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

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

Related to A124124...

Sloane's A077442
2*n^2 + 7 is a square.
http://oeis.org/A077442

Best,
Raphie

Source of Observation: an unusual identity for 2*pi accidentally derived earlier today from numbers associated with a) the Crystallographic Restriction Theorem b) phi, the Golden Ratio, c) n | n = 7*F_y = T_x for xy = n (Conjectured Solutions = 0, 13), and d) Division of a 3-Dimensional Space (Cake Numbers).

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

Related to A124124...

Sloane's A077442
2*n^2 + 7 is a square.
http://oeis.org/A077442

Best,
Raphie

Source of Observation: an unusual identity for 2*pi accidentally derived earlier today from numbers associated with a) the Crystallographic Restriction Theorem b) phi, the Golden Ratio, c) n | n = 7*F_y = T_x for xy = n (Conjectured Solutions = 0, 13), and d) Division of a 3-Dimensional Space (Cake Numbers).
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

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144x'+25 is a square just as 16x +9 is a square for the Sophie Germain triangular numbers.

An interesting curio to mention in relation to this observation:

16 + 9 = 5^2
144 + 25 = 169 = 13^2

13 is the greatest integer |
(n^(2+(0*2)) - 7)/(2^1) is square &
(2^(n+(1*2)) - 7)/(2^0) is also square
---------------
9^2 & 181^2

181 = 2^1*(9^(2^1) + (9*1)) + (1*1) = 2*(9^2 + 9) + 1
009 = 2^0*(9^(2^0) + (9*0)) + (1*0) = 9

There are two other integers for which this statement also holds, 3 & 5.

(3^2 - 7)/2 = 1^2 = 1 & 2^(3+2) - 7 = 5^2 = 25
(5^2 - 7)/2 = 3^2 = 9 & 2^(5+2) - 7 = 11^2 = 121
-----------------------------------------------------------------
(13^2 - 7)/2 = 9^2 = 81 & 2^(13+2) - 7 = 181^2 = 32761

144, 25,16 & 9 are all, of course, also squares (12^2, 5^2, 4^2 & 3^2)

Related Progression
Sloane's A077442
2*n^2 + 7 is a square.

http://oeis.org/A077442

- RF

P.S. OBSERVATION: Could be coincidence that 2, 3, 5, 7 & 13 are the unique prime factors of the Leech Lattice, and the 1st, 2nd, 3rd, 4th and 6th primes [Solutions to 2*cos ((2*pi)/n) is in Z (related to the Crystallographic Restriction Theorem)], but based on other observations, I suspect there is something meaningful lurking here e.g. A) the Leech Lattice is generated from a "light ray" in 26-D Lorentz Space, and the sum of squares of its coordinates is Pythagorean Theorem in nature: 0^2 + 1^2 +... 24^2 - 70^2 = 0; and B) 3*1 - 3 = 0 = K_0, 3*3 - 3 = 6 = K_2, 3*9 -3 = 24 = K_4, 3*81 - 3 = 240 = K_8, for K_n denotes a maximal Sphere Packing in Dimension n. And 4*(2^3 - 2) = 24 = K_4, 8*(2^5 - 2) = 240 = K_8, 24*(2^13 - 2) = 196560 = K_24; and C) The covering radius of the Leech Lattice is sqrt 2, which suggests a possible relationship with Pell-type number progressions; and D) The density of the Leech Lattice is pi^12/12!, the proper divisors of 12 being 1,2,3,4,6.

QUESTION: Can 9 + 0 = 3^2 = 1^3 + 2^3 be related in any manner to both Pell and Triangular Number Progressions?

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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.

Hi ramsey2879,

I did a search for the number progressions you posted and came up with nothing. Would I be correct in supposing you are posting original observations?

Also, to me, it is not surprising that Triangular Numbers and Pell Numbers are so intimately related when one considers the manner of their construction in summation terms...

PASCAL"S TRIANGLE-BASED
1A + 0B
[1, 1] 1, 1, 1, 1, 1, 1, 1
[1, 2] 3, 4, 5, 6, 7, 8, 9
[1, 3] 6, 10, 15, 21, 28, 36, 45

FIBONACCI SERIES-BASED
1A + 1B
[1, 1] 2, 3, 5, 8, 13, 21, 34
[1, 2] 4, 7, 12, 20, 33, 54, 88
[1, 3] 7, 14, 26, 46, 79, 133, 221

PELL SERIES-BASED
1A + 2B
[1, 1] 3, 7, 17, 41, 99, 239, 577
[1, 2] 5, 12, 29, 70, 169, 408, 985
[1, 3] 8, 20, 49, 119, 288, 696, 1681

That last row, for instance, relates to Indices of Triangular Numbers that are also square, and to Triangular Numbers that are twice Triangular Numbers, while the Fibonacci-Based Series can be related to Triangular Numbers as well, by dint of the "coincidence" that their first summation is equal to one less than a Fibonacci Number (F_(n+2)), meaning that all subsequent summations are embedding, in a sense, Pascal's Triangle.

