%I #24 Aug 15 2022 15:07:14
%S 1,0,1,0,-1,2,0,1,-6,6,0,-1,14,-36,24,0,1,-30,150,-240,120,0,-1,62,
%T -540,1560,-1800,720,0,1,-126,1806,-8400,16800,-15120,5040,0,-1,254,
%U -5796,40824,-126000,191520,-141120,40320,0,1,-510,18150,-186480,834120,-1905120,2328480,-1451520,362880
%N Coefficients of the signed Fubini polynomials in ascending order, F_n(x) = Sum_{k=0..n} (-1)^n*Stirling2(n,k)*k!*(-x)^k.
%C Signed version of A131689.
%C Integral_{x=0..1} F_n(x) = B_n(1) where B_n(x) are the Bernoulli polynomials.
%H Peter Luschny, <a href="/A164555/a164555.jpg">Illustration of the polynomials.</a>
%H Peter Luschny, <a href="http://luschny.de/math/zeta/The-Bernoulli-Manifesto.html">The Bernoulli Manifesto.</a>
%H Grzegorz Rządkowski, <a href="https://doi.org/10.1142/S1402925110000635">Bernoulli numbers and solitons - revisited</a>, Journal of Nonlinear Mathematical Physics, (2010) 17:1, 121-126.
%H J. Worpitzky, <a href="http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002158698">Studien über die Bernoullischen und Eulerschen Zahlen</a>, Journal für die reine und angewandte Mathematik, 94 (1883), 203-232.
%F T(n, k) = (-1)^(n-k) * Stirling2(n, k) * k!.
%F E.g.f.: 1/(1-x*(1-exp(-t))) = Sum_{n>=0} F_n(x) t^n/n!.
%F T(n, k) = k*(T(n-1, k-1) - T(n-1, k)) for 0 <= k <= n, T(0, 0) = 1, otherwise 0.
%F 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
%F 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
%F 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
%e Triangle of coefficients starts:
%e [1]
%e [0, 1]
%e [0, -1, 2]
%e [0, 1, -6, 6]
%e [0, -1, 14, -36, 24]
%e [0, 1, -30, 150, -240, 120]
%e [0, -1, 62, -540, 1560, -1800, 720]
%e [0, 1, -126, 1806, -8400, 16800, -15120, 5040]
%p F := (n,x) -> add((-1)^n*Stirling2(n,k)*k!*(-x)^k, k=0..n):
%p for n from 0 to 10 do PolynomialTools:-CoefficientList(F(n,x), x) od;
%t T[ n_, k_] := If[ n < 0 || k < 0, 0, (-1)^(n - k) k! StirlingS2[n, k]]; (* _Michael Somos_, Jul 08 2018 *)
%o (PARI)
%o {T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(n + i) * binomial(k, i) * i^n))};
%o /* _Michael Somos_, Jul 08 2018 */
%o (Julia)
%o function T(n, k)
%o if k < 0 || k > n return 0 end
%o if n == 0 && k == 0 return 1 end
%o k*(T(n-1, k-1) - T(n-1, k))
%o end
%o for n in 0:7
%o println([T(n,k) for k in 0:n])
%o end
%o # _Peter Luschny_, Mar 26 2020
%Y Row sums are A000012, diagonal is A000142.
%Y Cf. A131689 (unsigned), A019538 (n>0, k>0), A090582.
%Y 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).
%Y Cf. A163626, A173018.
%K sign
%O 0,6
%A _Peter Luschny_, Jan 09 2017