Venturings Angle of Vision Ready to travel the miles north, we two, about to part another time, twice kiss; i do not walk her to the door; but from the bedroom watch until she's back in view three stories below and sits upon her red suitcase so placed on the main walk the patio's post eclipses them, and with her await the maroon taxi & move the suitcase back a little & pace & check the time and open its door & lift onto its back seat the troubling case & get in & look up & smile & wave & wave & smile because we want to because we can. circa 1986 Brian Salchert
Thinking Lizard
About Me
- brian (baj) salchert
- Rhodingeedaddee is my node blog. See my other blogs and recent posts.
Saturday, May 31, 2008
sw00907v-54.poem52
sw00906d30-offline
What is it that suddenly motivates a person to do something that's been put off for years? Have not a thimble of evidence. Was offline most of today organizing my documents. Still need to clean up certain files, but that space looks 90% happier. 2008-05-31
Monday, May 26, 2008
sw00905math--find-c-with-straight-line
36 r = radius c = circumference Not expecting to finish this entry today, but have gotten into the substance of it far enough to show the confusion therein. This search had its origin in March of 1966. Shortly thereafter I abandoned it. Yesterday I began addressing it again. Circumference = 2 x pi x radius. Therefore, if the radius is 1/2 inch, the circumference = pi = 3.14159265, which is the pi value being used here. My project in 1966 was to see if the circumference of a circle could be determined without pi. This led me to considering the 90° arc with the notion that with the aid of a protractor I could draw interlocking circles and so establish a straight line that equalled the length of a circle's circumference divided by "4", or approximately did. I may scan my original calculations (which I uncovered several days back while looking for something else) and show them here, even though upon inspecting them then I became uneasy about my conclusions in 1966. Could be my conclusions were correct and I simply no longer understand them. Moot, moot. I now have new conclusions. Let r = one inch. [ Spotty storms are meandering in. ] c = 6.28318530 /4 = 1.57079633 = pi/2 I then found that 4/7 = .57142857 Today I found that 1.57079633 divided by 11 = .14279967 and that 7 x .14279967 = .99959766 which is quite close to one inch. However, I then proceeded to r = 2 (two inches) where c = 12.56637060 and /4 = pi but pi/11 = .28559933 and 7 x that = 1.99919532 which comes to a loss of .00040234 And so I then proceeded to r = 3 where c = 18.84955590 and /4 = 4.712388898 and 4.71238898/11 = .42839900 and 7 x that = 2.99879298 which comes to loss of .00080468 from the r = 1 result, meaning that there is a loss of .00040234 per inch and consequently a loss of .00005748 per 1/7 of an inch, and of .00063225 at 11/7. 5:50 PM - The .00040234 per-inch loss bothered me; so I ventured on. 1/7 = .14285714 1/56 = .01785714 whereat I tried x 9 and got .16071429 At this juncture I returned to my trial-by-error habit because I knew what I was seeking was far off. Finally upon dividing .16071429 by 399.45 (oh, yes) I got .00040234. So, I needed to know what 56 x 399.45 equalled. Hmmph!: how does 22369.2 strike you? Indeed it was far off. 1/22369.2 = .00004470 Next question: How important is the .2? Turns out that at eight out, not seemingly significant. 1/22369 = .00004470 But still I ventured, as I wanted to know how 22369 related to 7. It doesn't. I then tried 3, and it didn't either. This made me wonder if 22369 was either a prime number or a square; so I minused 1 from it and divided by 8. Result: 2796. At that point I stopped. And now I need to look at my earlier entry regarding terms in the natural number summation sequence. Hiatus. 8:47 PM - Fact: The remainder of an odd square - 1 into which 4 is divided always equals an even integer which is the product of two adjacent integers, and the sum of those two adjacent integers always equals the square root of the odd square. Example: 17 x 17 = 289 288/4 = 72 and 8 x 9 = 72 and 8 + 9 = 17 - Given this fact, 22369 is not an odd square/ because 4 divided into 22368 equals 699, which is not an even integer. It is still possible 22369 is a prime number since 22368 is evenly divisible by both "3" and "6" but I think that does not matter. 