Best,
Raphie

P.S.
Observation #1
-----------------
Summing Diagonals of the PELL SERIES-BASED summation series (e.g. 1+7+23+49+81+119+169+239 = 688) would seem to correspond with the following Progression: Sloane's A156664 Binomial transform of A052551 http://oeis.org/A156664, which is a convolution of two Pascal's Triangle related Integer Sequences (but I only verified the first several terms)

Observation #2
-----------------
1 + 6 + 17 + 32 + 49 + 70 + 99 = 274
(99-1)/2 = 49, (70-6)/2 = 32, (49-17)/2 = 17 - 1

1 + 7 + 23+ 49 + 81 + 119+ 169 + 239 = 688
(239-1)/2 = 119, (169-7)/2 = 81, (119-23)/2 = 49 - 1

1 + 8 + 30 + 72 + 130 + 200 + 288 + 408 + 577 = 1714
(577-1)/2 = 288, (408-8)/2 = 200, (288-30)/2 = 130 - 1, (200-72)/2 = 72 - 8

(99+1) + (577-1) = 10^2 + 24^2 = 26^2; and 10*24 = 239 + 1 = 240 --> K_8 = 16^2 - 16
Worth noting is that the number of solutions to Euler-Totient(n) = 24 is 10 (Range 35 --> 90 = 2*T_-10 - T_10 --> 2*T_-10)

Similarly, (1+1) + (3-1) = (sqrt 2)^2 + (sqrt 2)^2 = 2^2, and (sqrt 2)*(sqrt 2) = 1 + 1 = 2 --> K_1 = 2^2 - 2

Question: Are these unique cases in the sense that Pell-Lucas_n + Pell-Lucas_(n+2) is equal to a square?
Verified to sqrt (26102926097 + 152139002499)

Related Progression
Numerators of continued fraction convergents to sqrt(2) (aka "Pell-Lucas Numbers")
http://oeis.org/A001333

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Related Progression
Numerators of continued fraction convergents to sqrt(2) (aka "Pell-Lucas Numbers")
http://oeis.org/A001333

It would seem I slightly misspoke. The progression mentioned above (beginning 1, 1, 3, 7, 17, 41, 99, 239, 577, 1393...) is actually 1/2 the Pell-Lucas Numbers.

From Wolfram MathWorld (http://mathworld.wolfram.com/PellNumber.html)...
For n=0, 1, ..., the Pell-Lucas numbers are 2, 2, 6, 14, 34, 82, 198, 478, 1154, 2786, 6726, ... (Sloane's A002203 http://oeis.org/A002203). As can be seen, they are always even.

34 and 1154 are both in this series (the 5th & 9th terms, or 4th and 8th terms if one indexes from 0)...
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 Pell Series, for those unawares, begins 0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741... (Sloane's A000129 http://oeis.org/A000129) and the recursive addition rule is simply a(n) = a(n-2) + 2*a(n-1)

- RF

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In relation to Sophie Germain Triangular Numbers, the form 3*T_x + 1 is Triangular, Pell Numbers, the Ramanujan-Nagell Equation (and Kaprekar Numbers) ...

A surprising observation in relation to the following:

Also, 45 is a Sophie Germain Triangular Number (2x + 1 is triangular) partnered with 91 (and when summed, 91 + 45 equals another triangular number, T_9 + T_13 = T_(7+9) = T_16 = 136):
from: Observation: A Prime / Mersenne / (Ramanujan) Triangular Number Convolution

3*T_x + 1 is Triangular

This is a statement for which the following statement...

The sum of a Sophie Germain Triangular Number and it's partner is also a Triangular Number.

... is at least a subset since Sophie Germain Triangular Numbers are of the form T_x | 2*T_x + 1 is Triangular..

e.g.

3*00 + 1 = 3T_0 + 1 = T_01 == T_2^0 = 1
3*03 + 1 = 3T_2 + 1 = T_04 == T_2^2 = 10
3*45 + 1 = 3T_9 + 1 = T_16 == T_2^4 = 136
Any others?

