A117446 Triangle T(n, k) = binomial(L(k/3), n-k) where L(j/p) is the Legendre symbol of j and p, read by rows.

1, 0, 1, 0, 1, 1, 0, 0, -1, 1, 0, 0, 1, 0, 1, 0, 0, -1, 0, 1, 1, 0, 0, 1, 0, 0, -1, 1, 0, 0, -1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, -1, 0, 1, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 1, 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 0, 1, 1, 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 1

Offset

0,1

Links

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Formula

T(n, k) = binomial(L(k/3), n-k), where L(j/p) is the Legendre symbol of j and p.

Sum_{k=0..n} T(n, k) = A117447(n).

Sum_{k=0..floor(n/2)} T(n-k, k) = A117448(n) (diagonal sums).

Example

Triangle begins
  1;
  0, 1;
  0, 1,  1;
  0, 0, -1, 1;
  0, 0,  1, 0, 1;
  0, 0, -1, 0, 1,  1;
  0, 0,  1, 0, 0, -1, 1;
  0, 0, -1, 0, 0,  1, 0, 1;
  0, 0,  1, 0, 0, -1, 0, 1,  1;
  0, 0, -1, 0, 0,  1, 0, 0, -1, 1;
  0, 0,  1, 0, 0, -1, 0, 0,  1, 0, 1;
  0, 0, -1, 0, 0,  1, 0, 0, -1, 0, 1,  1;
  0, 0,  1, 0, 0, -1, 0, 0,  1, 0, 0, -1, 1;
  0, 0, -1, 0, 0,  1, 0, 0, -1, 0, 0,  1, 0, 1;
  0, 0,  1, 0, 0, -1, 0, 0,  1, 0, 0, -1, 0, 1, 1;

Mathematica

Table[Binomial[JacobiSymbol[k, 3], n-k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Jun 02 2021 *)

Prog

(Sage) flatten([[binomial(kronecker(k, 3), n-k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 02 2021

Crossrefs

Cf. A117447 (row sums), A117448 (diagonal sums), A117449 (inverse).

Sequence in context: A167371 A127241 A087748 * A187034 A101688 A155029

Adjacent sequences: A117443 A117444 A117445 * A117447 A117448 A117449

Keyword

easy,sign,tabl

Author

Paul Barry, Mar 16 2006

Status

approved