Conjecture: Sophie Germain Triangles & x | 2y^2 + 2y - 3 = z^2

 Quote by ramsey2879 True but I don't think that it's the sum of cubes that forms the rule.
In relation to your specific series of progressions, this may be the case, Ramsey2879, but 16*Sum of Cubes (or the square root of...) can definitely be related to, at the very least, the second to last term of even-indexed Pell Polynomials (See: Pell Polynomials @ http://mathworld.wolfram.com/PellPolynomial.html) in pretty simple manner, so I'm pretty sure there is a Triangular number related rule in play here, although it might well diverge from the "34- and 35-flavored" rule for your set of Triangular Numbers...

001 --> Sum = 1 = 3^0
002 001 --> Sum = 3 = 3^1
004 004 001 --> Sum = 9 = 3^2
008 012 006 001 --> Sum = 27 = 3^3
016 032 024 008 001 --> Sum = 81 = 3^4
032 080 080 040 010 001 --> Sum = 243 = 3^5
064 192 240 160 60 012 001 --> Sum = 729 = 3^6

The anti-diagonals of this triangle are Pell Numbers. Working right to left, simply insert values from the 3rd diagonal (4 times a Triangular Number) into the form n^2 and you get 16* the Sum of Cubes. Insert into the form (n^2)x + 2n + 1 and you get, in order...

16x + 9
144x + 25
576x + 49
1600x + 81
3600x + 121
etc.

Furthermore, the 9x + 1 form (Triangular where x is Triangular) can be iterated into the following equation that can be derived from the row sums of the above upper triangular matrix:

3^(2*n) + T_((3^n - 1)/2) = T_y

Thus, for x a Triangular number (Divide bolded values above by 4)...

9x + 1 = 3^2 + T_((3^1 - 1)/2) is Triangular
81x + 10 = 3^4 + T_((3^2 - 1)/2) is Triangular
729x + 91 = 3^2 + T_((3^3 - 1)/2) is Triangular
etc.

RELATED PROGRESSION
Triangle whose (i,j)-th entry is binomial(i,j)*2^(i-j).
http://oeis.org/A038207
As an upper right triangle, table rows give number of points, edges, faces, cubes, 4D hypercubes etc. in hypercubes of increasing dimension by column. - Henry Bottomley, Apr 14 2000. More precisely, the (i,j)-th entry is the number of j-dimensional subspaces of an i-dimensional hypercube (see the Coxeter reference). - Christof Weber, May 08 2009

Also worth noting is that the powers of 21 (1, 21, 441...) are embedded in the triangle above in similar manner to the way powers of 11 are embedded in Pascal's Triangle: 1/(100-21) = 1/79 gives the Pell Series overlaid in the decimal expansion, same as 1/(100-11) = 1/89 gives the Fibonacci Series. In general, the right to left diagonals are powers of 2 (far left "diagonal") times the n-hedral numbers. (e.g. 1*Unity Set, 2*Counting Numbers, 4*Triangular Numbers, 8*Tetrahedral Numbers, etc.).

Best,
RF

So, in other words, I guess I'm going to have to backtrack a tad on the following statement...
 Quote by Raphie I am inclined to agree with you, Ramsey2879, that I am going down a dead-end road in regards to Sum of Cubes and Pell Numbers
In principle, there should be one formula that generates (right to left) diagonal-specific Recursive rules for each entry in the above triangle. The 3rd diagonal relates to Triangular Numbers, the 4th diagonal to Tetrahedrals, etc.

 Quote by Raphie In principle, there should be one formula that generates (right to left) diagonal-specific Recursive rules for each entry in the above triangle.
If I'm reading correctly (and I'm not sure I am), the recurrence formula for your triangle entries should be
entry(n,k) = entry(n-1,k-1) + 2*entry(n-1,k)
The reason why they add up to 3^n is because that is the binomial expansion of
$(2+1)^n = \sum_{k=0}^n \binom n k 2^k$

Hope this helps, if this coincides with what you're going after.

 Quote by Dodo If I'm reading correctly (and I'm not sure I am), the recurrence formula for your triangle entries should beentry(n,k) = entry(n-1,k-1) + 2*entry(n-1,k)The reason why they add up to 3^n is because that is the binomial expansion of $(2+1)^n = \sum_{k=0}^n \binom n k 2^k$ Hope this helps, if this coincides with what you're going after.
I am well aware of the relationship you posted, Dodo, but thank you. One can actually generate triangles very easily wherein rows sum to any power desired. e.g. A 1, 3, 1 triangle with addition rule 1A + 3B sums to powers of 4. A 1, 4, 1 triangle with addition rule 1A + 4B sums to powers of 5. etc..

What is actually being addressed on this thread at the moment are the following recursion-based relationships uncovered by Ramsey2879.

