%I #26 Mar 14 2020 06:56:47
%S 0,1,2,5,7,8,9,10,12,13,16,17,18,20,21,22,26,28,30,31,34,35,38,41,43,
%T 45,47,48,52,55,58,59,61,62,63,65,66,67,68,70,71,73,75,77,80,82,85,86,
%U 92,93,98,101,103,107,108,110,111,113,116,118,120,121,127
%N Numbers that are the sum of three triangular numbers an odd number of ways.
%C Reduce the elements of A192717 by subtracting 3 and dividing by 8. This makes sense since the elements of A192717 are congruent to 3 (mod 8).
%C A positive integer n belongs to this sequence precisely when n can be written as t + 2u for triangular numbers t, u an odd number of times, equivalently, written as t + u + v for triangular numbers t, u, v, an odd number of times.
%H J. N. Cooper, D. Eichhorn, and K. O'Bryant, <a href="https://arxiv.org/abs/math/0506496">Reciprocals of binary power series</a>, arXiv:math/0506496 [math.NT], 2005.
%H J. N. Cooper, D. Eichhorn, and K. O'Bryant, <a href="https://doi.org/10.1142/S1793042106000693">Reciprocals of binary power series</a>, International Journal of Number Theory, 2 no. 4 (2006), 499-522.
%H J. N. Cooper and A. W. N. Riasanovsky, <a href="http://people.math.sc.edu/cooper/Sigma.pdf">On the Reciprocal of the Binary Generating Function for the Sum of Divisors</a>, 2012.
%e For n = 0, 1 representation: 0 + 0 + 0; so 0 belongs to this sequence.
%e For n = 1, 3 representations: 1 + 0 + 0, 0 + 1 + 0, 0 + 0 + 1; so 1 belongs.
%e For n = 2, 3 representations: 1 + 1 + 0, 1 + 0 + 1, 0 + 1 + 1; so 2 belongs.
%e For n = 3, 4 representations: 3 + 0 + 0, 0 + 3 + 0, 0 + 0 + 3, 1 + 1 + 1; so 3 does not belong.
%e For n = 4, 6 representations: 3 + 1 + 0, 3 + 0 + 1, 1 + 3 + 0, 1 + 0 + 3, 0 + 3 + 1, 0 + 1 + 3; so 4 does not belong.
%e ...
%o (Sage)
%o def BPS(n): #binary power series
%o return sum([q^s for s in n])
%o prec = 2^14
%o R = PowerSeriesRing(GF(2), 'q', default_prec = prec)
%o q = R.gen()
%o tList = [(n*(n+1))//2 for n in range(0, floor(-1+sqrt(8*prec+1))//2)]
%o tSeries = BPS(tList)
%o print((tSeries^3).exponents()[:128])
%Y Cf. A192717, A192628.
%K nonn
%O 1,3
%A _Alexander Riasanovsky_, Jan 20 2013