A117898 Number triangle 2^abs(L(C(n,2)/3) - L(C(k,2)/3))*[k<=n] where L(j/p) is the Legendre symbol of j and p.
1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1
Offset: 0
Examples
Triangle begins 1; 1, 1; 2, 2, 1; 1, 1, 2, 1; 1, 1, 2, 1, 1; 2, 2, 1, 2, 2, 1; 1, 1, 2, 1, 1, 2, 1; 1, 1, 2, 1, 1, 2, 1, 1; 2, 2, 1, 2, 2, 1, 2, 2, 1; 1, 1, 2, 1, 1, 2, 1, 1, 2, 1; 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1; 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1; 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1;
Links
- G. C. Greubel, Table of n, a(n) for the first 101 rows, flattened
Programs
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Mathematica
Flatten[CoefficientList[CoefficientList[Series[(1 +x(1+y) +x^2(2+2y+y^2) +x^3*y(1 +2y) +2x^4*y^2)/((1-x^3)(1-x^3*y^3)), {x,0,15}, {y,0,15}], x], y]] (* G. C. Greubel, May 03 2017 *) T[n_, k_]:= 2^Abs[JacobiSymbol[Binomial[n, 2], 3] - JacobiSymbol[Binomial[k, 2], 3]]; Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 27 2021 *) -
Sage
def T(n, k): return 2^abs(kronecker(binomial(n,2), 3) - kronecker(binomial(k,2), 3)) flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 27 2021
Formula
G.f.: (1 +x*(1+y) +x^2*(2+2*y+y^2) +x^3*y(1+2*y) +2*x^4*y^2)/((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)).
Comments