A035009 STIRLING transform of [1,1,2,4,8,16,32,...].
1, 1, 3, 11, 47, 227, 1215, 7107, 44959, 305091, 2206399, 16913987, 136823263, 1163490499, 10366252031, 96491364675, 935976996127, 9440144423875, 98800604237119, 1071092025420867, 12008090971866207, 139014305916844739, 1659578039401022079, 20405708646650507075
Offset: 0
Examples
Given the production matrix M, upper left term of M^5 = a(5)/2^4 = 227/16.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..558
- Paul Barry, Eulerian-Dowling Polynomials as Moments, Using Riordan Arrays, arXiv:1702.04007 [math.CO], 2017.
- Mikhail Khovanov, Victor Ostrik and Yakov Kononov, Two-dimensional topological theories, rational functions and their tensor envelopes, arXiv:2011.14758 [math.QA], 2020.
- Toufik Mansour and Mark Shattuck, A recurrence related to the Bell Numbers, Integers 11 (2011), #A67.
Programs
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Maple
A035009 := proc(n) local a,b,i; a := [seq(2,i=1..n-1)]; b := [seq(1,i=1..n-1)]; exp(-x)*hypergeom(a,b,x); round(evalf(subs(x=2,%), 10+2*n)) end: seq(A035009(n),n=0..19); # Peter Luschny, Mar 30 2011 # second Maple program: b:= proc(n, m) option remember; `if`(n=0, ceil(2^(m-1)), m*b(n-1, m)+b(n-1, m+1)) end: a:= n-> b(n, 0): seq(a(n), n=0..25); # Alois P. Heinz, Aug 03 2021
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Mathematica
1/(2*E^2)*Sum[(i + j)^n/(i!*j!), {i, 0, Infinity}, {j, 0, Infinity}] (* Starting from the 2nd term *) (* Vladimir Reshetnikov, Dec 31 2008 *) Join[{1}, Table[BellB[n, 2]/2, {n, 1, 25}]] (* Vaclav Kotesovec, Jun 26 2022 *)
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PARI
x='x+O('x^99); Vec(serlaplace((1 + exp(2*exp(x)-2))/2)) \\ Joerg Arndt, Apr 01 2011
Formula
a(n) = (1/2)*A001861(n), n > 0.
E.g.f.: (1 + exp(2*exp(x)-2))/2. - Emeric Deutsch, Feb 09 2002
a(n+1) = 1 + 2*Sum_{j=1..n} binomial(n, j)*a(j). - Jon Perry, Apr 25 2005
Define f_1(x), f_2(x), ... such that f_1(x)=e^x and for n=2,3,... f_{n+1}(x) = (d/dx)(x*f_n(x)). Then a(n) = e^(-2)*f_n(2). - Milan Janjic, May 30 2008
G.f.: 1 + x/(Q(0) - 2*x) where Q(k) = 1 - x*(k+1)/( 1 - 2*x/Q(k+1) ); (continued fraction ). - Sergei N. Gladkovskii, Mar 22 2013
G.f.: 1/Q(0), where Q(k)= 1 - x - 2*x/(1 - x*(2*k+1)/(1 - x - 2*x/(1 - x*(2*k+2)/Q(k+1)))); (continued fraction). - Sergei N. Gladkovskii, May 13 2013
G.f.: 1 + Sum_{k>=1} 2^(k-1)*x^k/Product_{j=1..k} (1 - j*x). - Ilya Gutkovskiy, Jun 19 2018
Comments