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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

%I #6 Jun 03 2021 03:20:28

%S 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,

%T 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,

%U 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

%N 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.

%H G. C. Greubel, <a href="/A117446/b117446.txt">Rows n = 0..50 of the triangle, flattened</a>

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

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

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

%e Triangle begins

%e 1;

%e 0, 1;

%e 0, 1, 1;

%e 0, 0, -1, 1;

%e 0, 0, 1, 0, 1;

%e 0, 0, -1, 0, 1, 1;

%e 0, 0, 1, 0, 0, -1, 1;

%e 0, 0, -1, 0, 0, 1, 0, 1;

%e 0, 0, 1, 0, 0, -1, 0, 1, 1;

%e 0, 0, -1, 0, 0, 1, 0, 0, -1, 1;

%e 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 1;

%e 0, 0, -1, 0, 0, 1, 0, 0, -1, 0, 1, 1;

%e 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 1;

%e 0, 0, -1, 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 1;

%e 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 0, 1, 1;

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

%o (Sage) flatten([[binomial(kronecker(k, 3), n-k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Jun 02 2021

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

%K easy,sign,tabl

%O 0,1

%A _Paul Barry_, Mar 16 2006