MathTwentieths Funky Bar Graph - and some nnss-related findings [ last modified: 2008-10-27 ] kestroke formula for bottom line: n + (n + 1) If you want to make a graph for twenty twentieths, you will need 20 + (20 + 1) bottom-line keystrokes. # 033107 insight which uses nnss terms: see at *** below and also an insight which uses non-nnss integers 032807 Note: n + (n + 1) is a natural number summation sequence formula when a term in the nnss is an odd integer. There are therefore only certain numbers which can be "n" numbers, and there is a pattern inherent in this set of numbers. 0 and 1 are the first pair of "n" numbers. 7 and 10 are the second pair of "n" numbers. 22 and 27 are the third pair of "n" numbers. 45 and 52 are the fourth pair of "n" numbers. What can be determined about these numbers? One: the difference between the numbers in each succeeding pair increases by a factor of 2. Two: the difference between the greater number in a given pair and the lesser number in the following pair increases by a multiple of 6. Three: notice how multiples of 8 relate to these "n" numbers. (8 x 0) - 0 = 0 / (8 x 1) - 7 = 1 (8 x 3) - 22 = 2 / (8 x 6) - 45 = 3 (8 x 10) - 76 = 4 / (8 x 15) - 115 = 5 - Stop. 22 - 7 = 15. (7 x 2) + 1 = 15. 45 - 22 = 23. (7 x 3) + 2 = 23. 76 - 45 = 31. (7 x 4) + 3 = 31. 115 - 76 = 39. (7 x 5) + 4 = 39. So, this difference group of numbers wherein the difference between terms is 8: 7, 15, 23, 31, 39, 47, 55, 63, 71, 79, n. - Now, (7 x 6) + 5 = 47. 47 + 115 = 162. (8 x 21) = 168. 168 - 162 = 6. - (7 x 7) + 6 = 55. 55 + 162 = 217. (8 x 28) = 224. 224 - 217 = 7. - (7 x 8) + 7 = 63. 63 + 217 = 280. (8 x 36) = 288. 288 - 280 = 8. - (7 x 9) + 8 = 71. 71 + 280 = 351. (8 x 45) = 360. 360 -351 = 9. - (7 x 10) + 9 = 79. 79 +351 = 430. (8 x 55) = 440. 440 -430 = 10. - 45 is both in the "n" number group and is a term in the nnss because 22 is in the "n" group and 22 + (22 + 1) = 45. - *** (7 x 0) + 0 = 0 (7 x 0) + 1 = 1 (7 x 1) + 0 = 7 (7 x 1) + 3 = 10 (7 x 3) + 1 =22 (7 x 3) + 6 = 27 (7 x 6) + 3 = 45 (7 x 6) + 10 = 52 (7 x 10) + 6 = 76 (7 x 10) + 15 = 85 (7 x 15) + 10 = 115 (7 x 15) + 21 = 126 # 0 + (0 + 1) = 1 | 1 + (1 + 1) = 3 7 + (7 + 1) = 15 | 10 + (10 + 1) = 21 22 + (22 + 1) = 45 | 27 + (27 + 1) = 55 45 + (45 + 1) = 91 | 52 + (52 + 1) = 105 - difference notes: pairs: 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50 < run: 14, 30, 46, n (beinning with 1): 1, 15, 45, 91, n > run: 18, 34, 50, n (beginning wirh 3): 3, 21, 55, 105, n ?: How can one tell from the "pairs" differences/ what the "< run" and "> run" integers are? Answer: Half of a pair difference times the pair difference equals an integer from which when the half of its pair difference is subtracted out = its "< run" # and from which when the half of its pair difference is added in = its "> run" #. Example: 14/2 = 7. 7 x 14 = 98. 98 - 7 = 91. 98 + 7 = 105. - 12:39 PM March 31, 2007 - This means that/ beginning with a positive "2" is a set of positive integers controlled by the positive even-number multiples of positive "8" from which the terms in the nnss (natural number summation sequense) can be derived. "98" is the fourth term in that set. Here is an initial look:2 18 50 98 162 242 338 45016 32 48 64 80 96 112 - How can one know that "7" is the essential positive integer associatedwith "98"?Enter termposition. 2 x 4 = 8, and and 8 - 1 = 7. How can one know that at term-position 4 the positive integer associated with it is "98"? This is a bit more difficult, and also contains a circular trap. This is the trap: 98 -2 = 96. 96/16 = 6. "6" is--if one accepts "0" as term 1 in the nnss--at term-position 4, with "1" being at term-position 2 and "3" being at term-position 3. So: time- out. [ Used my in-bold numbers in an advanced search and got a single result which is attributed to Joshua Zucker and the Castilleja School MathCounts club (joshua.zucher (AT) stanford alumni.org), Nov 07 2002, and has to do with: Maximum number of regions the plane can be divided into using n (concave) quadrilaterals. Do what I did to learn more. ] This doesn't settle my quandary, which is quite different. It may be that the trap I exposed is actually a proof, but I desire a less circuitous answer. 98 - 4 = 94. 50 - 3 = 47. 18 - 2 = 16. 2 - 1 = 1. 15, 31, 47, n which suggests "n" will be 63. 63 + 94 = 157, and 162 - 5 = 157. So now yet another set of differences begins, but to what purpose? Obviously, 16 or (8 x 2) controls this set, and (63 + 1)/16 = 4. Perhaps I will return here tomorrow. Can't let this go. 2 + (16 x 1) = 18; 18 + (16 x 2) = 50; 50 + (16 x 3) = 98; 98 + (16 x 4) = 162. The only answer I can come up with is the term-position match answer, aided by knowing that if 2 less than a number in my in-bold set is evenly divisble by 16, that number may be correct. I must go again. 4:53 PM - Who stole the magic? 3 - 1 = 2. 10 - 6 = 4. 21 - 15 = 6. 36 - 28 = 8. 55 - 45 = 10. I am trying to see something. Toward that end I have left out "0". 50 is at term-position (t-p) 3, and (50 - 2)/16 = 3.If I let 50 be at t-p 5, then its position would match its -/+ difference.My in-bold numbers do pertain only to the odd nnss terms, but it appears I simply ought to make believe the even nnss terms still matter/ and assign only odd term positions to my in-bold numbers. What thinkest thou? Doing so woud certainly simplify matters. I am going to do it. So: How can one know that at t-p 7 the positive integer associated with it is "98"? 98/7 = 14, which appears to be at t-p 4 in the "pairs" run, but from here on will be at t-p 7 in that run. There are times when making believe has its advantages. - see my nnss entry - ~ # twentieths funky bargraph

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# Brian A. J. Salchert

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