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

0,1

Comments

Row sums are A009947(n+2).

Diagonal sums are A117905.

Inverse is A117906.

Equals A117898 mod 2.

Links

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

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.

Example

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;

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 *)

Prog

(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

Crossrefs

Cf. A009947, A117898, A117905, A117906.

Sequence in context: A115361 A115358 A325898 * A259030 A212412 A185906

Adjacent sequences: A117901 A117902 A117903 * A117905 A117906 A117907

Keyword

easy,nonn,tabl

Author

Paul Barry, Apr 01 2006

Status

approved