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Conjecture: Sophie Germain Triangles & x  2y^2 + 2y  3 = z^2 
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#19
Mar2211, 11:28 AM

P: 688

entry(n,k) = entry(n1,k1) + 2*entry(n1,k)The reason why they add up to 3^n is because that is the binomial expansion of [itex](2+1)^n = \sum_{k=0}^n \binom n k 2^k[/itex] Hope this helps, if this coincides with what you're going after. 


#20
Mar2211, 02:04 PM

P: 153

What is actually being addressed on this thread at the moment are the following recursionbased relationships uncovered by Ramsey2879. 16x + 9 = (4^2)x + 2*4 + 1 144x + 25 = (12^2)x + 2*12 + 1 576x + 49 = (24^2)x + 2*24 + 1 1600x + 81 = (40^2)x + 2*40 + 1 etc. ... and that it ought to be possible to generalize his results to all of the Pell Triangle in some manner as yet TBD. For instance, Ramsey2879 is working with squares and triangles at the moment. If one were to "step over" one diagonal to the left, might there not perhaps be a way to come up with similar recursionbased formulas for multiples of tetrahedrals (perhaps in polynomial expanded form...) plus some constant that add up to cubes, pentagonal numbers or some other class of square number? Best, Raphie 


#21
Mar2211, 09:03 PM

P: 894

40^2 *T(x) + 81 is a square for T(x) = 0,1,488566,6349266,128600703 and 1671258205 60^2 *T(x) + 121 is a square for T(x) = 0,1,743590,11132121,110744403,and 1657929736. The x's increase in size rapidly and there is no clear recursive formula given only the above. If you want to try and solve each series, first put each in the order {[6],[4],[2],[1],[3],[5]}. 


#22
Mar2311, 05:44 PM

P: 688

Sorry for interrupting again with perhaps something obvious, but extending a result on triangular numbers to the other polygonal numbers may not be the proper thing, given that it's only the triangular numbers that appear as a diagonal of Pascal's triangle; subsequent diagonals in Pascal's triangle have no relation (no obvious one, anyway) with the squares, pentagonals, and so on. Why should they?
Edit: Hmm... actually, there IS a relation between polygonal and triangular numbers, given here: http://en.wikipedia.org/wiki/Polygonal_number#Formulae namely, the nth sgonal number is given by (s2)*T(n1)+n. That should lead to relations with the triangle in post #18; (the one which was Pascal's triangle but with diagonals multiplied by powers of 2). For example, taking the next diagonal of that triangle: 8,32,80,160,... call them R(n), with n=1,2,3... Then (R(n+1)R(n))/4+n+2 should produce squares. Examples: (328)/4+3=9; (8032)/4+4=16; (16080)/4+5=25; (280160)/4+6=36; (448280)/4+7=49; ... 


#23
Mar2411, 05:42 PM

P: 153

Yes, Dodo, All polygonal numbers are constructible as multiples of Triangular Numbers, plus a Counting Number. e.g. 22 = 3*T_3 + 4; 35 = 3*T_4 + 4; ; 51 = 3*T_5 + 4 > 4th, 5th and 6th Pentagonal Numbers. Pentagonal Numbers,specifically, also have the property that they are, each and all, 1/3 a Triangular Number.
Here is another way of looking at how Polygonal Numbers are constructed that better demonstrates how they and nHedral Numbers are (or can be...) "born" in very similar manner... Construction Rule: 1A + 2B: Start [1, n] Number of Summations (i.e. "Overlays"): 1 001,002,003,004,005,006,007,008,009,010... Counting Numbers: Start [1, 1] 001,003,006,010,015,021,028,036,045,055... Triangular Numbers: Start [1, 2] 001,004,009,016,025,036,049,064,081,100... Square Numbers: Start [1, 3] 001,005,012,022,035,051,070,092,117,145... Pentagonal Numbers: Start [1, 4] 001,006,015,028,045,066,091,120,153,190... Hexagonal Numbers: Start [1, 5] 001,007,018,034,055,081,112,148,189,235... Heptagonal Numbers: Start [1, 6] etc. e.g. Counting Numbers [01, 01], 01, 01, 01, 01, 01, 01...  "Zeroeth" Summation > (0n + 1) [01, 02], 03, 04, 05, 06, 07, 08...  First Summation > Counting Numbers e.g. Pentagonal Numbers [01, 04], 07, 10, 13, 16, 19, 22  "Zeroeth" Summation > (3n + 1) [01, 05], 12, 22, 35, 51, 70, 92  First Summation > Pentagonal Numbers compared to... PASCAL's TRIANGLE/SQUARE Construction Rule: 1A + 2B: Start [1, 1] or... Construction Rule: 0A + 1B: Start [1, 1] (amongst others formulations...) "Number" of Summations: Infinite 001,001,001,001,001,001,001,001,001,001... "Zeroeth" Summation > (0n + 1) 001,002,003,004,005,006,007,008,009,010... First Summation > Counting Numbers 001,003,006,010,015,021,028,036,045,055... Second Summation > Triangular Numbers 001,004,010,020,035,056,084,120,165,220... Third Summation > Tetrahedral Numbers etc. compared to... PELL NUMBERS Construction Rule: 1A + 2B: Start [1, 1] Number of Summations: 1 [1, 1], 3, 07, 17, 041, 099... "Zeroeth" Summation > (Half Companion Pell Numbers) [1, 2], 5, 12, 29, 070, 169... First Summation > (Pell Numbers) [1, 3], 8, 20, 49, 119, 288... Second Summation > (see links below...) Note, for instance, that (41 1)/2 = 20, (991)/2 = 49, etc. ============================================= Expansion of 1/(13x+x^2+x^3). http://oeis.org/A048739 Or, alternatively... a(n)th triangular number is a square: a(n+1) = 6*a(n)a(n1)+2, with a(0) = 0, a(1) = 1 If (X,X+1,Z) is a Pythagorean triple, then ZX1 and Z+X is in the sequence. http://oeis.org/A001108 alternating with... a(n) = 6a(n1)a(n2)+2 with a(0) = 0, a(1) = 3. Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X values. http://oeis.org/A001652 Best, Raphie 


