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

0,3

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

Cf. A117947, A117964 [= a(n) mod 2].

Sequence in context: A083817 A286222 A029273 * A321594 A112803 A124242

Adjacent sequences: A117960 A117961 A117962 * A117964 A117965 A117966

Keyword

easy,sign

Author

Paul Barry, Apr 05 2006

Status

approved