 Quote by ramsey2879 I can't answer your question yet but my research has shown a definite link with the Pell series every 4th term of a pell series forms a sequence S(n) = 34*S(n-1) - S(n-2) every 8th term of a pell series forms a sequence S(n) = 1154*S(n-1) -S(n-2) The Sophie Germain triangular numbers were already described A similar series of triangular numbers is x' = {0,1,171,1326,197506,1530375,227921925,1766051596,...} having the recursive formula S(n) = 1154*S(n-2) - S(n-4) + 172 144x'+25 is a square just as 16x +9 is a square for the Sophie Germain triangular numbers. Still More x" = {0,1,120,1275,138601,1471470,159945555,169807226...} having the recursive formula S(n) = 1154*S(n-2) - S(n-4) + 121 576x"+49 is a square just as 16x +9 is a square for the Sophie Germain triangular numbers. Interesting how triangular numbers relate to the Pythagorean triangle numbers and Pell series. Regards, Kenneth Ramsey
I was merely pointing out that the terms involved in his progressions can be derived from the Pell Triangle...

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 recursion-based 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

Blog Entries: 2
 Quote by Raphie I am well aware of the relationship you posted, Dodo, but thank you. One can actually generate triangles very easily wherein rows sum to any power desired. e.g. A 1, 3, 1 triangle with addition rule 1A + 3B sums to powers of 4. A 1, 4, 1 triangle with addition rule 1A + 4B sums to powers of 5. etc.. What is actually being addressed on this thread at the moment are the following recursion-based relationships uncovered by Ramsey2879. I was merely pointing out that the terms involved in his progressions can be derived from the Pell Triangle... 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 recursion-based 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
It appears difficult to generalize my result. Although it appears that for N = 4T, where T is a triangular number > 0, there is an infinite sequence of triangular numbers T(x) such that N^{2}*T(x) + 2N + 1 is a square beginning with x = 0,1,a,... The Sophie triangular numbers are the T(x)'s for T = 1. The sequences which I gave are for T = 3 and 6. However, I cannot determine a formula for the sequence of x's for T = 10 and 15 as I could for T = 3 and 6 so I am not so certain that the sequence of T(x)'s is infinite. My results for T = 10 and 15 are below:

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]}.
 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 n-th s-gonal number is given by (s-2)*T(n-1)+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: (32-8)/4+3=9; (80-32)/4+4=16; (160-80)/4+5=25; (280-160)/4+6=36; (448-280)/4+7=49; ...
 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 n-Hedral 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, (99-1)/2 = 49, etc. ============================================= Expansion of 1/(1-3x+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(n-1)+2, with a(0) = 0, a(1) = 1 If (X,X+1,Z) is a Pythagorean triple, then Z-X-1 and Z+X is in the sequence. http://oeis.org/A001108 alternating with... a(n) = 6a(n-1)-a(n-2)+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

 Quote by ramsey2879 It appears difficult to generalize my result. Although it appears that for N = 4T, where T is a triangular number > 0, there is an infinite sequence of triangular numbers T(x) such that N^{2}*T(x) + 2N + 1 is a square beginning with x = 0,1,a,... The Sophie triangular numbers are the T(x)'s for T = 1. The sequences which I gave are for T = 3 and 6. However, I cannot determine a formula for the sequence of x's for T = 10 and 15 as I could for T = 3 and 6 so I am not so certain that the sequence of T(x)'s is infinite. My results for T = 10 and 15 are below: 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]}.
The first hint of a pattern I see here is...

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
 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. Data-wise, 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

Blog Entries: 2
 Quote by Raphie Limiting ourselves to N... Conjecture: (sqrt (16x + 9) - 1)/2 = y | 2y^2 + 2y - 3 = z^2 for x = a Sophie Germain Triangular Number, which is recursively defined as: a(n)=34a(n-2)-a(n-4)+11 First 11 values (I have not checked further as I would be very surprised if this equivalency did not hold...) x = 0, 1, 10, 45, 351, 1540, 11935, 52326, 405450, 1777555, 13773376... y = 1, 2, 6, 13, 37, 78, 218, 457, 1273, 2666, 7422... Seems to me there ought to be a (at least a somewhat...) simple algebraic proof for this, if accurate, and my reasoned guess is that Pell Numbers would come in to play somewhere, but observation, research, and heuristically-based approaches to mathematics, not proofs, are my forte. RELATED PROGRESSIONS: Sloane's A124174 Sophie Germain triangular numbers tr: 2*tr+1 is also a triangular number. http://oeis.org/A124174 Sloane's A124124 Nonnegative integers n such that 2n^2+2n-3 is square. http://oeis.org/A124124
The proof goes as follows:
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.

 Tags 16x + 9, sophie germain, triangles