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%I #17 Mar 18 2023 08:49:14
%S 1,1,0,6,-4,0,120,-192,72,0,5040,-15840,13920,-3456,0,362880,-2096640,
%T 3306240,-1918080,345600,0,39916800,-413683200,1053803520,-1064448000,
%U 448519680,-62208000,0,6227020800,-114960384000,447866496000,-699342336000,506348236800,-164428185600,18289152000,0
%N 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 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 B_(n) = (1/2)*Sum_{k=0..n} (-1)^k*Eulerian2(n, k) / binomial(2*n - 1, k).
%C 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 An analogous representation based on the Eulerian numbers of first order is given in A356545.
%D 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 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 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 Pietro Majer, <a href="https://mathoverflow.net/a/46126">Expressions involving Eulerian numbers of the second kind</a>, MathOverflow, Nov 2010.
%H 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 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 Let p_n(x) = Sum_{k=0..n} Eulerian2(n, k)*k!*(2*n - k - 1)! * (-x)^k.
%F T(n, k) = [x^k] p_n(x).
%F T(n, k) = (-1)^k*Eulerian2(n, k)*k!*(2*n - k - 1)!.
%e The triangle T(n, k) of the coefficients, sorted in ascending order, starts:
%e [0] 1;
%e [1] 1, 0;
%e [2] 6, -4, 0;
%e [3] 120, -192, 72, 0;
%e [4] 5040, -15840, 13920, -3456, 0;
%e [5] 362880, -2096640, 3306240, -1918080, 345600, 0;
%e [6] 39916800, -413683200, 1053803520, -1064448000, 448519680, -62208000, 0;
%p E2 := proc(n, k) combinat:-eulerian2(n, k) end:
%p p := (n, x) -> `if`(n = 0, 1, add(E2(n, k)*k!*(2*n - k - 1)!*(-x)^k, k = 0..n)):
%p seq(print([n], seq(coeff(p(n, x), x, k), k = 0..n)), n = 0..7);
%p seq(`if`(n = 0, 1, p(n, 1)/(2*(2*n-1)!)), n = 0..14); # check Bernoulli numbers
%Y Cf. A201637 (Eulerian number 2nd order), A164555(n)/A027642(n) (Bernoulli numbers with B(1) = 1/2).
%Y Cf. A356545.
%K sign,tabl
%O 0,4
%A _Peter Luschny_, Aug 12 2022
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