A117963 Antidiagonal sums of a Legendre-binomial triangle for p = 3.
1, 1, 2, 0, 2, 2, 1, 3, 4, -2, 2, 0, 2, 2, 4, 0, 4, 4, -1, 3, 2, 2, 4, 6, 1, 7, 8, -6, 2, -4, 4, 0, 4, -2, 2, 0, 2, 2, 4, 0, 4, 4, 2, 6, 8, -4, 4, 0, 4, 4, 8, 0, 8, 8, -5, 3, -2, 4, 2, 6, -1, 5, 4, 0, 4, 4, 2, 6, 8, 2, 10, 12, -5, 7, 2, 6, 8, 14, 1, 15, 16, -14, 2, -12, 8, -4, 4, -6, -2, -8, 8, 0, 8, -4, 4, 0, 4, 4, 8, -6, 2
Offset: 0
Examples
The triangle mentioned in the name starts:
{1},
{1, 1},
{1,-1, 1},
{1, 0, 0, 1},
{1, 1, 0, 1, 1},
{1,-1, 1, 1,-1, 1},
{1, 0, 0,-1, 0, 0, 1}.
Links
- Antti Karttunen, Table of n, a(n) for n = 0..16384 (terms 0..5000 from Alois P. Heinz)
Programs
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Mathematica
a[n_] := Sum[JacobiSymbol[Binomial[n - k, k], 3], {k, 0, n/2}]; a /@ Range[0, 100] (* Jean-François Alcover, Oct 14 2019 *) -
PARI
{a(n)=local(A=1+x+x*O(x^n)); for(i=1,#binary(n), A=subst(A,x,x^3+x*O(x^n)) *(1-4*x^3-x^6)/(1-x-x^2+x*O(x^n))); polcoeff(A,n,x)} \\ Paul D. Hanna, Jul 11 2006 -
PARI
A117963list(upto_n) = { my(A=1+x+x*O(x^upto_n)); for(i=1, #binary(upto_n), print1(i,", "); A=subst(A, x, x^3+x*O(x^upto_n)) *(1-4*x^3-x^6)/(1-x-x^2+x*O(x^upto_n))); print(); vector(upto_n,n,polcoeff(A,n-1)); }; \\ Antti Karttunen, Jan 01 2023, after Paul D. Hanna's program above.
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PARI
A117963(n) = sum(k=0,n\2,kronecker(binomial(n-k,k),3)); \\ Antti Karttunen, Jan 01 2023
Formula
a(n) = a(3n+2)/a(2).
a(n) = Sum_{k=0..floor(n/2)} L(C(n-k,k)/3) where L(j/p) is the Legendre symbol of j and p.
From Paul D. Hanna, Jul 11 2006: (Start)
G.f. satisfies: A(x) = A(x^3)*(1 - 4*x^3 - x^6)/(1 - x - x^2).
a(n) == Fibonacci(n+1) (mod 3);
a(n) == a(n-1) + a(n-2) (mod 3). (End)
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