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)!.

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

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