A003575 Dowling numbers: e.g.f.: exp(x + (exp(b*x) - 1)/b) with b=3.
1, 2, 7, 35, 214, 1523, 12349, 112052, 1120849, 12219767, 143942992, 1819256321, 24526654381, 350974470746, 5308470041299, 84554039118383, 1413794176669942, 24745966692370607, 452277149756692105, 8612255652371171012, 170517319084490074405
Offset: 0
Keywords
Links
- Muniru A Asiru, Table of n, a(n) for n = 0..210
- Moussa Benoumhani, On Whitney numbers of Dowling lattices, Discrete Math., Vol. 159, No. 1-3 (1996), pp. 13-33.
- Thomas A. Dowling, A class of geometric lattices based on finite groups, Journal of Combinatorial Theory, Series B, Vol. 14, No. 1 (1973), pp. 61-86.
- 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. 8.
- Mahid M. Mangontarum and Jacob Katriel, On q-Boson Operators and q-Analogues of the r-Whitney and r-Dowling Numbers, J. Int. Seq., Vol. 18 (2015), Article 15.9.8.
Crossrefs
Programs
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Magma
m:=30; c:=3; R
:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( Exp(x+(Exp(c*x)-1)/c) )); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Feb 20 2019 -
Maple
seq(coeff(series(n!*exp(z+(1/3)*exp(3*z)-(1/3)),z,n+1), z, n), n=0..30); # Muniru A Asiru, Feb 19 2019
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Mathematica
With[{nn=20},CoefficientList[Series[Exp[x+Exp[3x]/3-1/3],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Jan 04 2019 *) Table[Sum[Binomial[n, k] * 3^k * BellB[k, 1/3], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 17 2020 *)
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PARI
my(x = 'x + O('x^30)); Vec(serlaplace(exp(x + exp(3*x)/3 - 1/3))) \\ Michel Marcus, Feb 09 2018
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Sage
b=3; def A003575_list(prec): P.
= PowerSeriesRing(QQ, prec) return P( exp(x +(exp(b*x)-1)/b) ).egf_to_ogf().list() A003575_list(30) # G. C. Greubel, Feb 20 2019
Formula
E.g.f.: exp(x + (exp(3*x) - 1)/3).
G.f.: 1/(1-x*Q(0)), where Q(k) = 1 + x/(1 - x + 3*x*(k+1)/(x - 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 19 2013
a(n) = exp(-1/3) * Sum_{k>=0} (3*k + 1)^n / (3^k * k!). - Ilya Gutkovskiy, Apr 16 2020
a(n) ~ 3^(n + 1/3) * n^(n + 1/3) * exp(n/LambertW(3*n) - n - 1/3) / (sqrt(1 + LambertW(3*n)) * LambertW(3*n)^(n + 1/3)). - Vaclav Kotesovec, Jun 26 2022
Extensions
Name clarified by G. C. Greubel, Feb 20 2019
Comments