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

[Andrew Weimholt]

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