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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.

%I A355257 #37 Apr 13 2025 01:46:03
%S A355257 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,
%T A355257 7560,0,1,13,110,714,3804,18204,91440,0,1,15,149,1154,7374,41028,
%U A355257 218868,1285200,0,1,17,194,1750,13144,85272,506064,3036144,20603520
%N 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.
%C A355257 Conjecture: For p prime, A(n, p) == -1 (mod p) for n >= 0.
%C A355257 Conjecture: Let n >= 0, k >= 1 and k != 4. Then k divides A(n, k) if and only if k is not prime.
%C A355257 From _Mélika Tebni_, Jul 04 2022: (Start)
%C A355257 Conjecture: The polynomials of A355259 generate the k+1 column of this array.
%C A355257 Conjecture: For p prime and n even, (A(n, p) / (p - 1)) == 1 (mod p). (End)
%F A355257 A(n, k) = k!*Sum_{j=0..k-1} binomial(k + n - 1, k - j - 1) / (j + 1).
%F A355257 A(n, k) = k!*Sum_{j=1..k} binomial(n + k - j - 1, n - 1)*(2^j - 1) / j.
%F A355257 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.
%e A355257 Table A(n, k) begins:
%e A355257   [0] 0, 1,  3,  14,   90,   744,   7560,    91440,   1285200, ... A029767
%e A355257   [1] 0, 1,  5,  29,  206,  1774,  18204,   218868,   3036144, ... A103213
%e A355257   [2] 0, 1,  7,  50,  406,  3804,  41028,   506064,   7084656, ... A355171
%e A355257   [3] 0, 1,  9,  77,  714,  7374,  85272,  1102968,  15908400, ... A355372
%e A355257   [4] 0, 1, 11, 110, 1154, 13144, 164136,  2251920,  33923760, ... A355407
%e A355257   [5] 0, 1, 13, 149, 1750, 21894, 295500,  4320420,  68487120, ... A355414
%e A355257   [6] 0, 1, 15, 194, 2526, 34524, 502644,  7838928, 131198544, ...
%e A355257   [7] 0, 1, 17, 245, 3506, 52054, 814968, 13543704, 239548176, ...
%p A355257 egf := n -> log((1 - x)/(1 - 2*x))/(1 - x)^n:
%p A355257 ser := n -> series(egf(n), x, 22):
%p A355257 row := n -> seq(k!*coeff(ser(n), x, k), k = 0..8):
%p A355257 seq(print(row(n)), n = 0..8);
%p A355257 # Alternative:
%p A355257 A := (n, k) -> add(k!*binomial(k + n - 1, k - j - 1)/(j + 1), j = 0..k-1):
%p A355257 seq(print(seq(A(n, k), k = 0..8)), n = 0..7);
%t A355257 A[0, 0] = 0; A[n_, k_] := k! * Binomial[n+k-1, k - 1] * HypergeometricPFQ[{1 - k, 1, 1}, {2, n + 1}, -1];
%t A355257 Table[A[n, k], {n, 0, 8}, {k, 0, 8}] // TableForm
%Y A355257 Cf. A029767, A103213, A355171, A355259, A355372, A355407, A355414.
%K A355257 nonn,tabl
%O A355257 0,6
%A A355257 _Peter Luschny_ and _Mélika Tebni_, Jul 01 2022