A355257 Array read by ascending antidiagonals. A(n, k) = k! * [x^k] log((1 - x) / (1 - 2*x)) / (1 - x)^n, for 0 <= k <= n.

0, 0, 1, 0, 1, 3, 0, 1, 5, 14, 0, 1, 7, 29, 90, 0, 1, 9, 50, 206, 744, 0, 1, 11, 77, 406, 1774, 7560, 0, 1, 13, 110, 714, 3804, 18204, 91440, 0, 1, 15, 149, 1154, 7374, 41028, 218868, 1285200, 0, 1, 17, 194, 1750, 13144, 85272, 506064, 3036144, 20603520

Offset: 0

Author

Peter Luschny and Mélika Tebni, Jul 01 2022

Keywords

• nonn
• tabl

Comments

Conjecture: For p prime, A(n, p) == -1 (mod p) for n >= 0.

Conjecture: Let n >= 0, k >= 1 and k != 4. Then k divides A(n, k) if and only if k is not prime.

From Mélika Tebni, Jul 04 2022: (Start)

Conjecture: The polynomials of A355259 generate the k+1 column of this array.

Conjecture: For p prime and n even, (A(n, p) / (p - 1)) == 1 (mod p). (End)

Examples

			Table A(n, k) begins:
  [0] 0, 1,  3,  14,   90,   744,   7560,    91440,   1285200, ... A029767
  [1] 0, 1,  5,  29,  206,  1774,  18204,   218868,   3036144, ... A103213
  [2] 0, 1,  7,  50,  406,  3804,  41028,   506064,   7084656, ... A355171
  [3] 0, 1,  9,  77,  714,  7374,  85272,  1102968,  15908400, ... A355372
  [4] 0, 1, 11, 110, 1154, 13144, 164136,  2251920,  33923760, ... A355407
  [5] 0, 1, 13, 149, 1750, 21894, 295500,  4320420,  68487120, ... A355414
  [6] 0, 1, 15, 194, 2526, 34524, 502644,  7838928, 131198544, ...
  [7] 0, 1, 17, 245, 3506, 52054, 814968, 13543704, 239548176, ...
		

Crossrefs

Cf. A029767, A103213, A355171, A355259, A355372, A355407, A355414.

Programs

Maple

egf := n -> log((1 - x)/(1 - 2*x))/(1 - x)^n:
ser := n -> series(egf(n), x, 22):
row := n -> seq(k!*coeff(ser(n), x, k), k = 0..8):
seq(print(row(n)), n = 0..8);
# Alternative:
A := (n, k) -> add(k!*binomial(k + n - 1, k - j - 1)/(j + 1), j = 0..k-1):
seq(print(seq(A(n, k), k = 0..8)), n = 0..7);

Mathematica

A[0, 0] = 0; A[n_, k_] := k! * Binomial[n+k-1, k - 1] * HypergeometricPFQ[{1 - k, 1, 1}, {2, n + 1}, -1];
Table[A[n, k], {n, 0, 8}, {k, 0, 8}] // TableForm

Formula

A(n, k) = k!*Sum_{j=0..k-1} binomial(k + n - 1, k - j - 1) / (j + 1).

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

A(n, k) = k!*binomial(n + k - 1, k - 1)*hypergeom([1, 1, 1 - k], [2, n + 1], -1) except for A(0, 0) = 0.