. One of these decades, 29 (2 + 9 = 11 = 2), has 17 x 17 (289) in it. ] For the above three rows of integers, the decades are: 2 5 8 11 14 17 20 23 26 29 - 32 35 38 41 44 47 50 53 56 59 - 62 65 68 71 74 77 80 83 86 89 Note 5 -- Whenever "3" divides evenly into a decade, it also divides evenly into the final-digit-1 integer in that decade. Some f-d-1 pown squares are in these decades. One is in decade 153: 39 x 39 (1521). 153 digit sums to 9, and so is not a decade in which "3" divides evenly into its f-d-5 integer. Note 6 -- More information on decades and "3": In decade 1, "3" divides evenly into 3 and 6 and 9; and does the same every third decade from thereon: that is, it divides evenly into an integer whose final digit is a 3 or 6 or 9. - In decade 3, "3" divides evenly into 21 and 24 and 27, integers whose final digits are 1, 4 and 7 respectively; and does the same every third decade from thereon. - In decade 2, "3" divides evenly into 12 and 15 and 18, integers whose final digits are 2, 5 and 8 respectively; and does the same every third decade from thereon. - From this it should be obvious why paired twin primes cannot be found in decades 1, 4, 7, 10, 13, 16, etc., or in decades 3, 6, 9, 12, 15, 18, etc. Note 7 -- Digit addition patterns for integers in decades > 1: final digit 9: 1 2 3 4 5 6 7 8 9 f-d-1: 2 3 4 5 6 7 8 9 1 f-d-2: 3 4 5 6 7 8 9 1 2 f-d-3: 4 5 6 7 8 9 1 2 3 f-d-4: 5 6 7 8 9 1 2 3 4 f-d-5: 6 7 8 9 1 2 3 4 5 f-d-6: 7 8 9 1 2 3 4 5 6 f-d-7: 8 9 1 2 3 4 5 6 7 f-d-8: 9 1 2 3 4 5 6 7 8 f-d-0: 1 2 3 4 5 6 7 8 9 - for even squares (if there is a pattern): 4 is 4 16 is 7 36 is 9 64 is 1 100 is 1 144 is 9 196 is 7 256 is 4 324 is 9 | 400 is 4 484 is 7 576 is 9 676 is 1 784 is 1 900 is 9 1024 is 7 1156 is 4 1296 is 9 4 7 9 1 1 9 7 4 9 - for odd squares (if there is a pattern): 9 is 9 25 is 7 49 is 4 81 is 9 121 is 4 169 is 7 225 is 9 289 is 1 361 is 1 9 7 4 9 4 7 9 1 1 - Curious related facts: 17 is 8; 17 x 17 = 289, a 1; 289 is in decade 29, a 2. 39 is 3; 39 x 39 = 1521, a 9; 1521 is in decade 153, a 9. 6:17 PM Addendum 1: Began this in Note 3; but moments ago, because I wanted to know the changes cycle for odd squares having "1" as a final digit, this: 1 x 1 = 1 9 x 9 = 81 (9) 11 x 11 = 121 (4) 19 x 19 = 361 (1) 21 x 21 = 441 (9) 29 x 29 = 841 (4) 31 x 31 = 961 (7) 39 x 39 = 1521 (9) 41 x 41 = 1681 (7) 49 x 49 = 2401 (7) 51 x 51 = 2601 (9) 59 x 59 = 3481 (7) 61 x 61 = 3721 (4) 69 x 69 = 4761 (9) 71 x 71 = 5041 (1) 79 x 79 = 6241 (4) 81 x 81 = 6561 (9) 89 x 89 = 7921 (1) 18-term cycle for which the decades are: 1 9 13 37 45 85 | 97 153 169 241 261 349 | 373 477 505 625 657 793 Because the final digit for each of these squares is a "1"/ each square/ digit sums to same number its decade does. This being so, f-d-1 squares do not occur in the decades where "3" divides evenly into an f-d-5 integer. Those decades digit sum to 2, 5 and 8 only. These decades digit sum to 1, 4, 7 and 9 only. Addendum 2: The f-d-9 squares occur in decades which digit sum to 1, 2, 5and 8 only, which is why the 289 square in decade 29 prevents that decade from containing paired twin primes. 8 PM - This is the 18-term changes cycle for f-d-9 squares: 3 x 3 = 9 (9) 7 x 7 = 49 (4) 13 x 13 = 169 (7) 1 5 8 17 x 17 = 289 (1) 23 x 23 = 529 (7) 27 x 27 = 729 (9) 2 8 1 33 x 33 = 1089 (9) 37 x 37 = 1369 (1) 43 x 43 = 1849 (4) 1 2 5 47 x 47 = 2209 (4) 53 x 53 = 2809 (1) 57 x 57 = 3249 (9) 5 2 1 63 x 63 = 3969 (9) 67 x 67 = 4489 (7) 73 x 73 = 5329 (1) 1 8 2 77 x 77 = 5929 (7) 83 x 83 = 6889 (4) 87 x 87 = 7569 (9) 8 5 1 See also: sw00878math/odd-primes-and-digit-addition sw00879math/digit-addition-families sw00881math/pown-and-pewn-squares-notes sw00882math/descension-by-squares-groups sw00883math/my-paired-twin-primes-conjecture sw00884math/more-views-of-my-conjecture # Brian A. J. Salchert
Tuesday, May 13, 2008
sw00885d29--crowded-day-yesterday
Yesterday began with a necessary fasting after which on a yellow legal pad I started rough-drafting my heuristically-created paired twin primes proof for finding them. About 9:30 my sister arrived to transport me to see a lawyer regarding important death-related matters. My sister had an appointment to see him today, but he provided services for both of us. Our experiences with him were brief but pleasant and definitely educational. Then it was on to an outpatient clinic where I was given a not- so-pleasant (mostly due to my broken nose and broken back and gag response) esophageal manometry exam. Next it was north to a supercenter for groceries and other items. On the way to my apartment we did the Wendy's windows tour for a nearly plain hamburger--their burgers are square and thin now--for me. Got to apt around 5:30. Later I returned to my math project. Not sure when it will be finished to my satisfaction. - Brian A. J. Salchert 2008-05-13
Sunday, May 11, 2008
sw00884math--more-views-of-my-conjecture
30 Recap: If "3" doesn't divide evenly into an integer having "2" or "8" as a final digit, the decade that integer's in does not contain paired twin primes. [ Note: 0 through 9 is the first decade. ] Continuance: 15 45 75 105 135 165 195 225 255 285 - Have decided to try this with "3". 12/3 = 4, 18/3 = 6; 42/3 = 14, 48/3 = 16; 72/3 = 24, 78/3 = 26; so on. 5, 15, 25, 35, 45, 55, 65, 75, 85, 95. If one multiplies every 30 of these integers by "3", the chance of encountering paired twin primes is better but not absolute. Much of this has to do with the integer squares. 5, 35, 65, 95, and on, are the every 30. In the first instance, 9 is in the prior decade and 25 is in the following decade. In the second instance, 81 is two decades down and 121 is two decades up. In the third instance, 169 is three decades down and 225 is three decades up. However, in the fourth instance, 17 x 17 (289) or (16 x 18) + 1 or (8 x 36) + 1 is in the decade. Is there anything about the integers from 280 through 289 that signal this? 1 x 1 = 1, 3 x 3 = 9, 5 x 5 = 25, 7 x 7 = 49, 9 x 9 = 81 11 x 11 = 121, 13 x 13 = 169, 15 x 15 = 225, 17 x 17 = 289, 19 x 19 = 361 Pattern of final digits: 1, 9, 5, 9, 1 49 - 9 = 40 and 289 - 169 = 120. Beware that pown sqrt whose only or final digit is "3" or "7". Am heading higher to see what the spread is there. 21 x 21 = 441, 23 x 23 = 529, 25 x 25 = 625, 27 x 27 = 729, 29 x 29 = 841 169 - 49 = 120 and 529 - 289 = 240. Internally: 40, 120, 200. During lunch break, these: 1. The first paired twin primes occur in decade two; and, since every nine decades above that decade are the best initial places to look, decades 11, 20, 29, 38, 47, 56, 65, 74, 83, 92, and so on, are the ones. 2. The following series of integers: 0 (0 x 1), 2 (1 x 2), 6 (2 x 3), 12 (3 x 4), 20 (4 x 5), 30 (5 x 6), 42 (6 x 7), 56 (7 x 8), 72 (8 x 9), 90 (9 x 10), 110 (10 x 11), and on. 3. If one of these integers is multiplied by "4" and "1" is added to the product, the sum will equal an odd square. 