[seqfan] Re: Numbers such that there exist a(n) consecutive triangle numbers which sum to a square.

[Max Alekseyev]

Alternatively this sequence can be defined as the values of d for which

binomial(x+d,3) - binomial(x,3) = square

has nonnegative solutions with respect to x.

Regards,
Max

On Sun, Apr 18, 2010 at 2:00 AM, Andrew Weimholt
<andrew.weimholt at gmail.com> wrote:
> Numbers such that there exist a(n) consecutive triangle numbers which
> sum to a square.
>
> 0, 1, 2, 3, 4, 11, 13, 22, 23, 25, 27, 32, 37, 39, 46, 47, 48, 49, 50,
> 52, 59, 66, 71, 73, 83, 94, 98, 100, 104, 107, 109, 111, 118, 121,
> 128, 143, 146, 147, 148, 157, 167, 176, 179, 181, 183, 191, 192,
> 193, 194, 200
>
> 0 is in the sequence because the sum of 0 consecutive triangle numbers
> is 0 (a square)
> 1 is in the sequence because there exists triangle numbers which are
> squares (Cf. A001110)
> 2 is in the sequence because ANY 2 consecutive triangle numbers sum to a square
> 3 is in the sequence because 15 + 21 + 28 = 64 (infinite number of solutions)
> 4 is in the sequence because 15 + 21 + 28 + 36 = 100 (infinite number
> of solutions)
>
> 5 is NOT in the sequence because no 5 consecutive triangle numbers sum
> to a square.
>
> N is in the sequence if there exists (non-degenerate) solutions to the
> diophantine
> equation
>
> 8x^2 - N*y^2 - A077415(N) = 0
>
> with the additional condition that y == N mod 2.
> (Not sure if solutions in which this is not true are actually possible)
>
> A degenerate solution is one in which relies on triangle numbers with
> negative indexes.
>
> For example, with N = 8, solutions to the diophantine equation exists, but
> they correspond to the consecutive "triangle numbers" starting at
> t(-2) or t(-6),
> where t(x) = x(x+1)/2 is the "triangle" function with domain = Z.
>
>    1 + 0 + 0 + 1 + 3 + 6 + 10 + 15 = 36
>    15 + 10 + 6 + 3 + 1 + 0 + 0 + 1 = 36
>
> There are no non-degenerate solutions for N=8.
> For this reason, I have not included 8 in the sequence above.
>
> It is also interesting to note that there are only a finite
> number of solutions for 32, 50 98, 128, and 200, but these
> numbers do have non-degenerate solutions.
>
> Andrew
>
>
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