For what x and y, does it hold that

((3*T_x + 1) - (3*T_y + 1))/2 = T_z

The first 4 examples I have found:

((3*T_n + 1) - (3*T_n + 1))/2 = ((3n + 1) - (3n + 1))/2 = T_00 = 0
((3*T_02 + 1) - (3*T_1 + 1))/2 = ((3(0003) + 1) - (3(01) + 1))/2 = (0010 - 004)/2 = T_02 = 3
((3*T_10 + 1) - (3*T_9 + 1))/2 = ((3(0055) + 1) - (3(45) + 1))/2 = (0166 - 136)/2 = T_05 = 15
((3*T_74 + 1) - (3*T_9 + 1))/2 = ((3(2775) + 1) - (3(45) + 1))/2 = (8326 - 136)/2 = T_90 = 4095

x = n, 2, 10, 74
y = n, 1, 09, 09

And why might I find that odd? Consider the following statement for which the only solutions are related to the Ramanujan-Nagell Triangular Numbers.

((p'_x * p'_2x) * (M_x - |T_x - 1|)) / ((T_(M_x) - T_|T_x - 1|) is in N

for...
|T_x - 1| = 1, 0, 2, 5, 90
x = 0, 1, 2, 3, 13

p'_x is an integer | 0 < d(n) < 3
M_x is a Mersenne Number (2^n - 1)

EXPANSION
((01 * 001) * (0000 - 01)) / ((0000^2 + 0000)/2 - (01^2 + 01)/2) = 01
((02 * 003) * (0001 - 00)) / ((0001^2 + 0001)/2 - (00^2 + 00)/2) = 06
((03 * 007) * (0003 - 02)) / ((0003^2 + 0003)/2 - (02^2 + 02)/2) = 07
((05 * 013) * (0007 - 05)) / ((0007^2 + 0007)/2 - (05^2 + 05)/2) = 10
((41 * 101) * (8191 - 90)) / ((8191^2 + 8191)/2 - (90^2 + 90)/2) = 01

It gets a bit "messy" at 0, thus the absolute value sign. If instead we restate:

((p'_x * p'_2x) * (2^x - T_x)) / ((T_(2^x) - T_(T_x)) is in N

... then the only solutions are x = 1, 2, 3, 13

T_00 = T_((T_01) - 1) = 0000 = 2^00 - 1
T_02 = T_((T_02) - 1) = 0003 = 2^02 - 1
T_05 = T_((T_03) - 1) = 0015 = 2^04 - 1
T_90 = T_((T_13) - 1) = 4095 = 2^12 - 1

If we set n from the expansion above to 0, then...

T_00 = ((3*T_00 + 1) - (3*T_0 + 1))/2 = 0000
T_02 = ((3*T_02 + 1) - (3*T_1 + 1))/2 = 0003
T_05 = ((3*T_10 + 1) - (3*T_9 + 1))/2 = 0015
T_90 = ((3*T_74 + 1) - (3*T_9 + 1))/2 = 4095

T_00 = 0000 & 00 = T_0
T_02 = 0003 & 01 = T_1
T_10 = 0055 & 45 = T_9
T_74 = 2775 & 45 = T_9

Now sum those y values...

45 + 45 + 1 + 0 = 91 = 2x + 1 for x = 45, a Sophie Germain Triangular Number.

And what does this all suggest?

A possible link between Pell Numbers, the Ramanujan-Nagell Equation and Sophie Germain Triangular Numbers, not to mention the Kaprekar Numbers, since 0, 1, & 45 (as well as 55) when squared...

00^2 = 0000 and 00 + 00 = 00
01^2 = 0001 and 00 + 01 = 01
45^2 = 2025 and 20 + 25 = 45
------------------------------------
55^2 = 3025 and 30 + 25 = 55

... can be reconstituted from the sum of their evenly partitioned parts (including leading zeroes). Ditto for their associated index numbers, 0, 1, 9, 10, if one incorporates a trailing zero, in tandem with, where necessary, a leading zero...

00^2 = 000.0 and 00 + 0.0 = 00
01^2 = 001.0 and 00 + 1.0 = 01
09^2 = 081.0 and 08 + 1.0 = 09
------------------------------------
10^2 = 100.0 and 10 + 0.0 = 10

Best,
RF

QUESTION:

For what Triangular Numbers greater than 45, does the following statement hold:

2*T_x + 1 is Triangular & 3*T_x + 1 is also Triangular?

e.g.
2*45 + 1 = 091 = T_13
3*45 + 1 = 136 = T_16

Conjecture: There are no others greater than 45.

A counter example would be more than welcome and, honestly, I have not checked for higher values...

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3*T_x + 1 is Triangular

This is a statement for which the following statement...

