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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.
4
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
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
KEYWORD
easy,sign,tabl
AUTHOR
Paul Barry, Mar 16 2006
STATUS
approved