OFFSET
0,3
COMMENTS
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
Row sums of triangle A137597 starting (1, 3, 11, 47, 227, ...). - Gary W. Adamson, Jan 29 2008
From Gary W. Adamson, Jul 22 2011: (Start)
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:
1, 1/2, 0, 0, 0, 0, ...
1, 1, 1/2, 0, 0, 0, ...
1, 2, 1, 1/2, 0, 0, ...
1, 3, 3, 1, 1/2, 0, ...
1, 4, 6, 4, 1, 1/2, ..., etc.
(End)
From Bruno Berselli, Mar 20 2013: (Start)
Note that, for t=A222391:
a(1)*t = Sum_{n >= 1} 1 /(Gamma(n/2)*Gamma((n+1)/2)),
a(2)*t = Sum_{n >= 1} n /(Gamma(n/2)*Gamma((n+1)/2)),
a(3)*t = Sum_{n >= 1} n^2/(Gamma(n/2)*Gamma((n+1)/2)),
a(4)*t = Sum_{n >= 1} n^3/(Gamma(n/2)*Gamma((n+1)/2)),
a(5)*t = Sum_{n >= 1} n^4/(Gamma(n/2)*Gamma((n+1)/2)),
a(6)*t = Sum_{n >= 1} n^5/(Gamma(n/2)*Gamma((n+1)/2)), etc.
(End)
Except for the initial term, the main diagonal of A129340. - Peter Bala, Apr 14 2017
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.
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
EXAMPLE
Given the production matrix M, upper left term of M^5 = a(5)/2^4 = 227/16.
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
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 *)
PROG
(PARI) x='x+O('x^99); Vec(serlaplace((1 + exp(2*exp(x)-2))/2)) \\ Joerg Arndt, Apr 01 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved