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A117963 Antidiagonal sums of a Legendre-binomial triangle for p = 3. 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Diagonal sums of A117947.

LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..5000

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-Fran├žois 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

CROSSREFS

Cf. A117947.

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

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Last modified June 2 14:12 EDT 2020. Contains 334782 sequences. (Running on oeis4.)