A278075 - OEIS

%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