72 x 4 = 288; 288 + 1 = 289. So 289 is an odd square, but of what?: 8 + 9. 4. 289 exists in the 29th decade. 5. Whatever decade is chosen, it is always wise to divide that decade's first integer by "4" and also to divide its ninth integer by "4" to see if the remainder is an integer which equals the product of two adjacent integers. Otherconsiderations: 2 x 0 = 0, 2 x 2 = 4, 2 x 4 = 8, 2 x 6 = 12, 2 x 8 = 16, 2 x 10 = 20, 2 x 12 = 24 2 x 6 = 4 x 3 2 x 10 = 4 x 5 2 x 36 = 4 x 18 = 8 x 9 = 72 - All the integers in the point 2 example are even integers. Their pattern is: 0 2 6 2 0 - In every decade where paired twin primes might be found, both that integer whose final digit is "2" and that integer whose final digit is "8" are evenly divisible by "6". - It might be that 288/6 = 48 is an indication 289 is a square since 49 is a square. - 1 + 8 = 9, 9 + 16 = 25, 25 + 24 = 49, 49 + 32 = 81, 81 + 40 = 121 9 - 3 = 6, 9 + 3 = 12; 25 - 5 = 20, 25 + 5 = 30; 49 - 7 = 42, 49 + 7 = 56 - # Brian A. J. Salchert
Saturday, May 10, 2008
sw00883math--my-paired-twin-primes-conjecture
29 Yesterday (Friday, May 09, 2008) I concluded my sw00882math/ descension-by-squares-groups/2841 with this conjecture: Regarding paired twin primes, the even integer between one of the twin primes is always divisible by "4" and the even integer between the other of the twin primes is always divisible by "6". Why did I posit this seemingly senseless conjecture? Because I wanted to use even integers since I had set up a descension-by-squares grouping controlled by "4". As it happens, the only way my groupings matter is that it was through them that I uncovered the "4" and "6" phenomenon. Early this after- noon, I was trying to find out if I was wrong, an effort which led to: "3" divides into every six integers, the final digits of which (excluding the special case: 3 x 3) are 5, 1, 7, 3, 9 as in 15, 21, 27, 33, 39. Here is a run with the above numbers in italics: 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 If the final digit is "1" or "7" or "3" or "9", there cannot be any paired twins primes because 1 + 4 = 5, 7 - 2 = 5, 3 + 2 = 5, 9 - 4 = 5. Therefore, the final digit must be "5" and the integer must be divisible by "3". 5 - 3 = 2 and 5 + 3 = 8. This then is my proof, such as it is. Also, where the integer whose final digit is "5" is not divisible by "3"/ neither the integer whose final digit is "2" nor the integer whose final digit is "8" is evenly divisible by "6". My effort began with this: 6 102 108 114 120 126 132 4 100 104 108 112 116 120 124 128 132 136 - - - 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 3 3x5x7 3 5 3 7 11 3 5 5:01 PM - Have uncovered a flaw. At the early going it involves "7". It is quite obvious, but I failed to check. "3" divides evenly into an integer whose final digit is "5" every thirty integers greater than "15" but it isn't until ninety above "15" that paired twin primes are encountered. [ This doesn't invalidate my conjecture as stated, but it certainly weakens it. ] One obstacle is the odd square. However, the (8 x an nnss term) + 1 method isn't the only way to find an odd square. These mulitplications: 2 x 4, 4 x 6, 6 x 8, 8 x 10, 10 x 20, and so on + 1 is another method. But I had mentioned "7". Every fourteen from forty-nine it's out to clip ya, and it doesn't follow an easy-to-detect routine the way "5" and "3" do. Perhaps "3" can help, or "5", or both. Some integers "7" divides evenly into: one above 55, two below 65, two above 75, one below 85, one above 90, two below 100, at 105, two above 110, one below 120, one above 125, two below 135, at 140, two above 145. Dizzy yet? Which one did I miss? Correct: at 70. Routine? Going back to 7 x 1, the routine begins at two above. So: two above, one below, one above, two below, at | two above, one below, one above, two below, at. I hate "7", which is to say nothing about the greater more difficult integers. What about how it relates to mulitiples of "3"? Which ones? How about just those that add to 3. four above 3, two above 12, at 21, two below 30, four below 39, three above 39, one above 48, one below 57, three below 66 | four above 66, two above 75, at 84, two below 93, four below 102, three above 102, one above 111, one below 120, three below 129. All this reminds of a poem I wrote in 2006. After I find it I'll place a link to it. It is 9:35 PM. Earlier a storm passed. Need to see if another one's coming. Gusts just thrust us. Satellite's showing an enlarging green patch over this city. Dost one blame it on the coal- fired power plant? Or is it merely a reflection of the city's lights? "Bone Ache" is the poem. It is the fourth one on its page. A quick six lines. It was written September 29, 2006. It fits this occasion. # Brian A. J. Salchert
Friday, May 9, 2008
sw00882math--descension-by-squares-groups
28 pewn = positive even whole number pown = positive odd whole number sqrt = square root resultant = remainder G = Group sq = square In a descension-by-squares/ pown squares are subtracted from a pewn square in a lesser to greater order, or pewn squares are subtracted from a pown sq in a lesser to greater order. Each subtraction produces a resultant integer which is always a pown. That pown is also the product of two integers which can be discovered by inspecting the sqrt values in a subtraction. If the values are "18" and "11", simply subtract the 11 from the 18 to get the multiplier and add the 11 to the 18 to get the multiplicand. As I will show, the construction of a d-b-s can quickly become cumbersome, tending to make such constructions less and less useful. The main use, however--finding the final resultant, is easily done. Here are some relevant formulas: (2 x G#) - 2 = pewn sqrt for that group G# = (pewn sqrt + 2)/2 (2 x G#) - 1 = pown sqrt for that group G# = (pown sqrt + 1)/2 (final resultant # + 1)/2 = sqrt (sqrt x 2) - 1 = final resultant # Example using the above: G28 pewn sqrt is 54 final resultant is 107 pown sqrt is 55 final resutant is 109 [ Note: Because they are at level one (the x 1 level), some final resultants are prime numbers. ] First ten groups: - G1 has one object: 1² - 0² = 1² = 1 - G2, wherein 2 + 1 = 3 and 3 + 2 = 5: 2 x 2 = 4 _ _ _ _ _ _ _ _ _ _ 3 x 3 = 9 2² - 1² = 3 _ _ _ _ _ _ _ _ _ 3² - 2² = 5 - G3 4 x 4 = 16 _ _ _ _ _ _ _ _ _ 5 x 5 = 25 4² - 1² = 15 _ _ _ _ _ _ _ _ 5² - 2² = 21 4² - 3² = 7 _ _ _ _ _ _ _ _ _5² - 4² = 9 - G4 6 x 6 = 36 _ _ _ _ _ _ _ _ _ 7 x 7 = 49 6² - 1² = 35 _ _ _ _ _ _ _ _7² - 2² = 45 6² - 3² = 27 _ _ _ _ _ _ _ _7² - 4² = 33 6² - 5² = 11 _ _ _ _ _ _ _ _7² - 6² = 13 - G5 8 x 8 = 64 _ _ _ _ _ _ _ _ 9 x 9 = 81 8² - 1² = 63 _ _ _ _ _ _ _ 9² - 2² = 77 8² - 3² = 55 _ _ _ _ _ _ _ 9² - 4² = 65 8² - 5² = 39 _ _ _ _ _ _ _ 9² - 6² = 45 8² - 7² = 15 _ _ _ _ _ _ _ 9² - 8² = 17 - G6 10 x 10 = 100 _ _ _ _ _ _ 11 x 11 = 121 10² - 1² = 99 _ _ _ _ _ _ 11² - 2² = 117 10² - 3² = 91 _ _ _ _ _ _ 11² - 4² = 105 10² - 5² = 75 _ _ _ _ _ _ 11² - 6² = 85 10² - 7² = 51 _ _ _ _ _ _ 11² - 8² =57 10² - 9² = 19 _ _ _ _ _ _ 11² - 10² = 21 - G7 12 x 12 = 144 _ _ _ _ _ _13 x 13 = 169 12² - 1² = 143 _ _ _ _ _ 13² - 2² = 165 12² - 3² = 135 _ _ _ _ _ 13² - 4² = 149 12² - 5² = 119 _ _ _ _ _ 13² - 6² = 129 12² - 7² = 95 _ _ _ _ __13² - 8² = 105 12² - 9² = 63 _ _ _ _ _ _13² - 10² = 69 12² - 11² = 23 _ _ _ _ _ 13² - 12² = 25 - G8 14 x 14 = 196 _ _ _ _ _ _ 15 x 15 = 225 14² - 1² = 195 _ _ _ _ _ _15² - 2² = 221 14² - 3² = 187 _ _ _ _ _ _15² - 4² = 209 14² - 5² = 171 _ _ _ _ _ _15² - 6² = 189 14² - 7² = 147 _ _ _ _ _ _15² - 8² = 161 14² - 9² = 115 _ _ _ _ _ _15² - 10² = 125 14² - 11² = 75 _ _ _ _ _ _15² - 12² = 81 14² - 13² = 27 _ _ _ _ _ _15² - 14² = 29 - G9 16 x 16 = 256 _ _ _ _ _ _ 17 x 17 = 289 16² - 1² = 255 _ _ _ _ _ _17² - 2² = 285 16² - 3² = 247 _ _ _ _ _ _17² - 4² = 273 16² - 5² = 231 _ _ _ _ _ _17² - 6² = 253 16² - 7² = 207 _ _ _ _ _ _17² - 8² = 225 16² - 9² = 175 _ _ _ _ _ _17² - 10² = 189 16² - 11² = 135 _ _ _ _ _ 17² - 12² = 145 16² - 13² = 87 _ _ _ _ _ _17² - 14² = 93 16² - 15² = 31 _ _ _ _ _ _17² - 16² = 33 - G10 18 x 18 = 324 _ _ _ _ _ _ 19 x 19 = 361 18² - 1² = 323 _ _ _ _ _ _19² - 2² = 357 18² - 3² = 315 _ _ _ _ _ _19² - 4² = 345 18² - 5² = 299 _ _ _ _ _ _19² - 6² = 325 18² - 7² = 275 _ _ _ _ _ _19² - 8² = 297 18² - 9² = 243 _ _ _ _ _ _19² - 10² = 261 18² -11² = 203 _ _ _ _ _ _19² - 12² = 217 18² - 13² = 155 _ _ _ _ _ 19² - 14² = 165 18² - 15² = 99 _ _ _ _ _ _19² - 16² = 105 18² - 17² = 35 _ _ _ _ _ _19² - 18² = 37 [ Note: I was surprised when 4 and 28 had prime pairs which do not match when digit addition is applied to them, but have since learned that there are congruities in two other ways. 28 - 4 = 24, which is divisible by both 3 and 8; 107 - 11 = 96, which is divisible by both 3 and 8. ] Guess what. Unless I've made an error someplace, I've just been thrown another dropping curve ball I cannot hit. 191, 193; 197, 199 are primes. G49 pewn sqrt is 96 final resultant is 191 pown sqrt is 97 final resultant is 193 Plainly put, if the even integer between two primes is not divisible by "4", those primes will not be in the same group. [ Note: What I have been calling prime pairs are usually called twin primes; so I guess one could call 11 13 and 17 19 paired twin primes. In every such pairing, only the even integer between one of the twins is divisible by "4". Not sure, but the integer between the twin primes which is not divisible by "4" may always be divisible by "6". ] 9:27 PM - With help from a table on page 211 of the paperback copy of The Mathematical Experience I own, just discovered that 9431 9433 and 9437 9439 are paired twin primes, and that 9432 is divisible by "4" and 9438 is divisible by "6". Conjecture: Regarding paired twin primes, the even integer between one of the twin primes is always divisible by "4" and the even integer between the other of the twin primes is always divisible by "6". # Brian A. J. Salchert
Wednesday, May 7, 2008
sw00881math--pown-and-pewn-squares-notes
27 pown = positive odd whole number da = digit addition sq = square tpo = term position DF = descension family pewn = positive even whole number sqrt = square root There are no integer squares in the 2, 3, 5, 6, 8 da families. [ The even squares run: 4 7 9 1 1 9 7 4 9 ] The sq of any pown/ divisible by "3" is in the 9da family. tpo pown sq da - 1 1 1 1 2 3 9 9 3 5 25 7 4 7 49 4 5 9 81 9 6 11 121 4 7 13 169 7 8 15 225 9 9 17 289 1 The da family rotation above is Group 1. 19 through 35 comprises Group 2. Here is a different view: pown 1 3 5 7 9 11 13 15 17 - sq da 1 9 7 4 9 4 7 9 1 -- pown 19 21 23 25 27 29 31 33 35 - pown 37 39 41 43 45 47 49 51 53 The goal I have may have already been reached, and it may be I cannot reach it; but that is neither here nor there, and at this moment I am reluctant to reveal it. - In the 1da family for pown squares, the powns are separated from each other in a 16 2 pattern. In the 4da family for pown squares, the pattern is 4 14; and in the 7da family for pown squares, the pattern is 8 10. In each case, 18 apart rules when going from a given column in one group to the same column in the group above or below it. - Also, there is a times-10 rule helper. Example: 49 + (7 x 10) = 119 (or 7 x 17). This is leading to final-digit math regarding both pown and pewn squares. I may need to change the title of this entry. Anyway, here's the info: Am skipping "1". Am skipping "5". 9 + (3 x 10) = 39 (or 3 x 13), 121 + (11 x 10) = 231 (or 11 x 21); 361 + (19 x 10) = 551 (or 19 x 29). So, where the final digit is "9" in the above, a 120 - 1 and a 40 -1 occur. Where, likewise, the final digit is "1", a 230 + 1 and a 550 + 1 occur. Similar outcomes occur with pewn squares. 4 + (2 x 10) = 24 (or 2 x 12) and 64 + (8 x 10) = 144(or 8 x 18). 16 + (4 x 10) = 56 (or 4 x 14) and 36 + (6 x 10) = 96 (or 6 x 16). So, where the final digit is "4", a 25 - 1 and a 65 - 1 occur. Where the final digit is "6", a 55 + 1 and a 95 + 1 occur. Obviously, pown squares revolve around "0" and pewn squares revolve around "5". May 08, 2008 Late last night before I lost my broadband connection, I subtracted 119 from 144. The 25 reminded me of descension-by-squares and Descension Families. Visit this entry. Moments ago I was there and I think it is better than what I had planned to write here. Still, I never know when a rethinking will move me to change terminology and/or how an idea is presented. For instance: there I let 3² - 2² be a DF of one level; but it could be a DF of two levels. The former is as above/ where the resultant integer is "5". Were I to show it as two levels, the second level would be 3 x 3 = 9, and would be there mainly for informational reasons. "5" would still be the resultant integer, the final resultant integer, that is. Do want to add some further thoughts. - It could be said that "3" and "5" are the only two-level (or one-level) DF primes. . . . After the 8DF (4 x 4 = 16) comes the 10DF (5 x 5 = 25). [ Haven't done an extensive check, but it appears the DF's rise in pairs. If so, they could be grouped so, and this might aid in understanding the secrets of the final resultants. Will probably be heading that direction. ] Proceeding with my changed thinking, these are the only three-level DF's, and the final resultant for the 10DF is not a prime. Here is the new look: 5 x 5 = 25 5² - 2² = 21 5² - 4² = 9 = 3² Points of interest: 5² descends to 3², 13² to 5², 25² to 7², 41² to 9², 61² to 11², 85² to 13², 113² to 15²; 145² to 17². This generates the following: 8 12 16 20 24 28 32 For going beyond 113² there is this, in which multiples of "36" reign: 5² 41² 113² 221² 365² 545² 761² These are for descensions to squares which are multiples of "3". The resulting squares are: 9, 81, 225, 441, 729, 1089; 1521. Here: 761² - 760² = 1521 = 39 x 39. # Brian A. J. Salchert