The sum of a Sophie Germain Triangular Number and it's partner is also a Triangular Number.

... is at least a subset since Sophie Germain Triangular Numbers are of the form T_x | 2*T_x + 1 is Triangular..

e.g.

3*00 + 1 = 3T_0 + 1 = T_01 == T_2^0 = 1
3*03 + 1 = 3T_2 + 1 = T_04 == T_2^2 = 10
3*45 + 1 = 3T_9 + 1 = T_16 == T_2^4 = 136
Any others?

For anyone interested, the check for the above would be this:

sqrt (8*(((n^2 + n)/2 - 1)/3) + 1) is in N

------------------------------------------------
sqrt (8*(((01^2 + 01)/2 - 1)/3) + 1) = 01 (= sqrt (4*T_00 + 1))
sqrt (8*(((04^2 + 04)/2 - 1)/3) + 1) = 05 (= sqrt (4*T_03 + 1))
sqrt (8*(((16^2 + 16)/2 - 1)/3) + 1) = 19 (= sqrt (4*T_13 + 1))
------------------------------------------------
Note: T_(0 + 3 + 13) = T_16 = 3*T_9 + 1 = (3*45 + 1) = 136

Additionally, the following pattern also emerges in relation to the sum of T_x and T_y for the solutions to the statement: ((3*T_x + 1) - (3*T_y + 1))/2 = T_z | T_z = 2^a - 1...

T_00 = T_(T_01 - 1) = ((3*T_00 + 1) - (3*T_0 + 1))/2 = 0000; T_00 + T_0 = 00^2
T_02 = T_(T_02 - 1) = ((3*T_02 + 1) - (3*T_1 + 1))/2 = 0003; T_02 + T_1 = 02^2
T_05 = T_(T_03 - 1) = ((3*T_10 + 1) - (3*T_9 + 1))/2 = 0015; T_10 + T_9 = 10^2
T_90 = T_(T_13 - 1) = ((3*T_74 + 1) - (3*T_9 + 1))/2 = 4095; T_74 + T_9 = 53^2 - (0 + 0 + 2 + 1 + 10 + 9)/2 = 53^2 - 11[/B]

Rather curiously, 0199 is the 11-th Lucas Number, and 11 the 5-th Lucas Number, 5 & 11 being a part of the primeth prime sequence e.g. pi (31) = 11, and pi (11) = 5. 11 & 31, of course, being associated with the recent findings of Ken Ono et al. in relation to Partition Numbers, Primes and Fractals, as is the 6-th prime, 13, leading to the following (equivalent) descriptive statements:

sigma (11) - sigma (11) = 0, sigma (11) - totient(11) = 2, and sigma (11) - d (11) = 10
totient (13) - sigma (pi (31)) = 0, totient (13) - totient(pi (31)) = 2, and totient (13) - d (pi (31)) = 10
12 - 12 = 0, 12 - 10 = 2, 12 - 2 = 10

... the respective differences between the parts of which can be slotted into the formula 2*n^2 + 7 = 2*(4*(x_1 - x_2) + 1)^2 + 7 = y^2

2*(4*(13 - 11) + 1) + 7 = 2*(4*(02) + 1) + 7 = 025 = 05^2
2*(4*(31 - 11) + 1) + 7 = 2*(4*(20) + 1) + 7 = 169 = 13^2

RELATED SEQUENCE
Sloane's A077442
2*n^2 + 7 is a square.
http://oeis.org/A077442

Alternatively...
2*n^2 + 7 = 2*(4*(2*T_z) + 1)^2 + 7 = y^2
for z = 1, 4 (1^2 & 2^2)

2*(4*(2*T_1) + 1) + 7 = 025 = 5^2
2*(4*(2*T_4) + 1) + 7 = 169 = 13^2

which simplifies to... 16*T_n + 9 for n = 1, 4

... which may or may not provide some insight in relation to the initial conjecture that is the reason for this thread...
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

... and which may or may not be relevant to the fact that 5 is the 1st prime associated with Hausdorff Dimension 0 in relation to partition numbers, while 13 is the first prime associated with Hausdorff Dimension 1 (11, & 31 are the last ones associated with those respective dimensions)

- RF

Background Research (Values of particular interest in bold)
-----------------------
Sophie Germain triangular numbers tr: 2*tr+1 is also a triangular number.
http://oeis.org/A124174
0, 1, 10, 45, 351, 1540, 11935, 52326, 405450, 1777555...

Associated Triangular Number Indices
0, 1, 04, 09, 019, 0053, 00309, 000647, 001801, 0003771...

