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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)!. 3
 1, 1, 0, 2, -1, 0, 6, -8, 2, 0, 24, -66, 44, -6, 0, 120, -624, 792, -312, 24, 0, 720, -6840, 14496, -10872, 2736, -120, 0, 5040, -86400, 285840, -347904, 171504, -28800, 720, 0, 40320, -1244880, 6181920, -11245680, 8996544, -3090960, 355680, -5040, 0 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS 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 B_(n) = (1/(n + 1)) * Sum_{k=0..n} (-1)^k * Eulerian(n, k) / binomial(n, k). 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. For a variant of this identity see the first formula in A356546. An analogous representation based on the Eulerian numbers of second order is given in A356547. REFERENCES 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.) LINKS Table of n, a(n) for n=0..44. Ira Gessel, Eulerian number identity, MathOverflow, Apr 2019. Peter Luschny, How are the Eulerian numbers of the first-order related to the Eulerian numbers of the second-order?, MathOverflow, Feb. 2021. Peter Luschny, Eulerian polynomials. Oskar Schlömilch, Ueber die Bernoulli'sche Funktion und deren Gebrauch bei der Entwickelung halbconvergenter Reihen, Zeitschrift fuer Mathematik und Pysik, vol. 1 (1856), p. 193-211. Julius Worpitsky, Studien über die Bernoullischen und Eulerschen Zahlen, Journal für die reine und angewandte Mathematik (Crelle), 94 (1883), 203-232. See page 22, first formula. FORMULA 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)!. T(n, k) = [x^k] p_n(x). T(n, k) = (-1)^k*Eulerian(n, k)*k!*(n - k)!. T(n, k) = k! * (n-k)! * Sum_{j=0..k} (-1)^(k-j)*(k-j+1)^n*binomial(n+1, j). T(n, k) = (-1)^k * A173018(n, k) * A098361(n, k). T(n, k) = (-1)^k * A123125(n, n - k) * A098361(n, n - k). EXAMPLE The table T(n, k) of the coefficients, sorted in ascending order, starts: [0] 1; [1] 1, 0; [2] 2, -1, 0; [3] 6, -8, 2, 0; [4] 24, -66, 44, -6, 0; [5] 120, -624, 792, -312, 24, 0; [6] 720, -6840, 14496, -10872, 2736, -120, 0; [7] 5040, -86400, 285840, -347904, 171504, -28800, 720, 0; [8] 40320, -1244880, 6181920, -11245680, 8996544, -3090960, 355680, -5040, 0; MAPLE E1 := proc(n, k) combinat:-eulerian1(n, k) end: p := (n, x) -> add(E1(n, k)*k!*(n - k)!*(-x)^k, k = 0..n): seq(print(seq(coeff(p(n, x), x, k), k=0..n)), n = 0..8); seq(p(n, 1)/(n + 1)!, n = 0..14); # check the Bernoulli representation MATHEMATICA 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 (* Diagonals: *) d[n_, k_] := k! * (n - k)! * Sum[(-1)^(n-k-j)*(n - j - k + 1)^n * Binomial[n + 1, j], {j, 0, n - k}]; CROSSREFS Cf. A173018 (Eulerian number), A164555(n)/A027642(n) (Bernoulli numbers with B(1) = 1/2), A129814 (row sums, but different sign for n = 1). Cf. A098361, A123125, A356546, A356547, A356601, A356602. Sequence in context: A281662 A163936 A288874 * A187555 A358188 A117651 Adjacent sequences: A356542 A356543 A356544 * A356546 A356547 A356548 KEYWORD sign,tabl AUTHOR Peter Luschny, Aug 11 2022 STATUS approved

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Last modified August 12 03:07 EDT 2024. Contains 375085 sequences. (Running on oeis4.)