sw00880d28--read-my-blogs-if-but
you are also welcome to use this one and my bajsalchert.blogspot.com one as gateways to other web logs. Here in the Favorite Places sidebar section are links to a variety of sites. Take a look. The one I visit the most is the Astronomy site, but a site such as the HTML ASCII site (www.lookuptables.com) often helps when I need to know special codes. My blogspot blog has more links to sites maintained by other writers (mostly poets). Its name is Rhodingeedaddee. Brian A. J. Salchert 2008-05-07
Tuesday, May 6, 2008
sw00879math--digit-addition-families
26 Confusion reigns even where there is order. 1, 10, 19, 28, 37, 46, 55, 64, 73, 82 | 91, 100, 109, 118, 127, 136, 145, 154, 163, 172 - 2, 11, 20, 29, 38, 47, 56, 65, 74, 83 | 92, 101, 110, 119, 128, 137, 146, 155, 164, 173 - 3, 12, 21, 30, 39, 48, 57, 66, 75, 84 | 93, 102, 111, 120, 129, 138, 147, 156, 165, 174 - 4, 13, 22, 31, 40, 49, 58, 67, 76, 85 | 94, 103, 112, 121, 130, 139, 148, 157, 166, 175 - 5, 14, 23, 32, 41, 50, 59, 68, 77, 86 | 95, 104, 113, 122, 131, 140, 149, 158, 167, 176 - 6, 15, 24, 33, 42, 51, 60, 69, 78, 87 | 96, 105, 114, 123, 132, 141, 150, 159, 168, 177 - 7, 16, 25, 34, 43, 52, 61, 70, 79, 88 | 97, 106, 115, 124, 133, 142, 151, 160, 169, 178 - 8, 17, 26, 35, 44, 53, 62, 71, 80, 89 | 98, 107, 116, 125, 134, 143, 152, 161, 170, 179 - 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 | 99, 108, 117, 126, 135, 144, 153, 162, 171, 180 # First: There cannot be a "0" digit-addition (0da) family. Second: In the 9da there is no prime number. Third: Each single-digit prime is also the da it is in (2, 3, 5, 7). Fourth: The first prime in the 1da is "19". Fifth: In the 6da there is no prime number. Sixth: The first prime in the 4da is "13". Seventh: The first prime in the 8da is "17". Eighth: In the 3da the only prime is "3". Ninth: Every integer in the 3da, 6da, and 9da is divisible by "3". Tenth: Each digit-addition family has a 10-term pattern. Eleventh: There are only 9 da patterns pertaining to powns and their pown multiples, and each of these patterns has 9 terms, and "9" is the 5th term in each pattern. Why? The da pattern repeats after every 9 terms because the da for x 1 equals the da for x 19; and, for this same reason, the da pattern repeats after every 18th higher pown. 19's da and 37's da is "1". Also, any integer times 9 equals a 9da integer. How crazy I am is an unknown; nonetheless, I'm pressing on. One interest is knowing how a prime relates to its square. Did a few calculations and it seems that in a given da family, both the primes and the squares of those primes, while each in a different da family, are respectively in the same da family. Thus, if the prime is in the 1da, its-- this is an exception--square is inthe 1da. If a prime is in the 2da, its squareis in the 4da. If a prime is in the 3da (and only one is), its square is in the 9da. If a prime is in the 4da, its square is in the 7da (because 4 x 4 = 16). If a prime is in the 5da, its square is in the 7da (because 5 x 5 = 25). If a prime is in the 7da, its square is in the 4da. If a prime is in the 8da, its square is in the 1da. So? Sew baritones on your underwear. Now, 11 (2da) times 17 (8da) equals 187 (7da) and 29 (2da) times 53 (8da) equals 1537 (7da). - What about 105, which is 3 x 5 x 7? - This is heading into combinatorics. - 29 (2da) times 19 (1da) equals 551 (2da), and 11 times 37 equals 407. 29 times 13 equals 377 (8da). 29 times 23 equals 667 (1da). 29 times 43 equals 1247 (5da). 29 times 47 equals 1363 (4da). - Conclusion 1: Every integer product is divisible by "1". Conclusion 2: The da of a pown product may be divisible by a pown of any other da. Had to fix an error in the da chart. After the correction, I noticed that "11" and "101" are both 2da, that "13" and "103" are both 4da, that "17" and "107" are both 8da, and that "19" and "109" are both 1da. This suggests that at multiples of 90 there may be other such prime pairs. In order to avoid being misunderstood, the pairs here are 11 and 13, 17 and 19, 101 and 103, 107 and 109. 90 up from 105 is 195 (3 x 5 x 13), and neither 7 nor 11 divide without remainders into 191, 193, 197, or 199. A prime does not assert its power until it attains its square. Primes > 13 do not, therefore, matter. 17 x 17 = 289. # Brian A. J. Salchert
Monday, May 5, 2008
sw00878math--odd-primes-and-digit-addition
25 Have off-and-on been trying for years to find a way to know by inspecting it if an integer is divisible by a prime number. It is easy to tell if it is divisible by "5", and only slightly harder to tell if it is divisible by "3"; but from "7" on up it is difficult. Tonight I decided to try numerological digit addition, and may have uncovered a way. digit addition = da times = x patterns: - 3 3 9 6 3 9 6 3 9 6 da 1 3 5 7 9 11 13 15 17 x 5 5 6 7 8 9 1 2 3 4 da 1 3 5 7 9 11 13 15 17 x 7 7 3 8 4 9 5 1 6 2 da 1 3 5 7 9 11 13 15 17 x 11 2 6 1 5 9 4 8 3 7 da 1 3 5 7 9 11 13 15 17 x 13 4 3 2 1 9 8 7 6 5 da 1 3 5 7 9 11 13 15 17 x 17 8 6 4 2 9 7 5 3 1 da 1 3 5 7 9 11 13 15 17 x 19 1 3 5 7 9 2 4 6 8 da 1 3 5 7 9 11 13 15 17 x 23 5 6 7 8 9 1 2 3 4 da 1 3 5 7 9 11 13 15 17 x 29 2 6 1 5 9 4 8 3 7 da 1 3 5 7 9 11 13 15 17 x - Interesting. There are 9 terms in each rotation. The greatest multiplier in each rotation is 18 less than the greatest multiplier in the next higher rotation. Every prime which is 18 greater than a lesser prime has the same digit-addition rotation as the lesser prime. This means there are da-rotation families. However, it would be better to use a prime's da. Example: 11's da is 2, 29's da is 2, 47's da is 2, but 65 is not a prime.