Triangular numbers of the form k^2 + k + 1
http://oeis.org/A069017
1, 3, 21, 91, 703, 3081, 23871, 104653, 810901, 3555111...

Associated Triangular Number Indices
1, 2, 06, 13, 075, 0157, 00437, 000915, 002547, 0005333...

First Differences of Indices
0, 1, 02, 04, 056, 0104, 00126, 000268, 000646, 0001562...

QUESTION: Can 9 + 0 = 3^2 = 1^3 + 2^3 be related in any manner to both Pell and Triangular Number Progressions?

I believe the answer to be "yes," and perhaps in more than a trivial way...

for x = 11, y = 13, z = 31
for b = 1, c = 2, a = 3, then...

(y - x)/2 = T_(b^2) = (0 + 01)
(z - x)/2 = T_(c^2) = (0 + 10)
(z - y)/2)^2 + (z - y)/2))/2 = T_(T_(c^2) - T_(b^2)) = T_(a^2) = (0 + 45) = T_(9 + 0)

pi (030) + pi (2) = pi (par_09) + pi (par_02) = x*b = 11
pi (101) + pi (1) = pi (par_13) + pi (par_01) = y*c = 26
pi (490) + pi (1) = pi (par_19) + pi (par_00) = z*a = 93

11 --> Sum of d (1, 2, 3, 5, 7, 13) --> (9 - 0^2)th - (9 - 1^2)th - (9 - 2^2)th) Cake Number = 130 - 93 - 26 --> x*b
26 --> Sum of totient (1, 2, 3, 5, 7, 13) --> (9 - 4)th Cake Number = 26 --> y*c
93 --> 3*Sum of (1, 2, 3, 5, 7, 13) --> (9 - 1)th Cake Number = 93 --> z*a

for 1, 2, 3, 5, 7, 13 is A) the complete set of divisors of the unique prime divisors of the Leech Lattice constructed from a "Light Ray" in 26-D Lorentz Space, or, alternatively, B) the complete set of integers at least to 10^1505 such that 1.5 Tetra_z = 2^x - 2 or 2*T_y = 2^n - 2 for some n in N, or, alternatively, C) the complete set of integers such that pi (n) is a solution to 2*cos (2*pi/(n/|n-1|!) [proof for Crystallographic Restriction Theorem], or, alternatively, D) the complete set of Integers such that 0 < d(n) < 3 and d (p_n -1) = n, or alternatively, E) the complete set of of divisors of the primes associated with anomaly cancellations in 26 Dimensional String Theory by Frampton & Kephart back in 1999. {1 U First Five Mersenne Primes}

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QUESTION:

For what Triangular Numbers greater than 45, does the following statement hold:

2*T_x + 1 is Triangular & 3*T_x + 1 is also Triangular?

e.g.
2*45 + 1 = 091 = T_13
3*45 + 1 = 136 = T_16

Conjecture: There are no others greater than 45.

A counter example would be more than welcome and, honestly, I have not checked for higher values...

This conjecture has now been checked to roughly 272 Quadrillion. As Pell Numbers and the form 2*T_x + 1 are related to sqrt 2, the form 3*T_x + 1 is related to sqrt 3.

More here:
2*T_x + 1 is Triangular & 3*T_x + 1 is also Triangular

Next up: 5*T_x + 1 is triangular, since 2,3,5 are the first 3 Sophie Germain Primes which pair with the safe primes 5, 7 & 11 of Ramanujan Congruence fame. At the very least, T_0, T_1, T_10, T_27 & T_493 map to T_1, T_3, T_23, T_61 & T_1103 . 5*0 + 1 = 1, 5*1 + 1 = 6, 5*55 + 1 = 276, 5*378 + 1 = 1891, 5*121771 + 1 = 608856 (1891 also equal to 3*T_35 + 1 = T_34 + T_35 + T_36).

Will be curious to see if sqrt 5 comes into play, particularly since for F_n a Fibonacci Number and L_n a Lucas Number:

(1 - 0) = 1 = F_1
(3 - 1) = 2 = F_3
(23 - 10) = 13 = F_7
(61 - 27) = 34 = F_9
(1103 - 493) = 610 = F_15

(2*1) + 1 = 3 = L_2
(2*3) + 1 = 7 = L_4
(2*23) + 1 = 47 = L_8
(2*61) + 1 = 123 = L_10
(2*1103) + 1 = 2207 = L_16

(2*0) + 1 = 1 = F_2
(2*1) + 1 = 3 = F_4
(2*10) + 1 = 21 = F_8
(2*27) + 1 = 55 = F_10
(2*493) + 1 = 987 = F_16

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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.

