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Array read by ascending antidiagonals: A(n, k) = Sum_{j = 0..k} binomial(k, j) * Euler(j, 0) *(2*n)^j.
4

%I #22 Nov 14 2024 14:43:12

%S 1,1,1,1,0,1,1,-1,-1,1,1,-2,-3,0,1,1,-3,-5,11,5,1,1,-4,-7,46,57,0,1,1,

%T -5,-9,117,205,-361,-61,1,1,-6,-11,236,497,-3362,-2763,0,1,1,-7,-13,

%U 415,981,-15123,-22265,24611,1385,1

%N Array read by ascending antidiagonals: A(n, k) = Sum_{j = 0..k} binomial(k, j) * Euler(j, 0) *(2*n)^j.

%H Peter Luschny, <a href="https://oeis.org/wiki/User:Peter_Luschny/EulerianPolynomialsGeneralized#Assorted_values_of_the_polynomials">Generalized Eulerian polynomials</a>. (See last row of the table.)

%F A(n, k) = n^k*(GHZeta(k, n, 4) - GHZeta(k, n, 2))) where GHZeta(k, n, m) = m^(k+1) * HurwitzZeta(-k, 1/(m*n)) for n > 0, and T(0, k) = 1.

%F A(n, k) = Im(P(n, k)) where P(n, k) = 2*i*(1 + Sum_{j=0..k} binomial(k, j)*polylog(-j, i)*n^j.

%F A(n, k) = substitute(x = -n, P(k, x)) where P(n, x) = (1/(n + 1)) * Sum_{j=0..n+1} binomial(n + 1, j) * Bernoulli(j, 1) * (4^j - 2^j)*x^(j-1).

%e Array A(n, k) starts:

%e [0] 1, 1, 1, 1, 1, 1, 1, ... A000012

%e [1] 1, 0, -1, 0, 5, 0, -61, ... A122045

%e [2] 1, -1, -3, 11, 57, -361, -2763, ... A212435

%e [3] 1, -2, -5, 46, 205, -3362, -22265, ... A225147

%e [4] 1, -3, -7, 117, 497, -15123, -95767, ... A156201

%e [5] 1, -4, -9, 236, 981, -47524, -295029, ... A377665

%e [6] 1, -5, -11, 415, 1705, -120125, -737891, ...

%e [7] 1, -6, -13, 666, 2717, -262086, -1599793, ...

%p GHZeta := (k, n, m) -> m^(k+1)*Zeta(0, -k, 1/(m*n)):

%p A := (n, k) -> ifelse(n = 0, 1, n^k*(GHZeta(k, n, 4) - GHZeta(k, n, 2))):

%p for n from 0 to 7 do lprint(seq(A(n, k), k = 0..7)) od;

%p # Alternative:

%p P := proc(n, k) local j; 2*I*(1 + add(binomial(k, j)*polylog(-j, I)*n^j, j = 0..k)) end:

%p A := n -> Im(P(n, k)): seq(lprint(seq(A(n, k), k = 0..7)), n = 0..7);

%p # Computing the transpose using polynomials P from A363393.

%p P := n -> add(binomial(n + 1, j)*bernoulli(j, 1)*(4^j - 2^j)*x^(j-1), j = 0..n+1)/(n + 1):

%p Column := (k, n) -> subs(x = -n, P(k)):

%p for k from 0 to 6 do seq(Column(k, n), n = 0..9) od;

%p # According to the definition:

%p A := (n, k) -> local j; add(binomial(k, j)*euler(j, 0)*(2*n)^j, j = 0..k):

%p seq(lprint(seq(A(n, k), k = 0..6)), n = 0..7);

%t A[n_, k_] := n^k (4^(k+1) HurwitzZeta[-k, 1/(4n)] - 2^(k + 1) HurwitzZeta[-k, 1/(2n)]);

%o (SageMath)

%o from mpmath import *

%o mp.dps = 32; mp.pretty = True

%o def T(n, k):

%o p = 2*I*(1+sum(binomial(k, j)*polylog(-j, I)*n^j for j in range(k+1)))

%o return int(imag(p))

%o for n in range(8): print([T(n, k) for k in range(7)])

%Y Rows: A000012, A122045, A212435, A225147, A156201, A377665.

%Y Cf. A377663 (column 3), A377664 (main diagonal), A363393 (column polynomials).

%K sign,tabl

%O 0,12

%A _Peter Luschny_, Nov 05 2024