A225437 Numbers of triples {x, y, z} such that z >= y > 0 and triangular(x) + triangular(y) * triangular(z) = 2^n.
1, 1, 2, 0, 4, 0, 5, 1, 7, 0, 4, 0, 18, 0, 2, 0, 17, 0, 16, 0, 15, 0, 9, 0, 39, 0, 9, 0, 61, 0, 10, 3, 27, 0, 18, 0, 56, 0, 8, 0, 80, 0, 48, 1, 41, 0, 12, 0, 118, 1, 10, 0, 90, 0, 30, 2, 24, 0, 24
Offset: 0
Examples
{0, 1, 1} is the only triple producing 2^0, so a(0) = 1.
{1, 1, 3} and {3, 1, 1} are the triples producing 2^2, so a(2) = 2.
Programs
-
C
#include
#include typedef unsigned long long U64; U64 isTriangular(U64 a) { // ! Must be a <= (1<<63) U64 s = sqrt(a*2); if (a>=(1ULL<<63)) { if (a==(1ULL<<63)) return 0; printf("Error: a = %llu\n", a), exit(1); } return (s*(s+1)/2 == a); } int main() { U64 c, n, x, tx, y, ty, z, prod; for (n = 1; n>0 && n <= (1ULL<<63); n+=n) { for (c = 0, x = tx = 0; tx <= n; ++x, tx+=x) for (z=prod=n-tx, y=ty=1; ty<=z; ++y, ty+=y, z=prod/ty) if ((z * ty == prod) && isTriangular(z)) c++; printf("%llu, ", c); } return 0; }