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

[Raphie]

e.g.

2 * 45 + 1 = 91 = T_13 = 2 * T_9 + 1

3 * 45 + 1 = 136 = T_16 = 3 * T_9 + 1

Anyone care to offer another integer > 45 ( = T_9) for which this is also the case?RF

 

[lpetrich]

Let triangular number T(x) = x(x+1)/2

Your problem is 2T(x) +1 = T(y) and 3T(x) + 1 = T(z)
for x,y,z being integers -- the first one is your earlier definition of a Sophie Germain number.

The inverse triangular numbers, the values of x that solve T(x) = y, are

x = (+- sqrt(8y+1) - 1)/2

I checked x up to 1 million, and I found that the only solution is
x = 9, y = 13, z = 16
what you had found.

I did some further investigation, and I found that the only possible solutions have
x = 9n
or
x = 9n-1
for integer n. Your solution is the first one with n = 1

 

[Raphie]

I just performed a quick check, inserting known Sophie Germain Triangular numbers into the following equation...

(sqrt ((8*(3*x) + 9)) - 1)/2 is in N

e.g.
(sqrt ((8*(2*45) + 9)) - 1)/2 = 13
(sqrt ((8*(3*45) + 9)) - 1)/2 = 16

... and there are no other Triangular Numbers up to 623 trillion and change (623094791644755), for which 2*T_x + 1 and 3*T_x + 1 are both triangular.

I'm skeptical that there will be another that can fulfill both cases (although I could well be wrong...), but, if this is correct, then the question becomes Why?

The easiest answer would be if it turns out that 136 is the greatest triangular number for which (T_z - 1)/3 is also triangular, but I have not yet checked into this...

UPDATE: Here is a counter-example to 136 being the greatest triangular number for which (T_z - 1)/3 is also triangular:
(61^2 + 61)/2 = 1891 = T_61
3*(35^2 + 35)/2 + 1 = 1891 = 3*T_35 + 1

I guess the next step would be to figure out a formula for 3*T_x + 1 is Triangular to see how it compares to the one for Sophie Germain Triangular Numbers.

- RF

==========================================================
P.S. What I find particularly interesting about this is that 136 comes into play in regards to the last two Ramanujan-Nagell Triangular Numbers in the following manner...

2^04 - 1 = ((3*T_10 + 1) - (3*T_9 + 1))/2 = ((3(0055) + 1) - (3(45) + 1))/2 = (0166 - 136)/2 = T_05 = 0015
2^12 - 1 = ((3*T_74 + 1) - (3*T_9 + 1))/2 = ((3(2775) + 1) - (3(45) + 1))/2 = (8326 - 136)/2 = T_90 = 4095

Delta (4, 12) = 0+8
Delta (10, 74) = 8^2
Delta (9, 9) = 0^2

... and 136 + 136 = 272, the maximal laminated lattice sphere packing for Dimension 9, the first dimension for which the Maximal known sphere packing (306) is greater than the maximal laminated lattice sphere packing (272 = 2^(0+8) + 2*(0+8)). And, of course, as I have noted previously, (4095*(2*12)) + (4095*(2*12)) = 196560 = K_24, (15*(2*4)) + (15*(2*4)) = 240 = K_08, proven lattice packings for Dimensions 8 (E8) and 24 (Lambda 24, the Leech Lattice). 240 = 2^(0+8) - 2*(0+8)

I have also found, just since I posted this yesterday, a number of suggestive numerical relationships in regards to 0,45 and the Ramanujan Congruences.

 

[Raphie]

I guess the next step would be to figure out a formula for 3*T_x + 1 is Triangular to see how it compares to the one for Sophie Germain Triangular Numbers.

Well, that turned out to be not too difficult to track down with a little research...