1/22370 also equals .00004470. 22370/2 = 11185. 1/22372 also equals .00004470. 22372/7 = 3196. 22372/4 = 5593. 3196 x 4 = 12784 and 12784 + 22372 = 35156 and 35156/11 = 3196. - .00040234/.00004470 = 9.00089486. .00004470 x 9 = .00040230 which puts the result off by .00000004. But what the heck is 1/22372 of an inch? The closest unimaginable ruler number is 1/16384. Then again, rulers don't bother with sevenths of an inch either. Whatever, 12784 + 9 = 12793 or 9/22372 more than 4/7 of an inch. Now what? Back to 57079633. ... Wait! I should be moving lower. 11:44 PM - Just discovered "13" divides into 57079633 evenly, getting 4390741. I should be in bed. Tomorrow perhaps. May 27, 2008 7:26 AM - Other findings included 12771/22374 = .57079646, but after truncating 2506218.849 to 2506219 and dividing that integer by 4390741, .57079636 arrived/ because (I surmise) that while 13 divides evenly into 57079633 it does not divide evenly into 4390741. 13 divides evenly into 4390737. At nine out, .570796364 arrives. My calculator does not show ten zeros beyond the decimal point, and since I'm done with hunting, somewhere between .00000003 and .00000004 sits the error. 1/4390741 = .00000023 or .000000228. 2506219 + 4390741 = 6896960; 1.57079633/6896960 = .00000023; .00000023 x 4390741 = .99999998 of an inch. At .000000228 it becomes 1.001088948. 1.001088948 - .99999998 equals .001088968. Under, over, the yard north of my bedroom win- dow is happily banded with white clover, but the dandelions which banded it last week are now nowhere to be seen. 1:48 PM - Among other activities, went out to take some photos and did find several d-lions. At r = 2 inches c = 12.56637060 and c/4 = pi = 3.14159265 3.14159265/6896960 = .00000046 and 4390741 x .00000046 = 1.99999995 loss per inch of .00000003 (2.5) At r = 3 inches c = 18.8495590 and c/4 = 4.71238975 4.71238975/6896960 = .00000068 and 4390741 x .00000068 = 2.99999993 loss per inch of .00000002 (2.33) - This is much closer, but had to go into outer space to get it. - Recap: 4/7 = .57142857 2506219/4390741 = .57079636 target .57079633 Sounds like the robin is practicing a new tune. It has a falsetto finishing note. May 29, 2008 2:12 PM - Went out to 9 on my calculator. 3.141592654/2 = 1.570796327 and after some t-and-e chose to use .570796330 because two integers, both of which happened to be divisible by 15, equalled it. So 250621905/439074135 became 16708127/29271609. Now, if I use the smaller integers, I first need to divide 1.570796330 by 15 which equals .104719755; and then divide this by 45979736 (16708127 + 29271609). Either way, the result is .000000002. 250621905 + 439074135 = 689696040. 1.570796330 divided by 689696040. [ Much of today's entry is calculator-related junk. My attempt to go out nine has failed. Am unable to verify calculations. ] - See this Wikipedia page on Pi. May 30, 2008 4:06 PM - About yesterday's numbers: Dividing 45979736 by 15 was wrong. The numbers used relate to a circle with a radius of one inch, making 2 x pi the circumference and division by 4 approximately equal to 1.57079633. After the divisions by 15, the essential integers are 16708127, 29271609, and 45979736. (However close these are to 4/7, 7/7, and 11/7, the 1.57079633 straight line represented by 45979736/29271609 can only be imagined.) Found a way to get the answers I needed from my calculator. I eliminated the decimal point in 1.57079633 when I divided it by 45979736. That gave me 3.416279576 which truncating to 3.41627958 proved best. This is a deceptive number in that it looks a bit like pi, but it has nothing to do with pi. When this number is multiplied by 29271609, my c-toy displays 100,000,000.1 or 1.000000001 after the decimal point is returned to its proper location. All these calculations and the need to verify them entailed more effort than it probably was worth. If 3.14159265 is allowed to equal pi, 6.2831853 (2 x pi) equals the circle's c. If 1.57079633 is allowed to equal c/4, 6.28318532 = the circle's c. # Brian A. J. Salchert
Saturday, May 24, 2008
sw00903math--squares-sets-from-4-divisor
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
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
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