Score one for the mathematics of observation and/or "The Babylonian Fishnet" approach to mathematics (as I came across one poster term it.). The proof for this is indeed quite simple and algebraic in nature. And the Generalized Pell equation is involved by implication since one of the intermediary steps requires that: 2*n^2 + 7 is a square http://oeis.org/A077442...

CONJECTURE
2y^2 + 2y + 3 = z^2
for...
y = (sqrt (16x + 9) - 1)/2
where...
x is a Sophie Germain Triangular Number

A) Restate: 2y^2 + 2y + 3 = z^2 as z^2 = 4*T_y - 3
for: y^2 + y = 2*T_y
B) T_y = (z^2 + 3)/4
C) y = (sqrt (8*(z^2 + 3)/4 + 1) - 1)/2

If Conjecture is true, then the following is an equivalency...

D) y = (sqrt (8*(z^2 + 3)/4 + 1) - 1)/2 = (sqrt (16x + 9) - 1)/2

Therefore, if true, by cancellation and rearrangement of terms...
E) 16x + 9 = 2z^2 + 7
F) (16x + 9 - 7)/2 = 8x + 1 = z^2

G) 8x + 1 is a square only if x is triangular

H) Restate: (sqrt (16x + 9)) as (sqrt (8(2x + 1) + 1))
I) (sqrt (8(2x + 1) + 1)) can be in N only if 2x + 1 is Triangular

I) Given that x is Triangular, 2x + 1 is Triangular, by definition, only if x is a Sophie Germain Triangular Number.

Therefore, by G) and I), in order for y = (sqrt (8*(z^2 + 3)/4 + 1) - 1)/2 = (sqrt (16x + 9) - 1)/2 to be a true statement, x must be a Sophie Germain Triangular Number.

Sophie Germain triangular numbers x (x | 2x+1 is also a triangular number): http://oeis.org/A124174
x = 0, 1, 10, 45, 351, 1540, 11935, 52326, 405450, 1777555, 13773376...
Nonnegative integers n such that 2y^2+2y-3 is square: http://oeis.org/A124124
y = 1, 2, 6, 13, 37, 78, 218, 457, 1273, 2666, 7422...
2*z^2 + 7 is a square: http://oeis.org/A077442.
z = 1, 3, 9, 19, 53, 111, 309, 647, 1801, 3771, 10497...

- RF

Cute Formula For Pi

Two (Inductive) Sources of Conjecture:
67092480 == 2^13 - 2*2^13 == p_1^p_6 - p_1*p_1^p_6 =8192^2 - 2*8192
for 1, 2, 6 & 13 the first 4 z's above [== (n^2 + 1) + T_(n-1) for n = 0, 1, 2, 3]
67092480 == Product [Cake_n]; (n = 1 --> 7) = 2*4*8*15*26*42*64; totient (7) = 6, & totient (1) = 1 (Cake_n --> Division of 3-Space)
67092480 == (p'_0*par_0)*(M_0 - |T_0 - 1|)+(T_(M_0) - T_|T_0 - 1|) + (p'_13*par_13)*(M_13 - |T_13 - 1|) + (T_(M_13) - T_|T_13 - 1|)
for p'_n = {1 U Primes}, par_n = nth partition number, T_n = nth Triangular Number, and M_n = nth Mersenne Number

Also...
2*pi = (sqrt (4*|T_13 - 1| + 1) + 1)/2))*arccos (sqrt (4*|T_0 - 1| + 1) + 1)/2))/(sqrt (4*|T_2 - 1| + 1) + 1)/2))
= (3^2 + 1) arccos (((2^2 + 1)^(1/2) + (0^2 + 1)^(1/2)) /(1^2 + 1)^(2/1)))
= 10 arccos ((sqrt 5 + sqrt 1)/2^2)
= 10 arccos (Golden Ratio /2)

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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 now notice two more facts about each of the above series. Each series also have the recursive formula S(n) = 1155(S(n-2)-S(n-4)) + S(n-6) similar to the Sophie-Germain formula S(n) = 35(S(n-2)-S(n-4)) - S(n-6). Also similar to the fact that each of Sophie Germain Triangular numbers times 2 plus 1 is a Triangular number, each of the above triangular numbers times 9 plus 1 is also a Triangular number( except "169807226" should have been --1698075226-- ). Haven't checked for other numbers of the series. In answer to the question about these series, the properties of these series were discovered by me when I tried to generalize the properties of the Sophie Germain triangular numbers using the apparent relation between the Pell series and triangular numbers. Still working on the question of the origional post.