=========================================
A001571 a(0) = 0, a(1) = 2, a(n) = 4a(n-1) - a(n-2) + 1.
0, 2, 9, 35, 132, 494, 1845, 6887, 25704, 95930, 358017
http://oeis.org/A001571

Second member of the Diophantine pair (m,k) that satisfies 3(m^2+m)=k^2+k: a(n)=k. - Bruce Corrigan, Nov 04 2002

a(n)=-(1/2)-(1/4)*sqrt(3)*[2-sqrt(3)]^n+(1/4)*sqrt(3)*[2+sqrt(3)]^n+(1/4)*[2-sqrt(3)]^n+(1/4) *[2+sqrt(3)]^n, with n>=0 [From Paolo P. Lava, Jul 31 2008]

A133161 Indices of the triangular numbers which are also centered triangular number.
1, 4, 16, 61, 229, 856, 3196, 11929, 44521, 166156
http://oeis.org/A133161

Also, indices of the triangular numbers which are sums of three consecutive triangular numbers (see A129803).

a(n+2)=4*a(n+1)-a(n)+1, a(n+1)=2*a(n)+0.5+0.5*(12*a(n)^2+12*a(n)-15)^0.5.

a(n)=-(1/2)-(1/4)*sqrt(3)*[2-sqrt(3)]^n+(1/4)*sqrt(3)*[2+sqrt(3)]^n+(3/4)*[2-sqrt(3)]^n+(3/4) *[2+sqrt(3)]^n, with n>=0 - Paolo P. Lava, Jul 30 2008
=========================================

(sqrt (8*((3*(0^2 + 0)/2) + 1)+1) - 1)/2 = 1
(sqrt (8*((3*(2^2 + 2)/2) + 1)+1) - 1)/2 = 4
(sqrt (8*((3*(9^2 + 9)/2) + 1)+1) - 1)/2 = 16
(sqrt (8*((3*(35^2 + 35)/2) + 1)+1) - 1)/2 = 61
(sqrt (8*((3*(132^2 + 132)/2) + 1)+1) - 1)/2 = 229
(sqrt (8*((3*(494^2 + 494)/2) + 1)+1) - 1)/2 = 856
(sqrt (8*((3*(1845^2 + 1845)/2) + 1)+1) - 1)/2 = 3196

Sophie Germain Triangular Numbers are associated with Pell Numbers (sqrt 2), and the 3*T_x + 1 form is associated with (sqrt 3).

Note that 16 is a member of A133161, which then seems to diverge away from powers of 4 thereafter.

1, 4, 16, 61, 229, 0856, 3196, 11929...
1, 4, 16, 64, 256, 1024, 4032, 16384...
0, 0, 00, 03, 027, 0168, 0836, 04455... = First Differences (3 = 2^2 - 1 and 168 = 13^2 - 1)

The form: (sqrt (8*((3*(x^2 + x)/2) + 1)+1) - 1)/2 simplifies to:

(sqrt ((12*(x^2 + x)) + 9) - 1)/2

For x = a(n) = 4a(n-1) - a(n-2) + 1

e.g.
4*2 - 0 + 1 = 9
4*9 - 2 + 1 = 35
4*35 - 9 + 1 = 132

The first several triangular numbers for which (T_x - 1)/3 is also triangular are:

1, 10, 136, 1891, 26635, 366796, 5108806, 71156485, 991081981, 13803991246, 19226479546, 2677903145191, 37298379237211, 519499406175760, 7235693307223426, 100780206894952201, 1403687203222107385, 19550840638214551186, 272308081731781609216

Up to 19550840638214551186, and excluding 1 (the sum T_-1 + T_0 + T_1), these values can all be found here:

A129803 Triangular numbers which are the sum of three consecutive triangular numbers.
http://oeis.org/A129803

136, for instance, is the sum of 36 + 45 + 55 = T_8 + T_9 + T_10

Since none of these values is in N...

(sqrt (8*((2*(100780206894952201 - 1)/3) + 1)+1) - 1)/2
(sqrt (8*((2*(1403687203222107385 - 1)/3) + 1)+1) - 1)/2
(sqrt (8*((2*(19550840638214551186 - 1)/3) + 1)+1) - 1)/2
(sqrt (8*((2*(272308081731781609216 - 1)/3) + 1)+1) - 1)/2

... we can now safely state that if there is a counter-example to the proposition:

There is no Triangular Number T_x > 45 | 2*T_x + 1 and 3*T_x + 1 are also triangular

... it would have to be greater than roughly 272 Quintillion.