#24
Mar3111, 07:28 PM

P: 153

128600703/488566 = 263.220738 [5]/[3] 1671258205/6349266 = 263.220694 [6]/[4] 110744403/743590 = 148.932077 [5]/[3] 1657929736/11132121= 148.932062 [6]/[4] But I don't have enough data to extrapolate further, other than to suggest... 3.38503663 × 10^10 = (128600703*263.220694) 1.64933923 × 10^10 = (110744403*148.932062) ... as likely lower bounds for the next number in each series.  RF 


#25
Apr811, 12:43 PM

P: 153

Ramsey2879, here is a secondary observation that clearly (and cleanly) relates the first differences of the square number indices associated with the 2nd, 3rd and 6th terms of the series you posted  for 1600(T_x) + 81 and 3600(T_x) + 121 are square  to the Pell Numbers...
sqrt (3600*1+121)  sqrt (1600*1+81) = (61  41) = 20 = 2* 10 sqrt (3600*743590+121)  sqrt (1600*488566+81) = (51739  27959) = 23780 = 2378 * 10 sqrt (3600*1657929736+121)  sqrt (1600*1671258205+81) = (2443061  1635241) = 807820 = 80782 * 10 A000129 Pell Numbers 0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428, 38613965, 93222358, 225058681, 543339720, 1311738121, 3166815962, 7645370045, 18457556052, 44560482149 http://oeis.org/A000129 Presumably, a second pattern TBD will link the 1st, 4th and 5th terms to the Pell numbers as well. sqrt (3600*0+121)  sqrt (1600*0+81) = (11  9) = 2 sqrt (3600*11132121+121)  sqrt (1600*6349266+81) = (200189  100791) = 99398 sqrt (3600*110744403+121)  sqrt (1600*128600703+81) = (631411  453609) = 177802 Best, Raphie P.S. Datawise, below are the index numbers of the Triangular Numbers associated with the two series of squares. (sqrt ((8*0) + 1)  1) = 0 (sqrt ((8*1) + 1)  1) = 1 (sqrt ((8*488566) + 1)  1) = 1976 (sqrt ((8*6349266) + 1)  1) = 7126 (sqrt ((8*128600703) + 1)  1) = 32074 (sqrt ((8*1671258205) + 1)  1)/2 = 57814 (sqrt ((8*0) + 1)  1) = 0 (sqrt ((8*1) + 1)  1) = 1 (sqrt ((8*743590) + 1)  1) = 2438 (sqrt ((8*11132121) + 1)  1) = 9436 (sqrt ((8*110744403) + 1)  1) = 29764 (sqrt ((8*1657929736) + 1)  1) = 115166 


#26
Jun2311, 02:23 PM

P: 894

A) 2T(n) + 1 = n^2 + n + 1 = (y^2+y)/2 B) 16T(n) + 9 = 4y^2 + 4y^2 + 1 (Multiply A by 8 and add 1) C) 4n^2 + 4n + 1 = 2y^2 + 2y 3 (Mulyiply A by 4 and subtract 3) Equation B shows that 16T(n) + 9 is (2y+1)^2 from which you derived your y series. Equation C shows that 2y^2 + 2y 3 is (2n+1)^2 QED I recently revisited this problem and the proof just stared at my face. You posted a lot of added comments re Sloanes sequences A124124 and 124174 to which I have a lot to add. I invite you to coauthor some edits to the comments on these sequences with me. 


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