A278075 Coefficients of the signed Fubini polynomials in ascending order, F_n(x) = Sum_{k=0..n} (-1)^n*Stirling2(n,k)*k!*(-x)^k.

1, 0, 1, 0, -1, 2, 0, 1, -6, 6, 0, -1, 14, -36, 24, 0, 1, -30, 150, -240, 120, 0, -1, 62, -540, 1560, -1800, 720, 0, 1, -126, 1806, -8400, 16800, -15120, 5040, 0, -1, 254, -5796, 40824, -126000, 191520, -141120, 40320, 0, 1, -510, 18150, -186480, 834120, -1905120, 2328480, -1451520, 362880

Offset

0,6

Comments

Signed version of A131689.

Integral_{x=0..1} F_n(x) = B_n(1) where B_n(x) are the Bernoulli polynomials.

Links

Table of n, a(n) for n=0..54.

Peter Luschny, Illustration of the polynomials.

Peter Luschny, The Bernoulli Manifesto.

Grzegorz Rządkowski, Bernoulli numbers and solitons - revisited, Journal of Nonlinear Mathematical Physics, (2010) 17:1, 121-126.

J. Worpitzky, Studien über die Bernoullischen und Eulerschen Zahlen, Journal für die reine und angewandte Mathematik, 94 (1883), 203-232.

Formula

T(n, k) = (-1)^(n-k) * Stirling2(n, k) * k!.

E.g.f.: 1/(1-x*(1-exp(-t))) = Sum_{n>=0} F_n(x) t^n/n!.

T(n, k) = k*(T(n-1, k-1) - T(n-1, k)) for 0 <= k <= n, T(0, 0) = 1, otherwise 0.

Bernoulli numbers are given by B(n) = Sum_{k = 0..n} T(n, k) / (k+1) with B(1) = 1/2. - Michael Somos, Jul 08 2018

Let F_n(x) be the row polynomials of this sequence and W_n(x) the row polynomials of A163626. Then F_n(1 - x) = W_n(x) and Integral_{x=0..1} U(n, x) = Bernoulli(n, 1) for U in {W, F}. - Peter Luschny, Aug 10 2021

T(n, k) = [z^k] Sum_{k=0..n} Eulerian(n, k)*z^(k+1)*(z-1)^(n-k-1) for n >= 1, where Eulerian(n, k) = A173018(n, k). - Peter Luschny, Aug 15 2022

Example

Triangle of coefficients starts:
[1]
[0,  1]
[0, -1,    2]
[0,  1,   -6,    6]
[0, -1,   14,  -36,    24]
[0,  1,  -30,  150,  -240,   120]
[0, -1,   62, -540,  1560, -1800,    720]
[0,  1, -126, 1806, -8400, 16800, -15120, 5040]

Maple

F := (n, x) -> add((-1)^n*Stirling2(n, k)*k!*(-x)^k, k=0..n):
for n from 0 to 10 do PolynomialTools:-CoefficientList(F(n, x), x) od;

Mathematica

T[ n_, k_] := If[ n < 0 || k < 0, 0, (-1)^(n - k) k! StirlingS2[n, k]]; (* Michael Somos, Jul 08 2018 *)

Prog

(PARI)
{T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(n + i) * binomial(k, i) * i^n))};
/* Michael Somos, Jul 08 2018 */
(Julia)
function T(n, k)
    if k < 0 || k > n return 0 end
    if n == 0 && k == 0 return 1 end
    k*(T(n-1, k-1) - T(n-1, k))
end
for n in 0:7
    println([T(n, k) for k in 0:n])
end
# Peter Luschny, Mar 26 2020

Crossrefs

Row sums are A000012, diagonal is A000142.

Cf. A131689 (unsigned), A019538 (n>0, k>0), A090582.

Let F(n, x) = Sum_{k=0..n} T(n,k)*x^k then, apart from possible differences in the sign or the offset, we have: F(n, -5) = A094418(n), F(n, -4) = A094417(n), F(n, -3) = A032033(n), F(n, -2) = A004123(n), F(n, -1) = A000670(n), F(n, 0) = A000007(n), F(n, 1) = A000012(n), F(n, 2) = A000629(n), F(n, 3) = A201339(n), F(n, 4) = A201354(n), F(n, 5) = A201365(n).

Cf. A163626, A173018.

Sequence in context: A085845 A138106 A131689 * A114329 A361956 A241011

Adjacent sequences: A278072 A278073 A278074 * A278076 A278077 A278078

Keyword

sign

Author

Peter Luschny, Jan 09 2017

Status

approved