%I #48 Mar 14 2024 15:31:38

%S 1,1,2,1,8,4,1,26,36,8,1,80,232,128,16,1,242,1320,1360,400,32,1,728,

%T 7084,12160,6320,1152,64,1,2186,36876,99288,81200,25312,3136,128,1,

%U 6560,188752,768768,929376,440832,91392,8192,256,1,19682,956880,5758880

%N 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).

%C Row sums are A126390.

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

%C 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

%C From _Peter Bala_, May 21 2023: (Start)

%C 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!.

%C 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)

%H Khristo N. Boyadzhiev, <a href="https://arxiv.org/abs/2011.03101">New identities with Stirling, hyperharmonic, and derangement numbers, Bernoulli and Euler polynomials, powers, and factorials</a>, arXiv:2011.03101v3 [math.NT], 2020-2021.

%H Paweł Hitczenko, <a href="https://arxiv.org/abs/2403.03422">A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality</a>, arXiv:2403.03422 [math.CO], 2024. See p. 9.

%H Wolfdieter Lang, <a href="https://arxiv.org/abs/1708.01421">On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles</a>, arXiv:1708.01421 [math.NT], August 2017.

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/DobinskisFormula.html">Dobiński's formula</a>

%F From _Peter Bala_, Oct 28 2011: (Start)

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

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

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

%F 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)

%F 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

%F 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

%e {1},

%e {1, 2},

%e {1, 8, 4},

%e {1, 26, 36, 8},

%e {1, 80, 232, 128, 16},

%e {1, 242, 1320, 1360, 400, 32},

%e {1, 728, 7084, 12160, 6320, 1152, 64},

%e {1, 2186, 36876, 99288, 81200, 25312, 3136, 128},

%e {1, 6560, 188752, 768768, 929376, 440832, 91392, 8192, 256},

%e {1, 19682, 956880, 5758880, 9901920, 6707904, 2069760, 305664, 20736, 512},

%e {1, 59048, 4823764, 42225920, 100635040, 93590784, 40322688, 8724480, 963840, 51200, 1024}

%e ...

%e 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

%t p[x_, n_] = Sum[(2*m + 1)^n*x^m/m!, {m, 0, Infinity}]/(Exp[x]);

%t Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}]

%t Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}];

%t Flatten[%]

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

%K nonn,easy,tabl

%O 0,3

%A _Roger L. Bagula_, Jan 11 2009

%E Edited by _N. J. A. Sloane_, Jan 12 2009