%I #12 Jan 24 2016 17:38:38
%S 8,10,19,26,44,58,79,84,91,105,111,121,135,140,154,172,184,188,195,
%T 203,208,212,217,222,230,240,246,265,276,286,292,316,322,329,338,364,
%U 390,426,429,462,490,498,506,548,550,605,704,714,715,763,770,780,782,807
%N Numbers whose square can be written as sum of at least 3 consecutive triangular numbers (A000217).
%C Any square can trivially be written as sum of two consecutive triangular numbers T = A000217, since T(n-1) + T(n) = n(n-1)/2 + n(n+1)/2 = n*2n/2 = n^2. But it seems nontrivial to determine the squares that can be written as sum of more than 2 consecutive triangular numbers.
%C Some of these have two different decompositions of this form, e.g., 286^2 = T(13)+...+T(78) = T(75)+...+T(96), 826^2 = T(13)+...+T(159) = T(43)+...+T(160). What is the sequence of these numbers?
%C The terms > 2 of A129445, numbers k > 0 such that k^2 is a centered triangular number, form a subsequence: they correspond to k^2 = T(n-2) + T(n-1) + T(n), with n in sequence A129444: numbers n such that centered triangular number A005448(n) = 3n(n-1)/2 + 1 is a perfect square.
%C Terms > 2 of sequence A075870 also form a subsequence, namely the numbers whose square is the sum of four triangular numbers T(n-3)+...+T(n), with n given by twice the terms > 1 of A046090 or A182435.
%H Chai Wah Wu, <a href="/A257274/b257274.txt">Table of n, a(n) for n = 1..10000</a>
%e 8^2 = T(5)+T(6)+T(7), 10^2 = T(5)+T(6)+T(7)+T(8), 19^2 = T(14)+T(15)+T(16), 26^2 = T(3)+...+T(15), 44^2 = T(13)+...+T(23), ...
%o (PARI) {a=[];(S(n)=binomial(n+2,3)); for(n=1,999,for(k=1,n-3,issquare(S(n)-S(k))&&a=concat(a,sqrtint(S(n)-S(k))))); Set(a)[1..50]}
%K nonn
%O 1,1
%A _M. F. Hasler_, May 02 2015
%E a(14), a(43)-a(54) from _Chai Wah Wu_, Jan 20 2016