OFFSET
0,3
COMMENTS
Diagonal sums of A117947.
LINKS
Antti Karttunen, Table of n, a(n) for n = 0..16384 (terms 0..5000 from Alois P. Heinz)
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)
EXAMPLE
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}.
MATHEMATICA
a[n_] := Sum[JacobiSymbol[Binomial[n - k, k], 3], {k, 0, n/2}];
a /@ Range[0, 100] (* Jean-François Alcover, Oct 14 2019 *)
PROG
(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.
(PARI) A117963(n) = sum(k=0, n\2, kronecker(binomial(n-k, k), 3)); \\ Antti Karttunen, Jan 01 2023
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Paul Barry, Apr 05 2006
STATUS
approved