A117944 Triangle related to powers of 3 partitions of n.

1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Author

Paul Barry, Apr 05 2006

Keywords

• easy
• nonn
• tabl

Comments

Inverse is A117945.

Row sums of inverse are A039966.

Examples

			Triangle begins
  1;
  0, 1;
  1, 0, 1;
  0, 0, 0, 1;
  0, 0, 0, 0, 1;
  0, 0, 0, 1, 0, 1;
  1, 0, 0, 0, 0, 0, 1;
  0, 1, 0, 0, 0, 0, 0, 1;
  1, 0, 1, 0, 0, 0, 1, 0, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1;
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Crossrefs

Cf. A039966, A117939, A117943, A117945.

Programs

Mathematica

T[n_, k_]:= Mod[Sum[JacobiSymbol[Binomial[n, j], 3]*JacobiSymbol[Binomial[n-j, k], 3], {j,0,n}], 2];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 29 2021 *)

Sage

def A117944(n, k): return ( sum(jacobi_symbol(binomial(n, j), 3)*jacobi_symbol(binomial(n-j, k), 3) for j in (0..n)) )%2
flatten([[A117944(n, k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Oct 29 2021

Formula

Triangle T(n,k) = Sum_{j=0..n} L(C(n,j)/3)*L(C(n-j,k)/3) mod 2, where L(j/p) is the Legendre symbol of j and p.

T(n, k) = A117939(n,k) mod 2.

T(n, k) = A117939^(-1)(n,k) mod 2.

Sum_{k=0..n} T(n, k) = A117943(n).