A364298 Square array read by ascending antidiagonals: T(n,k) = [x^k] 1/(1 + x) * Legendre_P(k, (1 - x)/(1 + x))^(-n) for n >= 1, k >= 0.

1, 1, 1, 1, 3, 19, 1, 5, 73, 721, 1, 7, 163, 3747, 49251, 1, 9, 289, 10805, 329001, 5370751, 1, 11, 451, 23623, 1179251, 44127003, 859748023, 1, 13, 649, 43929, 3100001, 190464755, 8405999785, 190320431953, 1, 15, 883, 73451, 6751251, 589050007, 42601840975, 2160445363107

Offset: 1

Author

Peter Bala, Jul 18 2023

Keywords

• nonn
• tabl
• easy

Comments

In the square array A364113, the k-th entry in row n is defined as [x^k] 1/(1 - x) * Legendre_P(k, (1 + x)/(1 - x))^n. Here we essentially extend A364113 to negative values of n.

The two types of Apéry numbers A005258 and A005259 are related to the Legendre polynomials by A005258(k) = [x^k] 1/(1 - x) * Legendre_P(k, (1 + x)/(1 - x)) and A005259(k) = [x^k] 1/(1 - x) * Legendre_P(k, (1 + x)/(1 - x))^2 and thus form rows 1 and 2 of the array A364113

Both types of Apéry numbers satisfy the supercongruences

1) u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r))

and the shifted supercongruences

2) u(n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))

for all primes p >= 5 and positive integers n and r.

We conjecture that each row sequence of the present table satisfies the same pair of supercongruences.

Examples

			 Square array begins
 n\k|  0    1     2       3          4            5               6
  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1 |  1    1    19     721      49251      5370751       859748023
  2 |  1    3    73    3747     329001     44127003      8405999785
  3 |  1    5   163   10805    1179251    190464755     42601840975
  4 |  1    7   289   23623    3100001    589050007    152184210193
  5 |  1    9   451   43929    6751251   1479318759    434790348679
  6 |  1   11   649   73451   12953001   3219777011   1062573281785
		

Crossrefs

A364299 (row 1), A364300 (row 2), A364301 (main diagonal), A364302 (first subdiagonal). Cf. A005258, A005259, A143007, A364113.

Programs

Maple

T(n,k) := coeff(series(1/(1+x)* LegendreP(k,(1-x)/(1+x))^(-n), x, 11), x, k):
# display as a square array
seq(print(seq(T(n, k), k = 0..10)), n = 1..10);
# display as a sequence
seq(seq(T(n-k, k), k = 0..n-1), n = 1..10);