A335333 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2*(2*k+1)*x + x^2).

1, 1, 1, 1, 3, 1, 1, 5, 13, 1, 1, 7, 37, 63, 1, 1, 9, 73, 305, 321, 1, 1, 11, 121, 847, 2641, 1683, 1, 1, 13, 181, 1809, 10321, 23525, 8989, 1, 1, 15, 253, 3311, 28401, 129367, 213445, 48639, 1, 1, 17, 337, 5473, 63601, 458649, 1651609, 1961825, 265729, 1

Offset: 0

Author

Seiichi Manyama, Jun 02 2020

Keywords

• nonn
• tabl
• changed

Examples

			Square array begins:
  1,    1,     1,      1,      1,       1, ...
  1,    3,     5,      7,      9,      11, ...
  1,   13,    37,     73,    121,     181, ...
  1,   63,   305,    847,   1809,    3311, ...
  1,  321,  2641,  10321,  28401,   63601, ...
  1, 1683, 23525, 129367, 458649, 1256651, ...
		

Links

• Seiichi Manyama, Antidiagonals n = 0..139, flattened
• Eric Weisstein's World of Mathematics, Legendre Polynomial.

Crossrefs

Columns k=0..4 give A000012, A001850, A006442, A084768, A084769.

Rows n=0..6 give A000012, A005408, A003154(n+1), A160674, A144124, A335338, A144126.

Main diagonal gives A331656.

T(n,n-1) gives A331657.

Cf. A307883, A307884.

Programs

Mathematica

T[n_, k_] := LegendreP[n, 2*k + 1]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 03 2021 *)

PARI

T(n, k) = pollegendre(n, 2*k+1);

Formula

T(n,k) is the coefficient of x^n in the expansion of (1 + (2*k+1)*x + k*(k+1)*x^2)^n.

T(n,k) = Sum_{j=0..n} k^j * (k+1)^(n-j) * binomial(n,j)^2.

T(n,k) = Sum_{j=0..n} k^j * binomial(n,j) * binomial(n+j,j).

n * T(n,k) = (2*k+1) * (2*n-1) * T(n-1,k) - (n-1) * T(n-2,k).

T(n,k) = P_n(2*k+1), where P_n is n-th Legendre polynomial.

From Seiichi Manyama, Aug 30 2025: (Start)

T(n,k) = (-1)^n * Sum_{j=0..n} (1/(2*(2*k+1)))^(n-2*j) * binomial(-1/2,j) * binomial(j,n-j).

T(n,k) = Sum_{j=0..floor(n/2)} (k*(k+1))^j * (2*k+1)^(n-2*j) * binomial(n,2*j) * binomial(2*j,j).

E.g.f. of column k: exp((2*k+1)*x) * BesselI(0, 2*sqrt(k*(k+1))*x). (End)