-
Will need to continue this another day. # Brian A. J. Salchert
Sunday, May 4, 2008
sw00877math--1-3-5-7-9
Certain groups with 5 terms. Not sure where this will go. 1 3 5 7 9 3 9 15 21 27 5 15 25 35 45 7 21 35 49 63 9 27 45 63 81 Two times in the above is the only odd integer prime triad: 3 5 7 11 33 55 77 99 13 39 65 91 117 15 45 75 105 135 17 51 85 119 153 19 57 95 133 171 21 63 105 147 189 23 69 115 161 207 Not much to behold. 11 13 17 19 23 So what is of interest? x 1 x 3 x 5 x 7 x 9 The first integer in each group gets the group name, but not the gpo. A group's position is always its (first integer + 1)/2. Thus, the gpo of the 23 Group is 12. It happens that "23" is a prime number and that therefore 24 (3 x 8) - 1 equals a prime number. So what? I am not certain, but (1 x 8) - 1 equals a prime number too. What other single-digit integers work? (9 x 8) - 1, but neither (5 x 8) - 1 nor (7 x 8) - 1. (2 x 8) - 1? No. (4 x 8) - 1? Yes. (6 x 8) - 1? Yes. (8 x 8) - 1? No. Messy. Yet maybe not. The "Yes" ones are: 7, 23, 31, 47, 71. The "No" ones are: 15, 39, 55, 63. It appears that multiples of "3" matter extensively. Something else. The what's missing game. This involves powns only after 3 5 7, and forget integers that have "5" for a final digit. In the 3 Group the missing powns are 11, 13, 17, 19, 23. Beyond this group the search toughens/ because some of the missing integers show up in higher groups. My interest in this method is weakening. Something other. The spreads, and where they occur. Of course, a higher strangeness exists in the higher realms, so I may not get far here either. 3 [2] 5 [2] 7 [4] 11 [2] 13 [4] 17 [2] 19 [4] 23 [6] 29 [2] 31 [6] 37 [4] 41 [2] 43 [4] 47 [6] 53 [6] 59 [2] 61 [6] 67 [4] 71 [2] 73 [6] 79 [4] 83 [6] 89 [8] 97 [4] 101 [2] 103 [4] 107 [2] 109 [4] 113 [14] 127 Notice that before we encounter the two prime pairs above 100 we hit the first 8-apart, and after 113 we hit a stunning 14-apart dead zone. What's with 115, 117, 119 121, 123, 125? 5, 3, 7, 11, 3, 5. "119", being two less than an odd square, should be (according to previous experiences) a prime number, but 7 x 17 invalidates that assumption. The immediate realization is that more of the same can be expected. 083 085 5 x 17 087 3 x 29 089 091 7 x 13 093 3 x 31 095 5 x 19 097 099 3 x 3 x 11 101 103 105 3 x 5 x 7 (105 is a major node integer) 107 109 111 3 x 37 113 115 5 x 23 117 3 x 3 x 13 119 7 x 17 121 11 x 11 123 3 x 41 125 5 x 5 x 5 127 # Brian A. J. Salchert
Saturday, May 3, 2008
sw00875math--even-squares-to-odd-squares
This is another revisit to the natural number summation sequence and its relation to "8". Excluding "0", the first twelve nnss terms are: 1 3, 6 10, 15 21, 28 36, 45 55, 66 78 These can be approached variously; but how even squares are progenitors of them, and how odd squares are derived from them is valuable knowledge. As can be seen, this sequence can be shown in 2-term groups because the first two are odd integers and the second two are even integers, and so on in that alternating manner. Things to note: 1 + 3 = 4; 6 + 10 = 16; 15 + 21 = 36; 45 + 55 = 100; and so on in that manner. Obviously, the sums are even squares, meaning that hidden in any even square are two terms of the natural number summation sequence. Therefore, if 12 x 12 (or 144), is chosen for inspection, how are the nnss terms in it exposed? If 144 is divided by 2, the quotient is 72. If 12 is divided by 2, the quotient is 6. 72 - 6 = 66, and 72 + 6 = 78. As per above, 66 and 78 constitute a group of two even terms in the natural number summation sequence (nnss). - How do odd squares relate to this? Every odd square is 1 > any nnss term (including "0") multiplied by "8". So, say the target is 13 x 13 (or 169). How is it discovered? One way is to subtract "1" from "169" and then divide "168" by "8". Having done this, the next question is: What is the proof that "21" is an nnss term? There are several ways, but I want to try something. In the nnss there are odd groups and even groups. Since the first group is an odd group, I'm calling the odd groups the minor sequence, making the even groups the major sequence. Now 1 + 3 = 4, so "4" is a number which can be used as a divisor. Here's a run: .25, .75; 3.75, 5.25; 11.25, 13.75; 22.75, 26.25 Here's a run using even nnss terms: 1.5, 2.5; 7, 9; 16.5, 19.5; 28, 36 I'm sure you see the difference and also the nuances. 4 into 21 = 5.25, but--. .25 represents "1"; therefore, if 5.25 is divided by .25, the quotient should be 21, and it is. "21" is the sixth term in the nnss, which means that it is the sum of 1 + 2 + 3 + 4 + 5 + 6. If term position (tpo) is used as a guide, the results of dividing each term by its position creates an incontrovertible sequence: 1, 1.5, 2, 2.5, 3, 3.5, n Each of these numbers can also be obtained by adding "1" to an nnss tpo and then dividing by 2. From this it is also possible to get back to an nnss term's value by subtracting .5 from the (tpo + 1)/2 quotient and then multiplying that by the (tpo + 1). 3.5 - .5 = 3, and 3 x 7 = 21. [ By these methods any unknown nnss term can be found. Example: Let 63 be the tpo for an nnss term. 63 + 1 = 64. 64/2 = 32. 64 x 31.5 = 2016. ] So sayeth my calculator. What then is this?: (8 x 2016) + 1. 16129. What is the sqrt of that? Three calculator tries got me "127". - 3² 5² 7² 9² 11² 13² 15² 17² 19² 21² 23² 25² Now, the square root (sqrt) of 4 is 2; of 16 is 4; of 36 is 6; of 64 is 8; of 100 is 10; of 144 is 12 and 2 x 2 = 4, 2 x 4 = 8, 2 x 6 = 12, 2 x 8 = 16, 2 x 10 = 20; 2 x 12 = 24 Remember 1 + 3 = 4, 6 + 10 = 16, 15 + 21 = 36, 28 + 36 = 64, 45 + 55 = 100; 66 + 78 = 144 and 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 4 x 4 = 16, 4 x 5 = 20; 4 x 6 = 24 So, if the 2-term group position is 32, the integer between the odd square roots of each term in that group must be 4 x 32, which is 128. See "127" above. If the spread between the nnss terms in a group is known, then that value divided by 2 will equal that group's position (gpo). Therefore, (32 x 2) + 2016 must equal a term whose value is 2080; and (8 x 2080) + 1 must equal an odd integer which is 129 x 129. According to my calculator/ both the former and the latter is 16641. Of course, then, the difference between the odd squares divided by the spread between the related nnss terms will always equal "8". # Brian A. J. Salchert
Friday, May 2, 2008
sw00874math-3at4
