A224928 Numbers of pairs {x, y} such that x <= y and triangular(x) + triangular(y) = 2^n.

1, 1, 1, 0, 2, 0, 1, 0, 3, 0, 2, 0, 4, 0, 1, 0, 8, 0, 2, 0, 4, 0, 4, 0, 8, 0, 2, 0, 24, 0, 2, 0, 8, 0, 8, 0, 8, 0, 2, 0, 32, 0, 4, 0, 16, 0, 4, 0, 32, 0, 4, 0, 32, 0, 4, 0, 4, 0, 8, 0, 16, 0, 2, 0, 32, 0, 6, 0, 48, 0, 16, 0, 16, 0, 8, 0, 384, 0, 4, 0, 16, 0, 16, 0, 16, 0, 8, 0, 768, 0, 2, 0, 8, 0, 4, 0, 32, 0, 32, 0, 256

Offset: 0

Author

Alex Ratushnyak, May 08 2013

Keywords

• nonn

Comments

Conjectures:

1. a(n) = 0 for odd n > 1.

2. a(n) is even for even n > 14.

Examples

			2^1 = 1 + 1, the only representation of 2 as a sum of two triangular numbers, so a(1)=1.
2^4 = 16 = 1+15 = 6+10, two representations, so a(4) = 2.
2^8 = 256 = 3+253 = 66+190 = 120+136, so a(8) = 3.
2^12 = 4096 = 1+4095 = 91+4005 = 1540+2556 = 2016+2080, so a(12) = 4.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..329

Crossrefs

Cf. A000217, A052343, A006307, A225437.

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;
  for (n = 1; n; n+=n) {
    for (c = x = tx = 0; tx*2 <= n; ++x, tx+=x)
      if (isTriangular(n - tx))
        ++c;//, printf("(%llu+%llu) ", tx, n-tx);
    printf("%llu, ", c);
  }
  return 0;
}

PARI

A008441(n) = if(!n,n,sumdiv(4*n + 1, d, (d%4==1) - (d%4==3)));
A052343(n) = if(!n,1,my(u=A008441(n)); ((u\2)+(u%2)));
A224928(n) = A052343(2^n); \\ Antti Karttunen, May 24 2021

Formula

a(n) = A052343(2^n).

Extensions

More terms from Antti Karttunen, May 24 2021