A117904 Number triangle [k<=n]*0^abs(L(C(n,2)/3) - L(C(k,2)/3)) where L(j/p) is the Legendre symbol of j and p.

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

Offset: 0

Author

Paul Barry, Apr 01 2006

Keywords

• easy
• nonn
• tabl

Comments

Row sums are A009947(n+2).

Diagonal sums are A117905.

Inverse is A117906.

Equals A117898 mod 2.

Examples

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

Links

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

Crossrefs

Cf. A009947, A117898, A117905, A117906.

Programs

Mathematica

T[n_, k_]:= If[Abs[JacobiSymbol[Binomial[n, 2], 3] - JacobiSymbol[Binomial[k, 2], 3]]==0, 1, 0];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 20 2021 *)

Sage

def A117904(n,k): return 1 if abs(jacobi_symbol(binomial(n,2), 3) - jacobi_symbol(binomial(k,2), 3))==0 else 0
flatten([[A117904(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Oct 20 2021

Formula

G.f.: (1 +x*(1+y) +x^2*y^2 +x^3*y)/((1-x^3)*(1-x^3*y^3)).

T(n, k) = [k<=n] * 2^abs(L(C(n,2)/3) - L(C(k,2)/3)) mod 2.