22 There are two ways three adjacent integers can be related to "4" that are of interest to me. The one way I am not choosing first begins at zero. In the 6n Elimination Table the odd squares always occur in the +1 column, the column beginning with "7". In today's entry, "4" controls the distribution, but the first n is "1", and each n is a pown integer which is 2 > the preceding n, and "2" is the multiplier, and the odd squares always occur in the -1 column, and there is only one row. # 1 2 3 5 6 7 9 10 11 13 14 15 17 18 19 21 22 23 25 26 27 29 30 31 33 34 35 37 38 39 41 42 43 45 46 47 49 50 51 53 54 55 57 58 59 61 62 63 65 66 67 69 70 71 73 74 75 77 78 79 81 82 83 85 86 87 89 90 91 93 94 95 97 98 99 101 102 103 105 106 107 109 110 111 113 114 115 117 118 119 121 122 123 125 126 127 129 130 131 133 134 135 137 138 139 141 142 143 145 146 147 149 150 151 153 154 155 157 158 159 - final-digit rotations: 1 5 9 3 7; 2 6 0 4 8; 3 7 1 5 9 Since this row consists of groups of three adjacent integers, the first group having a prime pair is: 5 6 7 Beginning with 5 6 7, every group which is greater by "12" than the group preceding it/ might--except for those groups in which "4" or "6" is the final digit--contain a prime pair. The final-digit rotation is: 6 8 0 2 4. Here are the first four: 5 6 7 17 18 19 29 30 31 41 42 43 Beginning with 53 54 55, problems: 65 66 67 77 78 79 89 90 91 do not contain prime pairs because of "5" in the first two, and of "7" in the second two. In both cases "11" and "13" are the complicit multiplicands. Then comes 101 102 103, wherein 101 and 103 is a prime pair. Then comes 137 138 139, wherein 137 and 139 is a prime pair. Then comes 149 150 151, wherein 149 and 151 is a prime pair. Then comes 161 162 163, but/ 7 x 23 = 161. Notice: 23 - 11 = 12; 25 - 13 = 12. Okay, backtrack. 5 x 23 = 115 5 x 25 = 125 7 x 23 = 161 7 x 25 = 175 Further: 175 - 70 = 105; and 175 - 14 = 161, and 105 - 14 = 91 Also: 105 + 14 = 119, and 175 + 14 = 189 So these: 3 x 7, 5 x 7, 7 x 7; 13 x 7, 15 x 7, 17 x 7; 23 x 7, 25 x 7, 27 x 7 [ This means that "5" controls distribution--downwards and upwards-- for every pown including itself via a 1 x, 3 x, 5 x, 7 x, 9 x rotation of single digits and final digits. ] The node number "105" is a good small example of this/ because it isdivisible by 3 and 5 and 7, forcing "3" to 102 and 108; forcing "7" to 98 and 112; and thereby allowing 101, 103, 107, and 109 to be prime numbers. - "107" and "109" bring me to the second way, which I will begin at "4" instead of at "0". The rotation is: 4 8 2 6 0 3 4 5 7 8 9 11 12 13 15 16 17 19 20 21 23 24 25 27 28 29 31 32 33 35 36 37 39 40 41 43 44 45 47 48 49 51 52 53 55 56 57 59 60 61 63 64 65 67 68 69 71 72 73 75 76 77 79 80 81 83 84 85 87 88 89 91 92 93 95 96 97 99 100 101 103 104 105 107 108 109 111 112 113 115 116 117 119 120 121 123 124 125 127 128 129 131 132 133 135 136 137 139 140 141 143 144 145 147 148 149 151 152 153 155 156 157 159 160 161 163 164 165 167 168 169 In this view the pown squares always occur in the +1 column. After 3 4 5, wherever a "4" or a "6" is a final digit in a group, the powns in it cannot be a prime pair. Beginning with 11 12 13, it appears that only where the pewn of a group is divisible by "12" can its powns possibly be a prime pair. In this view it also appears there are fewer prime pairs. In the first view the first prime pair is in the 5 6 7 group. "6" is 6 x 1. In this view "12" is 6 x 2. I am deducing from this that in the first view the pewn in a group containing a prime pair will always be an odd multiple of "6", whereas in the second view the pewn in a group containing a prime pair will always be an even multiple of "6". # Brian A. J. Salchert
Thursday, May 1, 2008
Wednesday, April 30, 2008
sw00872ut-38.2poems
The Undulant Trees Opportunity I had a life once, a life/ handed to me on a golden matter: more, more, & more; but did not see it, and let it pass. April's End 'Tis the last day of April. 'Tis, 'tis, 'tis. Oh how quickly 30 whizz. And here I sit, sullen and somber, where undulant evil is. Yo, rise/ from your stool. There are reams to conquer, twirling fool. - Brian A. J. Salchert
Monday, April 28, 2008
sw00871d26-melodies
Lately melodies have been streaming through my body in strings that go on for an hour or more. It got so tonight that I felt I was getting to be like (ha ha) Franz Schubert. This led to a Franz Schubert search and to Wikipedia. That man, sometimes known as the little mushroom, was amazing! He wrote over 600 lieder, a fact alone, given that he died at the age of 31, enough to assure his fame; but he also composed chamber music, operas, and was working on (I believe) his 10th symphony when he died. These melodies in me come out as modulated exhalings. I need to get some recording equipment. It could be that these melodies already exist and I am just pulling them out of the ether somehow. Sometimes, as now, I'll get stuck on one and repeat it over and over. At other times one or two notes in one melody will lead me into another melody, one which may or not/ be related to the progenitor melody. It's strange, but it can be enjoyable. Years ago there was a period when I was composing parts of symphonies in a similar manner, but I was not able to transcribe them. I still am not able to, and the only instrument I can play is my breath instrument. Brian A. J. Salchert 2008-04-28
Friday, April 25, 2008
sw00870math-3zs5zs7zs.part2
Other integer knowns pertaining to primes and nonprimes: - 1. What I call Elimination Tables can be constructed/ wherein 2 x the pown (positive odd whole number) whose mulitples are being eliminated generates the first pewn divisible by that pown. That pewn and the integer multiples of it become C2 of a three-column table/ wherein C1 contains all the powns which are 1 less than each C2 pewn, and C3 contains all the powns which are 1 greater than each C2 pewn. If "3" and its multiples are the powns being eliminated, all the powns not divisble by "3" will be among the minor or major powns in the table. Many of these powns will be prime numbers, but note that all the squares of powns will be major powns. 2. Some facts about squares, powns, and pewns: a) Subtracting 1 from any pewn square whose final digit is 6 equals a pown which is divisible by "5". b) Subtracting 2 from any pown square often equals a pown which is a prime number, but there are some interesting exceptions. 119, or 7 x 17, is one of these. c) The recurring final-digit sequence for pewn squares is: 4, 6, 6, 4, 0; and the recurring final-digit sequence for pown squares is: 1, 9, 5, 9, 1. d) The Elimination Table for multiples of "3" is the 6n table. At each 6n (C2), the C1 pown and the C3 pown constitute a pair such as 5 7, 11 13, 17 19, 23 25, 29 31, 35 37, 41 43, 47 49, 53 55, 59 61, 65 67, 71 73, 77 79, 83 85, 89 91, 95 97, 101 103, 107 109, 113 115, 119 121. If this table is taken into infinity, all the possible prime pairs exist in it. e) Any integer whose final digit is 5, being divisible by "5", is automatically a nonprime. Therefore, for all powns > 9 only those powns whose final digits are 1 or 3 or 7 or 9 need be inspected for primality. f) 2 is the only pewn prime, and the only pown prime triad is: 3 5 7. g) An integer's power begins at its square. This is why "2" is the only pewn prime. This is also why the lesser of two pown integers is always the ruling eliminator of integers greater than itself. Brian A. J. Salchert
Thursday, April 24, 2008
sw00869math-3zs5zs7zs
( 3) 1+2 4+5 7+8 10+11 13+14 16+17 19+20 22+23 ( 5) 2+3 7+8 12+13 17+18 22+23 27+28 32+33 37+38 ( 7) 3+4 10+11 17+18 24+25 31+32 38+39 45+46 52+53 ( 9) 4+5 13+14 22+23 31+32 40+41 49+50 58+59 67+68 ( 11) 5+6 16+17 27+28 38+39 49+50 60+61 71+72 82+83 ( 13) 6+7 19+20 32+33 45+46 58+59 71+72 84+85 97+98 ( 15) 7+8 22+23 37+38 52+53 67+68 82+83 97+98 112+113