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I now notice two more facts about each of the above series. Each series also have the recursive formula S(n) = 1155(S(n-2)-S(n-4)) + S(n-6) similar to the Sophie-Germain formula S(n) = 35(S(n-2)-S(n-4)) - S(n-6).

Possibly related:

1155 = 11* T_14 = 11*105
2*105 = 210 = T_20
Thus, (11*T_20)/2 = 1155

That first index number, 14, is a Pell-Lucas Number (as are 34 and 1154...). In fact, the indices of all Triangular Numbers that are square or twice a triangular number can be related to the Pell-Lucas Series.

Also, possibly coincidence, but 11+14 = 25 = 5^2 and 20 - 11 = 9 = 3^2 (and 14 + 20 = 34). Curious if this could be connected in some manner to the forms (16x + 9) = (4^2x + 3^2) and (144x + 25) = (16*9)x + 25 = (4^2*3^2)x + 5^2.

Also...

A) (1+ 2 + 5 + 12 + 29 + 70 + 169)/2 = 288/2 = 144 == 8*T_8/2 for n*T_n is a Pentagonal Pyramid number.
B) 9 * 11 = (10-1)*(10+1) = 10^2 - 1^2 = 99 = 198/2 for 198 a Pell-Lucas Number
C) 1155 is also expressible as 33*35 = (34-1)*(34+1) = 34^2 - 1^2

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Hi Ramsey2879,

I took a closer look at the progressions you posted and am coming to the opinion that you may well be on to (the beginnings of...) a general formula. Consider:

016x' + 09 = z'^2
144x''+ 25 = z''^2
576x'''+ 49 = z'''^2

(016*01)x' + 09 = z'^2
(016*09)x'' + 25 = z''^2
(016*36)x''' + 49 = z'''^2

1, 9 & 36 are the first 3 Sum of Cubes (= T_n^2)

Restatement #1
(4^2*1^2)x' + (2*1 + 1)^2 = z'^2
(4^2*3^2)x'' + (2*2 + 1)^2 = z''^2
(4^2*6^2)x''' + (2*3 + 1)^2 = z'''^2

FORMULA:
(4^2*T_n^2)*x + (2*n+1)^2 = z^2

Restatement #2
8(02x' + 1) + 1 = z'^2
8(18x'' + 3) + 1 = z''^2
8(72x''' + 6) + 1 = z'''^2

FORMULA:
8(2*((T_n)^2)*x + T_n) + 1 = z^2

The x values in each case are Triangular Numbers and my first thought/best thought is that all x values can be related to either A) Pell Numbers, B) Linear Recurrences of the form 1*a_(n-1) + 2*((T_n)^2)*a_(n) = a_(n+1), or C) possibly A & B both

For n = 1, we have the "Sophie Germain" Triangular Numbers
2x + 1 == (1^3 + 1)x + 1 == (T_2 - 1)x + 1 a Triangular Number

For n = 2, we have the "Unnamed_(2)" Triangular Numbers
9x + 1 == (2^3 + 1)x + 1 == (T_4 - 1)x + 1 a Triangular Number

For n = 3, we have the "Unnamed_(3)" Triangular Numbers?
First Guess(es)
A) 28x + 1 = (3^3 + 1)x + 1 a Triangular Number UPDATE: [WRONG GUESS]
OR
B) 20x + 1 = (T_6 - 1)x + 1 a Triangular Number UPDATE: [WRONG GUESS]

Will have to run some of the values you posted and see what comes of it...

Oh, and the recursive rule? All x's may well be related to powers of 34 and/or 35 with some manner of offset? e.g. 34^3 - 2 = 39302, the 12th Pell-Lucas Number.

Best,
RF

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UPDATE:
25x''' + 3 is Triangular for x'''|576x'"+49 is square

25 + 3 = 3^3 + 1 = T_7

25*0 + 3 = T_2
25*1 + 3 = T_7
25*120 + 3 = T_77
25*1275 + 3 = T_252
25*138601 + 3 = T_2632
25*1471470 + 3 = T_8577
25*159945555 + 3 = T_89427
25*1698075226 + 3 = T_291382

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UPDATE:
25x''' + 3 is Triangular for x'''|576x'"+49 is square

25 + 3 = 3^3 + 1 = T_7

25*0 + 3 = T_2
25*1 + 3 = T_7
25*120 + 3 = T_77
25*1275 + 3 = T_252
25*138601 + 3 = T_2632
25*1471470 + 3 = T_8577
25*159945555 + 3 = T_89427
25*1698075226 + 3 = T_291382
That is well known to me
If x is a triangular number then x*(2n+1)^2 + T(n) is always a triangular number for any integer n. Your posting is simply the case for n = 2 applied to my set of triangular (x) numbers. I am not sure it applies to all x|576 x + 49 is a perfect square, as it is only certain where x is also triangular.

