A154537 Triangle T(n,m) read by rows: let p(n,x) = exp(-x) * Sum_{m >= 0} (2*m + 1)^n * x^m/m!; then T(n,m) = [x^m] p(n,x).

1, 1, 2, 1, 8, 4, 1, 26, 36, 8, 1, 80, 232, 128, 16, 1, 242, 1320, 1360, 400, 32, 1, 728, 7084, 12160, 6320, 1152, 64, 1, 2186, 36876, 99288, 81200, 25312, 3136, 128, 1, 6560, 188752, 768768, 929376, 440832, 91392, 8192, 256, 1, 19682, 956880, 5758880

Offset

0,3

Comments

Row sums are A126390.

These numbers are related to Stirling numbers of the second kind as MacMahon numbers A060187 are related to Eulerian numbers.

Let p and q denote operators acting on a function f(x) by pf(x) = x*f(x) and qf(x) = d/dx(f(x)). Let A be the anticommutator operator qp + pq. Then A^n = Sum_{k = 0..n} T(n,k) p^k q^k. For example, A^3(f) = f + 26*x*df/dx + 36*x^2*d^2(f)/dx^2 + 8*x^3*d^3(f)/dx^3. - Peter Bala, Jul 24 2014

From Peter Bala, May 21 2023: (Start)

Compare the definition of the polynomial p(n,x) with Dobiński's formula for the Bell polynomials (row polynomials of A008277 for n >= 1): Bell(n,x) = exp(-x) * Sum_{m >= 0} m^n * x^m/m!.

Boyadzhiev has shown that Bell(n,x) = d/dx( exp(-x) * Sum_{m >= 0} (1^n + 2^n + ... + (m-1)^n) * x^m/m! ). The corresponding result for this table is that the n-th row polynomial p(n,x) = d/dx( exp(-x) * Sum_{m >= 0} (1^n + 3^n + ... + (2*m-1)^n) * x^m/m! ). (End)

Links

Table of n, a(n) for n=0..48.

Khristo N. Boyadzhiev, New identities with Stirling, hyperharmonic, and derangement numbers, Bernoulli and Euler polynomials, powers, and factorials, arXiv:2011.03101v3 [math.NT], 2020-2021.

Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 9.

Wolfdieter Lang, On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.

Eric Weisstein's World of Mathematics, Dobiński's formula

Formula

From Peter Bala, Oct 28 2011: (Start)

T(n,k) = 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.

Recurrence relation: T(n,k) = 2*T(n-1,k-1) + (2*k+1)*T(n-1,k).

T(n,k) = (2^k)*A039755(n,k).

E.g.f.: exp(x + y*(exp(2*x) - 1)) = 1 + (1 + 2*y)*x + (1 + 8*y + 4*y^2)*x^2/2! + .... (End)

T(n, k) = Sum_{m=0..n} binomial(n, m)*2^m*Stirling2(m, k), 0 <= k <= n, where Stirling2 is A048993. - Wolfdieter Lang, Apr 13 2017

Boas-Buck recurrence for column sequence m: T(n,k) = (1/(n - k))*[n*(1 + m)*T(n-1,k) + k*Sum_{p=m..n-2} binomial(n,p)(-2)^(n-p)*Bernoulli(n-p)*T(p,k)], for n > m >= 0, with input T(m,m) = 2^m. See a comment in A282629, also for references, and an example below. - Wolfdieter Lang, Aug 11 2017

Example

{1},
{1, 2},
{1, 8, 4},
{1, 26, 36, 8},
{1, 80, 232, 128, 16},
{1, 242, 1320, 1360, 400, 32},
{1, 728, 7084, 12160, 6320, 1152, 64},
{1, 2186, 36876, 99288, 81200, 25312, 3136, 128},
{1, 6560, 188752, 768768, 929376, 440832, 91392, 8192, 256},
{1, 19682, 956880, 5758880, 9901920, 6707904, 2069760, 305664, 20736, 512},
{1, 59048, 4823764, 42225920, 100635040, 93590784, 40322688, 8724480, 963840, 51200, 1024}
...
Boas-Buck recurrence for column m = 2, and n = 4: T(4,2) =(1/2)*[4*3*T(3, 2) + 2*6*(-2)^2*Bernoulli(2)*T(2,2))] = (1/2)*(12*36 + 12*4*(1/6)*4) = 232. - Wolfdieter Lang, Aug 11 2017

Mathematica

p[x_, n_] = Sum[(2*m + 1)^n*x^m/m!, {m, 0, Infinity}]/(Exp[x]);
Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}]
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}];
Flatten[%]

Crossrefs

Cf. A008277, A000110, A039755, A048993, A060187.

Sequence in context: A142075 A366173 A110107 * A201641 A110446 A109979

Adjacent sequences: A154534 A154535 A154536 * A154538 A154539 A154540

Keyword

nonn,easy,tabl

Author

Roger L. Bagula, Jan 11 2009

Extensions

Edited by N. J. A. Sloane, Jan 12 2009

Status

approved