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A356547 Triangle read by rows. T(n, k) are the coefficients of polynomials p_n(x) based on the Eulerian numbers of second order representing the Bernoulli numbers as B_n = p_n(1) / (2*(2*n - 1)!).

%I A356547 #17 Mar 18 2023 08:49:14
%S A356547 1,1,0,6,-4,0,120,-192,72,0,5040,-15840,13920,-3456,0,362880,-2096640,
%T A356547 3306240,-1918080,345600,0,39916800,-413683200,1053803520,-1064448000,
%U A356547 448519680,-62208000,0,6227020800,-114960384000,447866496000,-699342336000,506348236800,-164428185600,18289152000,0
%N A356547 Triangle read by rows. T(n, k) are the coefficients of polynomials p_n(x) based on the Eulerian numbers of second order representing the Bernoulli numbers as B_n = p_n(1) / (2*(2*n - 1)!).
%C A356547 The Bernoulli numbers with B(1) = 1/2 can be represented as the weighted sum of Eulerian numbers of second order, where we use the definition as given by Graham et al., Eulerian2(n, k) = A201637(n, k). For n >= 1 we have
%C A356547    B_(n) = (1/2)*Sum_{k=0..n} (-1)^k*Eulerian2(n, k) / binomial(2*n - 1, k).
%C A356547 Although this representation looks classical it was apparently first proved by Majer in 2010; later Fu and recently O'Sullivan gave an alternative proof (see links).
%C A356547 An analogous representation based on the Eulerian numbers of first order is given in A356545.
%D A356547 R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270. (Since the thirty-fourth printing, Jan. 2022, with B(1) = 1/2.)
%H A356547 Amy M. Fu, <a href="https://arxiv.org/abs/2104.09316">Some Identities Related to the Second-Order Eulerian Numbers</a>, arXiv:2104.09316 [math.CO], Apr. 2021.
%H A356547 Peter Luschny, <a href="https://mathoverflow.net/q/384146">How are the Eulerian numbers of the first-order related to the Eulerian numbers of the second-order?</a>, MathOverflow, Feb. 2021.
%H A356547 Pietro Majer, <a href="https://mathoverflow.net/a/46126">Expressions involving Eulerian numbers of the second kind</a>, MathOverflow, Nov 2010.
%H A356547 G. Rzadkowski, M. Urlinska, <a href="http://arxiv.org/abs/1612.06635">A Generalization of the Eulerian Numbers</a>, arXiv:1612.06635 [math.CO], 2016
%H A356547 Cormac O'Sullivan, <a href="https://arxiv.org/abs/2208.02898">Stirling's approximation and a hidden link between two of Ramanujan's approximations</a>, arXiv:2208.02898 [math.NT], Aug. 2022.
%F A356547 Let p_n(x) = Sum_{k=0..n} Eulerian2(n, k)*k!*(2*n - k - 1)! * (-x)^k.
%F A356547 T(n, k) = [x^k] p_n(x).
%F A356547 T(n, k) = (-1)^k*Eulerian2(n, k)*k!*(2*n - k - 1)!.
%e A356547 The triangle T(n, k) of the coefficients, sorted in ascending order, starts:
%e A356547 [0]        1;
%e A356547 [1]        1,          0;
%e A356547 [2]        6,         -4,          0;
%e A356547 [3]      120,       -192,         72,           0;
%e A356547 [4]     5040,     -15840,      13920,       -3456,         0;
%e A356547 [5]   362880,   -2096640,    3306240,    -1918080,    345600,         0;
%e A356547 [6] 39916800, -413683200, 1053803520, -1064448000, 448519680, -62208000, 0;
%p A356547 E2 := proc(n, k) combinat:-eulerian2(n, k) end:
%p A356547 p := (n, x) -> `if`(n = 0, 1, add(E2(n, k)*k!*(2*n - k - 1)!*(-x)^k, k = 0..n)):
%p A356547 seq(print([n], seq(coeff(p(n, x), x, k), k = 0..n)), n = 0..7);
%p A356547 seq(`if`(n = 0, 1, p(n, 1)/(2*(2*n-1)!)), n = 0..14); # check Bernoulli numbers
%Y A356547 Cf. A201637 (Eulerian number 2nd order), A164555(n)/A027642(n) (Bernoulli numbers with B(1) = 1/2).
%Y A356547 Cf. A356545.
%K A356547 sign,tabl
%O A356547 0,4
%A A356547 _Peter Luschny_, Aug 12 2022