I am not sure it applies to all x|576 x + 49 is a perfect square, as it is only certain where x is also triangular.

I believe we are taking it as a given, within the context of this thread, that all x's are triangular as per the following restatement:
Restatement #2
8(02x' + 1) + 1 = z'^2
8(18x'' + 3) + 1 = z''^2
8(72x''' + 6) + 1 = z'''^2

FORMULA:
8(2*((T_n)^2)*x + T_n) + 1 = z^2
Do you happen to already have a rule for 1600x'''' + 81 is square, or 3600x''''' + 121 is square? I am thinking in terms of 16 times the Sum of Cubes (T_n^2) for the first term. Obviously, for any equation of this form, we can be sure that the first two "x" terms are 0 and 1 (T_0 and T_1). From this fact, it follows also that for kx + z is a Triangular number, then k + z must also be Square[/edit].

- RF

UPDATE #1: Other than for x = 0 and 1, (assuming I checked correctly...) there are no squares of the form 1600x'''' + 81 less than 242556 (=T_696).

UPDATE #2: sqrt ((1600*488566)+81) = 27959; 488566 = T_988 (--> 1 (modulo 7)) and sqrt (8*(9*488566 + 1) + 1) = 5931.

P.S. Ramsey2879, I see what you were getting at now. Any x such that 9x + 1 is Triangular ((3^2)x + T_1) will also follow the form 25x + 3 is Triangular ((5^2)x + T_2). Thus, for example, sqrt (8*(7^2*488566 + T_3) + 1) = 13839 is also triangular. Where x is Triangular, then 9x + 1 will always be triangular.

Unfortunately, sqrt (8*(39302*488566 - 0 + 488567) + 1) = sqrt (8*(34^3 - 2)*488566 - 0 + 488567) + 1) is nowhere even close to being in N, although sqrt (8*((34 -2)*488566 + (0+2) + 488567) + 1) = 11357; (32 +2 = 34)

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I believe we are taking it as a given, within the context of this thread, that all x's are triangular as per the following restatement:

Do you happen to already have a rule for 1600x'''' + 81 is square, or 3600x''''' + 121 is square? I am thinking in terms of 16 times the Sum of Cubes (T_n^2) for the first term. Obviously, for any equation of this form, we can be sure that the first two "x" terms are 0 and 1 (T_0 and T_1). From this fact, it follows also that for kx + z is a Triangular number, then k + z must also be Triangular.

- RF

.
True but I don't think that it's the sum of cubes that forms the rule. My assumption was that since each of these series is based on the Pythagorean triplet A^2 + B^2 = C^2 where A is odd and C-B = 1 that the general form of the Sophie-Germain triamgles was B^2*x + A^2 = a square in such a case. Indeed, my investigation shows a sequence x of triangular numbers for each of the odd A that I checked. However, for A > 7 the recursive formula for x is not so simple. For A = 5 the rule you used for y|2y^2 + 2y -3 is a square also works and the y repeats from your series. But this is not the case for A = 7. Note that the recursive series form for x is the same for A = 5 and 7. While the form 2y^2 + 2y -3 as a square is the same for A = 3 and 5. I suppose that there is a new form for the square that works for A = 7 and 9 while there is a new recursive form for x that works for A = 9 and 11, etc.

Actually, in relation to the below, my bad, but k + z are square, not triangular.

e.g. 1600 + 81 = 41^2 and 3600+121 = 61^2

The form being n^2 + 2n + 1 = (n+1)^2
Do you happen to already have a rule for 1600x'''' + 81 is square, or 3600x''''' + 121 is square? I am thinking in terms of 16 times the Sum of Cubes (T_n^2) for the first term. Obviously, for any equation of this form, we can be sure that the first two "x" terms are 0 and 1 (T_0 and T_1). From this fact, it follows also that for kx + z is a Triangular number, then k + z must also be Triangular.
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, although I see it as an interesting progression in its own right. If there's a relationship, I think it would have to do with linear recurrences of the form a(n-2) + k*a(n-1), with Pell Numbers being the special case of k = 2.

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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...
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.

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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.

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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.

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 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

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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 {,,,,,}.

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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; ...

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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

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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 {,,,,,}.

The first hint of a pattern I see here is...

128600703/488566 = 263.220738 /
1671258205/6349266 = 263.220694 /

110744403/743590 = 148.932077 /
1657929736/11132121= 148.932062 /

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

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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.