login
STIRLING transform of [1,1,2,4,8,16,32,...].
22

%I #88 Jun 26 2022 02:21:52

%S 1,1,3,11,47,227,1215,7107,44959,305091,2206399,16913987,136823263,

%T 1163490499,10366252031,96491364675,935976996127,9440144423875,

%U 98800604237119,1071092025420867,12008090971866207,139014305916844739,1659578039401022079,20405708646650507075

%N STIRLING transform of [1,1,2,4,8,16,32,...].

%C Numerators of sequence that shifts left one place under 1/2 order binomial transform. (Denominators are 2^(n-1) for n > 0.) - _Franklin T. Adams-Watters_, Jul 31 2005

%C Row sums of triangle A137597 starting (1, 3, 11, 47, 227, ...). - _Gary W. Adamson_, Jan 29 2008

%C From _Gary W. Adamson_, Jul 22 2011: (Start)

%C a(n)/2^(n-1) = upper left term in M^n, M = an infinite square production matrix in which a column of (1/2, 1/2, 1/2, ...) is appended to the right of Pascal's triangle, as follows:

%C 1, 1/2, 0, 0, 0, 0, ...

%C 1, 1, 1/2, 0, 0, 0, ...

%C 1, 2, 1, 1/2, 0, 0, ...

%C 1, 3, 3, 1, 1/2, 0, ...

%C 1, 4, 6, 4, 1, 1/2, ..., etc.

%C (End)

%C From _Bruno Berselli_, Mar 20 2013: (Start)

%C Note that, for t=A222391:

%C a(1)*t = Sum_{n >= 1} 1 /(Gamma(n/2)*Gamma((n+1)/2)),

%C a(2)*t = Sum_{n >= 1} n /(Gamma(n/2)*Gamma((n+1)/2)),

%C a(3)*t = Sum_{n >= 1} n^2/(Gamma(n/2)*Gamma((n+1)/2)),

%C a(4)*t = Sum_{n >= 1} n^3/(Gamma(n/2)*Gamma((n+1)/2)),

%C a(5)*t = Sum_{n >= 1} n^4/(Gamma(n/2)*Gamma((n+1)/2)),

%C a(6)*t = Sum_{n >= 1} n^5/(Gamma(n/2)*Gamma((n+1)/2)), etc.

%C (End)

%C Except for the initial term, the main diagonal of A129340. - _Peter Bala_, Apr 14 2017

%H Seiichi Manyama, <a href="/A035009/b035009.txt">Table of n, a(n) for n = 0..558</a>

%H Paul Barry, <a href="https://arxiv.org/abs/1702.04007">Eulerian-Dowling Polynomials as Moments, Using Riordan Arrays</a>, arXiv:1702.04007 [math.CO], 2017.

%H Mikhail Khovanov, Victor Ostrik and Yakov Kononov, <a href="https://arxiv.org/abs/2011.14758">Two-dimensional topological theories, rational functions and their tensor envelopes</a>, arXiv:2011.14758 [math.QA], 2020.

%H Toufik Mansour and Mark Shattuck, <a href="https://www.emis.de/journals/INTEGERS/papers/l67/l67.Abstract.html">A recurrence related to the Bell Numbers</a>, Integers 11 (2011), #A67.

%F a(n) = (1/2)*A001861(n), n > 0.

%F E.g.f.: (1 + exp(2*exp(x)-2))/2. - _Emeric Deutsch_, Feb 09, 2002

%F a(n+1) = 1 + 2*Sum_{j=1..n} binomial(n, j)*a(j). - _Jon Perry_, Apr 25 2005

%F 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

%F 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

%F 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

%F G.f.: 1 + Sum_{k>=1} 2^(k-1)*x^k/Product_{j=1..k} (1 - j*x). - _Ilya Gutkovskiy_, Jun 19 2018

%e Given the production matrix M, upper left term of M^5 = a(5)/2^4 = 227/16.

%p A035009 := proc(n) local a,b,i;

%p a := [seq(2,i=1..n-1)]; b := [seq(1,i=1..n-1)];

%p exp(-x)*hypergeom(a,b,x); round(evalf(subs(x=2,%), 10+2*n)) end:

%p seq(A035009(n),n=0..19); # _Peter Luschny_, Mar 30 2011

%p # second Maple program:

%p b:= proc(n, m) option remember;

%p `if`(n=0, ceil(2^(m-1)), m*b(n-1, m)+b(n-1, m+1))

%p end:

%p a:= n-> b(n, 0):

%p seq(a(n), n=0..25); # _Alois P. Heinz_, Aug 03 2021

%t 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 *)

%t Join[{1}, Table[BellB[n, 2]/2, {n, 1, 25}]] (* _Vaclav Kotesovec_, Jun 26 2022 *)

%o (PARI) x='x+O('x^99); Vec(serlaplace((1 + exp(2*exp(x)-2))/2)) \\ _Joerg Arndt_, Apr 01 2011

%Y Cf. A000110, A011782, A137597, A222391, A129340.

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_