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A356545 Triangle read by rows. T(n, k) are the coefficients of polynomials p_n(x) based on the Eulerian numbers of first order representing the Bernoulli numbers as B_n = p_n(1) / (n + 1)!.

%I A356545 #43 Aug 11 2025 07:22:21
%S A356545 1,1,0,2,-1,0,6,-8,2,0,24,-66,44,-6,0,120,-624,792,-312,24,0,720,
%T A356545 -6840,14496,-10872,2736,-120,0,5040,-86400,285840,-347904,171504,
%U A356545 -28800,720,0,40320,-1244880,6181920,-11245680,8996544,-3090960,355680,-5040,0
%N A356545 Triangle read by rows. T(n, k) are the coefficients of polynomials p_n(x) based on the Eulerian numbers of first order representing the Bernoulli numbers as B_n = p_n(1) / (n + 1)!.
%C A356545 The Bernoulli numbers with B(1) = 1/2 can be represented as the weighted sum of Eulerian numbers, where we use the definition as given by Graham et al., Eulerian(n, k) = A173018(n, k). For n >= 0 we have
%C A356545    B_(n) = (1/(n + 1)) * Sum_{k=0..n} (-1)^k * Eulerian(n, k) / binomial(n, k).
%C A356545 The formula was given by Worpitsky in 1883 (see link) as an example for the application of a formula of Schlömilch from 1856. In 2019 the identity was proved in the modern fashion by Gessel on MathOverflow.
%C A356545 For a variant of this identity see the first formula in A356546.
%C A356545 An analogous representation based on the Eulerian numbers of second order is given in A356547.
%D A356545 R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 268. (Since the thirty-fourth printing, Jan. 2022, with B(1) = 1/2.)
%H A356545 Ira Gessel, <a href="https://mathoverflow.net/a/328534">Eulerian number identity</a>, MathOverflow, Apr 2019.
%H A356545 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 A356545 Peter Luschny, <a href="http://www.luschny.de/math/euler/EulerianPolynomials.html">Eulerian polynomials</a>.
%H A356545 Oskar Schlömilch, <a href="https://gdz.sub.uni-goettingen.de/id/PPN599415665_0001?tify=%7B%22pages%22%3A%5B196%2C197%5D%2C%22pan%22%3A%7B%22x%22%3A1.202%2C%22y%22%3A0.817%7D%2C%22view%22%3A%22info%22%2C%22zoom%22%3A0.344%7D">Ueber die Bernoulli'sche Funktion und deren Gebrauch bei der Entwickelung halbconvergenter Reihen</a>, Zeitschrift fuer Mathematik und Pysik, vol. 1 (1856), p. 193-211.
%H A356545 Julius Worpitsky, <a href="https://gdz.sub.uni-goettingen.de/id/PPN243919689_0094">Studien über die Bernoullischen und Eulerschen Zahlen</a>, Journal für die reine und angewandte Mathematik (Crelle), 94 (1883), 203-232. See page 22, first formula.
%F A356545 Let p_n(x) = Sum_{k=0..n} Eulerian(n, k)*k!*(n - k)! * (-x)^k. For x = 1 these polynomials give rise to the representation Bernoulli(n) = p_n(1) / (n + 1)!.
%F A356545 T(n, k) = [x^k] p_n(x).
%F A356545 T(n, k) = (-1)^k*Eulerian(n, k)*k!*(n - k)!.
%F A356545 T(n, k) = k! * (n-k)! * Sum_{j=0..k} (-1)^(k-j)*(k-j+1)^n*binomial(n+1, j).
%F A356545 T(n, k) = (-1)^k * A173018(n, k) * A098361(n, k).
%F A356545 T(n, k) = (-1)^k * A123125(n, n - k) * A098361(n, n - k).
%e A356545 The table T(n, k) of the coefficients, sorted in ascending order, starts:
%e A356545 [0]     1;
%e A356545 [1]     1,        0;
%e A356545 [2]     2,       -1,       0;
%e A356545 [3]     6,       -8,       2,         0;
%e A356545 [4]    24,      -66,      44,        -6,       0;
%e A356545 [5]   120,     -624,     792,      -312,      24,        0;
%e A356545 [6]   720,    -6840,   14496,    -10872,    2736,     -120,      0;
%e A356545 [7]  5040,   -86400,  285840,   -347904,  171504,   -28800,    720,     0;
%e A356545 [8] 40320, -1244880, 6181920, -11245680, 8996544, -3090960, 355680, -5040, 0;
%p A356545 E1 := proc(n, k) combinat:-eulerian1(n, k) end:
%p A356545 p := (n, x) -> add(E1(n, k)*k!*(n - k)!*(-x)^k, k = 0..n):
%p A356545 seq(print(seq(coeff(p(n, x), x, k), k=0..n)), n = 0..8);
%p A356545 seq(p(n, 1)/(n + 1)!, n = 0..14); # check the Bernoulli representation
%t A356545 T[n_, k_] := k! * (n-k)! * Sum[(-1)^(k-j) * (k-j+1)^n * Binomial[n+1, j], {j, 0, k}]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // TableForm
%t A356545 (* Diagonals: *)
%t A356545 d[n_, k_] := k! * (n - k)! * Sum[(-1)^(n-k-j)*(n - j - k + 1)^n * Binomial[n + 1, j], {j, 0, n - k}];
%Y A356545 Cf. A173018 (Eulerian number), A164555(n)/A027642(n) (Bernoulli numbers with B(1) = 1/2), A129814 (row sums, but different sign for n = 1).
%Y A356545 Cf. A098361, A123125, A356546, A356547, A356601, A356602.
%K A356545 sign,tabl
%O A356545 0,4
%A A356545 _Peter Luschny_, Aug 11 2022