# About 1+2, 2+3, 3+4, and: The first # = the row #, and the second # indicates the column # where the square of the ( #) is. # Next: If the two adjacent #s which sum to a ( #) exists in a prior row, that ( #) is not a prime #. Example: 4+5 is in row 1 column 2, and 4+5 is also in row 4 column 1. # Next: Column 1 is the x1 column, C2 is the x3 column, C3 the x5, C4 the x7, C5 the x9, C6 the x11, C7 the x13; C8 the x15. So, (column # x2) - 1 = the times # for that column. # Next: Since 4+5 is in C2 of R1, which is 3s row, 4+5 = 3x3. # Next: That # which is the second # in R1C1 is the first in R2C1. This means that there are three distinct numbers C1 of any two adjacent rows. These three #s in C1 of R1 and R2 sum to 6. These three #s in C1 of R2 and R3 sum to 9. These three #s in R5 and R6 sum to 18, or 3x6. The ( #) in R5 is 11, and the ( #) in R6 is 13. 11+13=24, or 3x8. 24 - 18 = 6. Go back. 3+5=8; 8 - 6 = 2. 5+7=12; 12 - 9 = 3. So, the distinct #s in adjacent rows rise by 3, and ( #)s in adjacent rows rise by 4 when they are summed as terms in a set of adjacent rows. I am not yet certain what the value of knowing these facts is, but I suspect there is an interpolation value. # Next: A known about primes is that there is a prime number between n and 2n for every integer n > 1. If n = 2, 2n = 4. 3 is between 2 and 4. If n = 3, 2n = 6. 5 is between 3 and 6. If n = 4, 2n = 8. Both 5 and 7 are between 4 and 8. If n = 5, 2n = 10. Both 7 and 9 are between 5 and 10, but we know that 9, or (10 - 1), is not a prime. Between 8 and 16 are 9, 11, 13, and 15. Question: Why do 5+6 and 6+7 equal numbers which are prime numbers, while 4+5 and 7+8 do not? Brian A. J. Salchert
Wednesday, April 23, 2008
sw00868d25
Yesterday I joined Goodreads, a free connectivity service for people who read. Anyone interested in a recipe for a large batch of really hot chili should go to my Other Journals section and click on a place in Ohio. - [ 2008-10-23 note: the place in Ohio is in limbo and I'm not sure when it will return from there, if it does return. ] Brian A. J. Salchert 2008-04-23
Tuesday, April 22, 2008
Monday, April 21, 2008
sw00866d23
d = diddlings Do not particularly like this Verdana size 3, but I tried Times New Roman and Courier New and did not like them. I had a wrong impression about the former and, as I have known, the latter is like typewriter type--which I like--but is too light. The jump from 2 to 3 in Verdana is too great. I would prefer a size between those choices.
This is Verdana size 2.
This is Georgia size 3. a b c d e f g h i j k l m n o p q r s t u v w x y z 0 1 2 3 4 5 6 7 8 9 - Though it's a bit heavy, I may go to this.
This is Trebuchet MS size 3. a b c d e f g h i j k l m n o p q r s t u v w x y z
This is Courier New size 4.
This is Perpetua size 5. a b c d e f g h i j k l m n o p q r s t u v w x y z
- - Should get back to entering my old online diaries but I just don't feel up to it. Keep floating around the Internet for other web logs of interest to me. Have found several. Spring has sprang, and I took some photos over the weekend of things in bloom and of bushes displaying variant colors. Today it warmed up enough for me to go to my mailbox without a jacket. - - An interesting serendipity occurred this evening which relates to a poem by Sir Thomas Wyatt. Read what Boyd Tonkin writes about himself in his "Why poetry still matters" feature and then go K. Silem Mohammad's site where he shows and discusses the same poem. see Monday, April 14, 2008 Brian A. J. Salchert 2008-04-21
Saturday, April 19, 2008
sw00865d22
d = design two articles from Technology Review "Nano Drugs to Starve Tumors" By Jocelyn Rice - Friday April 11, 2008 (about fighting cancer more safely) "Corn Primed for Making Biofuel" By Alexandra M. Goho - Wednesday April 16, 2008 (about less expensive and cleaner ethanol) Brian A. J. Salchert 2008-04-19
sw00864sl13-this.lyrics
This Being I Am Wherever I wander; wherever I flee / the Lord is always after me. Wherever I wander; wherever I flee / the Lord is walking ahe-ead of me. It keeps its mystery however I ponder. It keeps its mystery however I plead. It keeps its mystery however I wonder. It keeps its mystery however I weave. Brian A. J. Salchert 2008-04-19
Friday, April 18, 2008
sw00863d21
d = disposition Death to any human who does not believe as I believe is an evil perception fanned by desperation and anger allowed to heat to the level of hate. Such, whether based in a theology, a philosophy, a psychology, or a whatever, is self-serving, and cares not about the common good. If all you can do is pray, do that. - Over at Common Dreams: "The Unmaking of the Market" by Sally Kohn Read it and also scroll down to the Galen April 16th, 2008 12:04 pm comment and read its final paragraph. "Carter Calls Gaza Blockade a Crime and Atrocity" This is a Reuters article by Jonathan Wright. Brian A. J. Salchert 2008-04-18
Wednesday, April 16, 2008
sw00862v-43.poem41
Venturings World Without What it was and why it was no one knew. For unknown years it had been there. Unbothered, it seemed/ it could stay/ centuries more. Building, mountain, boulevard, trail, the sun rolls round while over snow south of its range a Snowy Owl rises/ off a fence post/ and floats beyond. Neither time, nor reason, nor an empty heart jostles the neurons of desire in this fantasm. What lingers, lingers; what departs, departs. Nod. - Brian A. J. Salchert
Tuesday, April 15, 2008
sw00861d20-campaign
d = decision If no changes that matter have occurred, when voting day for the next president of the these endangered United States comes, Senator John McCain (a politician I have liked at times over the years, but for whom I/ will not vote then) will be the Republican candidate; and either Senator Barrack Obama or Senator Hillary Clinton (neither of whom is a politician who sells what I would prefer to buy then) will be the Democratic candidate; and there will be outsider candidates. If someone could prove to me that an outsider candidate I would be willing to vote for then/ will have at that time a reasonable chance of winning, I would vote for that person. Yesterday over at Common Dreams I read Obama Is Right by Katrina vanden Heuvel and many of the comments. In one of those comments was the conclusion that a vote for an outsider candidate would in effect be a vote for McCain, that is if/ that vote/ would have been cast for the Democratic candidate otherwise. I agree. Therefore, aside from not voting, I will vote for whichever senator is the Democratic candidate. The central reason men gathered to form this nation was the tyranny of King George III. What besets this nation now, and indeed the world, is oligarchic tyrrany based on the so-called free market system. President Eisenhower, who had he felt as comfortable in his statesman role as he felt in his military role (difficult as that was) would likely have made some significant salutary changes, warned us about this. Given how insidiously the ruling powers have trapped those being ruled, it is hard to know what to do about our incarceration. After all, it is not totally odious. One could make, little by little, moves toward self-sufficiency, and wait for the overlord system to collapse; but then what? One could hasten its downfall by supporting only the clean and green, of which ethanol is not a part. See Care2 but avoid its petitions area. One could help promote, organize, and then convene a Peoples Congress. I, for one, would like to know what those who do not vote and those who've been disenfranchised think, even if as a group their opinions mirror the opinions of those whose votes do count. Two aspects of collusional oligarchic power which sadden me most are how easily it can eliminate what threatens it and/or how easily it can co-opt an adversarial creation. Alexander Hamilton was killed by Aaron Burr in a duel, but Hamilton's spirit (wherever it is) must be so excited it's about to pop/ out of its hiding place. At Wikipedia is this quick biography of Hamilton. Brian A. J. Salchert 2008-04-15
