A333303 T(n, k) = [x^k] (-2)^n*(B(n, x/2) - B(n, (x+1)/2)) where B(n, x) are the Bernoulli polynomials. Triangle read by rows, for 0 <= k <= n.
0, 1, 1, -2, 0, -3, 3, -1, 0, 6, -4, 0, 5, 0, -10, 5, 3, 0, -15, 0, 15, -6, 0, -21, 0, 35, 0, -21, 7, -17, 0, 84, 0, -70, 0, 28, -8, 0, 153, 0, -252, 0, 126, 0, -36, 9, 155, 0, -765, 0, 630, 0, -210, 0, 45, -10, 0, -1705, 0, 2805, 0, -1386, 0, 330, 0, -55, 11
Offset: 0
Examples
B*(8, z) = 1024*(Zeta(-7, (z+1)/2) - Zeta(-7, z/2))
= -17 + 84*z^2 - 70*z^4 + 28*z^6 - 8*z^7.
Triangle starts:
[ 0] [ 0]
[ 1] [ 1]
[ 2] [ 1, -2]
[ 3] [ 0, -3, 3]
[ 4] [ -1, 0, 6, -4]
[ 5] [ 0, 5, 0, -10, 5]
[ 6] [ 3, 0, -15, 0, 15, -6]
[ 7] [ 0, -21, 0, 35, 0, -21, 7]
[ 8] [-17, 0, 84, 0, -70, 0, 28, -8]
[ 9] [ 0, 153, 0, -252, 0, 126, 0, -36, 9]
[10] [155, 0, -765, 0, 630, 0, -210, 0, 45, -10]
[11] [ 0, -1705, 0, 2805, 0, -1386, 0, 330, 0, -55, 11]
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..11325 (rows 0 <= n <= 150, flattened)
- Digital Library of Mathematical Functions, Lerch’s Transcendent.
- Peter H. N. Luschny, An introduction to the Bernoulli function, arXiv:2009.06743 [math.HO], 2020.
- Eric Weisstein's World of Mathematics, Lerch Transcendent.
Crossrefs
Programs
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Mathematica
B[n_, x_] := (-2)^n (BernoulliB[n, x/2] - BernoulliB[n, (x + 1)/2]); Prepend[Table[CoefficientList[B[n, x], x], {n, 1, 11}], 0] // Flatten -
SageMath
def Bstar(n,x): return (-2)^n*(bernoulli_polynomial(x/2,n) - bernoulli_polynomial((x+1)/2,n)) print(flatten([expand(Bstar(n, x)).list() for n in (0..11)]))
Formula
Let B*(n, x) denote the alternating Bernoulli rational polynomial functions defined by Z*(s, x) = Phi(-1, s, x) and B*(s, x) = -s Z*(1 - s, x). Here Phi(z, s, x) is the Hurwitz-Lerch transcendent defined as an analytic continuation of Sum_{k>=0} z^k/(k+x)^s. Then T(n, k) = (-1)^n [x^k] 2 B*(n, x).
T(n, 0) = 2*(2^n - 1)*Bernoulli(n, 1) = n*Euler(n - 1, 1) = -A226158(n).
Main diagonal is (-1)^(n+1)*n = A181983(n).
Comments