(sqrt (8*((2*(y - 1)/3) + 1)+1) - 1)/2, by the way, simplifies to...
(sqrt ((16*((y - 1)/3)) + 9) - 1)/2 = (sqrt ((16*x + 9) - 1)/2

RELATED CONJECTURE

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

Thus, by substitution:
2*((sqrt (16*45 + 9) - 1)/2)^2 + 2*(sqrt (16*45 + 9) - 1)/2 - 3
2*((sqrt (16*T_9 + 9) - 1)/2)^2 + 2*(sqrt (16*T_9 + 9) - 1)/2 - 3

2*((sqrt (16*((91 - 1)/2) + 9) - 1)/2)^2 + 2*(sqrt (16*((91 - 1)/2) + 9) - 1)/2 - 3
2*((sqrt (16*((T_13 - T_1)/2) + 9) - 1)/2)^2 + 2*(sqrt (16*((T_13 - T_1)/2) + 9) - 1)/2 - 3

2*((sqrt (16*((136 - 1)/3) + 9) - 1)/2)^2 + 2*(sqrt (16*((136 - 1)/3) + 9) - 1)/2 - 3
2*((sqrt (16*((T_16 - T_1)/3) + 9) - 1)/2)^2 + 2*(sqrt (16*((T_16 - T_1)/3) + 9) - 1)/2 - 3

= 8*T_9 + 1
= 2*T_16 + 89; 89 == F_11 == 24-th Prime
= 19^2
= 361

Combine with the only other known solution... T_0 = 0, 2*T_0 + 1 = T_1, 3*T_0 + 1 = T_1...

2*((sqrt (16*T_0 + 9) - 1)/2)^2 + 2*(sqrt (16*T_0 + 9) - 1)/2 - 3
2*((sqrt (16*((T_1 - 1)/2) + 9) - 1)/2)^2 + 2*(sqrt (16*((T_1 - 1)/2) + 9) - 1)/2 - 3
2*((sqrt (16*((T_1 - 1)/3) + 9) - 1)/2)^2 + 2*(sqrt (16*((T_1 - 1)/3) + 9) - 1)/2 - 3

= 8*T_0 + 1
= 2*T_1 - 1; 1 == F_1 or F_2 == "0-th" Prime (first positive integer with less than 3 divisors)
= 1^2
= 1

And the difference between the two, in geometric terms, can be thought of as one full rotation around a circle.

 

[Raphie]

Ipetrich,

I know you are good with UFD's, so let me simply make mention of what I see as a possible relationship between them and the above. Here are two simple ways to relate 0, 1 and 13, 9

floor [2*|(((sqrt ((0^2 - 7)/2) + 1)/2)^7)|^(1/7) - 1] = 1
floor [2*|(((sqrt ((13^2 - 7)/2) + 1)/2)^7)|^(1/7) - 1] = 9

2*T_9 + 1 = T_13
(T_1 - 1)/2 = T_0

Also...

First and last positive integer solutions to 2^n - 1 is Triangular (Ramanujan-Nagell)

2^(01 - 1) - 1 = 0000 = T_(10*0) = T_(T_01 - 1)
2^(13 - 1) - 1 = 4095 = T_(10*9) = T_(T_13 - 1)

1 and 13 are the only two integers for which that statement can be made. In other words, both...

2^(x - 1) - 1 = T_(T_x - 1); x = 1, 13
and...
2^(x - 1) - 1 = T_(10*y); y = 0, 9

... are unique statements.

And to sum up the above: If there is a Triangular Number T_x > 45 | 2*T_x + 1 and 3*T_x + 1 are also triangular, then it would have to be greater than roughly 272 Quintillion.Raphie