Sunday, April 13, 2008
Saturday, April 12, 2008
sw00859d18-USA
d = deliberation Since I can't do it in person, am thinking about touring online. Have no plan as yet. May start with overviews. Satellite images of some U.S. cities. Thanks to a 4-11-08 link on Linh Dinh's Detainees to an article by one Ted Gup, I'm changing direction. This is a timeline history of the U.S. Civil War. - current U.S. cabinet - U.S. Supreme Court - U.S. Senate - U.S. House of Representatives - 1952 U.S. presidential election With thanks again to Linh Dinh, who placed a link to the following in a comment to me on his blog. Read this: "The Great American Media Mind Warp" by Joe Bageant. Brian A. J. Salchert 2008-04-12
Wednesday, April 9, 2008
sw00858ut-37.poem
The Undulant Trees [ Note: One should not laugh at another's misfortune. The content below popped into my head several days back, and does not relate to any familial event in my life. ] Out There I laughed when my fa-ather lost his shoe. I laughed when my fa-ather lost his shoe. I laughed, and laughed, and laughed, and laughed. I laughed when my fa-ather lost his shoe. I laughed when my fa-ather lost his shoe. I laughed when my fa-ather lost his shoe. I laughed, and laughed, and laughed, and laughed. I laughed when my fa-ather lost his shoe. - Brian A. J. Salchert
Tuesday, April 8, 2008
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"The Ideology of American Empire" "737 U.S. Military Bases = Global Empire" PBS Frontline: Understanding The Book of Revelation - Brian A. J. Salchert 2008-04-08
Monday, April 7, 2008
Sunday, April 6, 2008
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d = diary 11 steps to a better brain at NewScientist is-- even though from 5-28-05--a worthwhile read. "Why the demise of civilization may be inevitable" by Deborah MacKenzie is also at NewScientist, but you need to subscribe if you want to read the full article. Here is one available sentence: "It appears that once a society develops beyond a certain level of complexity it becomes increasingly fragile." Brian A. J. Salchert 2008-04-06
Saturday, April 5, 2008
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Partly because it may not be available much longer, visit this journal: Isaac Stolzfuts' Journal Its author has been missing since about August third of 2007. He was 87 at that time. Have been involved with a math project based on Pascal's Triangle, its fourth row to be exact. It's a different way of looking at the 6n Elimination Table, a table which begins with the 5 7 pair, rising through all the odd natural numbers which are not divisible by "3". Thus the next three pairs are: 11 13, 17 19, 23 25. The second row of this triangle is comprised of all the natural numbers except "1". 1 is the number which borders two sides of the triangle, and "0" is not in the triangle. The triangle's third row is comprised of numbers in the natural number summation sequence, beginning with "3". I have done what I set out to do with these numbers. This is all I'm saying about it. Brian A. J. Salchert 2008-04-05
Friday, April 4, 2008
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Other than a brief post at Rho-, today has been an online reading day. Went to Robert Peake's site--though I'd been there late last night--to read his "Post-Postmodernism And Hope" entry again. Because his thoughts speak to me aesthetically and as a human through his words about Paul Celan and Umberto Saba (a poet new to me but one who was able to choose life over death against the ravagings of WW2), I am encouraged by them. Also revisited the "Performatism, or the End of Postmodernism" essay by Raoul Eshelman which relates to Robert Peake's entry, but which I searched for separately. At Silliman's Blog for April 04, 2008, is a link to an interview of Rachel Blau DuPlessis by CA Conrad I had to read in two sittings. Course work, but of an essential nature. I then went to Jacket 35 and read her Draft 88 and Draft 89. Revelatory. Read a few other poems and parts of poems in that issue, but as has often happened during the last 12+ months, I was not moved by them other than toward ennui, a state I suspect reading many of my poems would induce: which says less about the poems than it does about the reader. Sometimes I wish I lived in Philadelphia instead of most of the places I have lived in. But complaining is such a deadening act. Was at Latta's Isola too, which today was extra terrestrial. Brian A. J. Salchert 2008-04-04
Thursday, April 3, 2008
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Just had enough time this morning to change the dates in All About Me and to check the weather map. It is now 4:30 PM and overcast, but rains seem to be angling in from Oklahoma City. Could go mostly south and east of Springfield. So far, most of my day was away from here. Had to get ready to go shopping, and when that was done, had to eat while putting the items I bought in their proper places; and also deal with other needs. Best benefit I got was exercise. Brian A. J. Salchert 2008-04-03
Wednesday, April 2, 2008
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Rest of today and all of tomorrow are expected to be stormy. May not be online much, if at all. Plan to go grocery shopping tomorrow. If that happens, beets and lima beans will be at the center of my food changes. Will be pulling away from dairy products and processed meats. Brian A. J. Salchert 2008-04-02
Tuesday, April 1, 2008
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Partly cloudy with brisk cool winds. Took two photos of the blooming sturdy bush. My sister then took me to where my general practitioner is so I could have blood drawn for two tests. We also went to the closest library where I requested two books. After that we drove around a Wendy's because I wanted one plain baked potato. And then, and then, and then, and then, and then, and then, and then, and this. Did you get all that? This is a good site to read whatever of if you want information about healthy foods such as lima beans. Brian A. J. Salchert 2008-04-01