login
This site is supported by donations to The OEIS Foundation.

 

Logo

Thanks to everyone who made a donation during our annual appeal!
To see the list of donors, or make a donation, see the OEIS Foundation home page.

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A035009 STIRLING transform of [1,1,2,4,8,16,32,...]. 11
1, 1, 3, 11, 47, 227, 1215, 7107, 44959, 305091, 2206399, 16913987, 136823263, 1163490499, 10366252031, 96491364675, 935976996127, 9440144423875, 98800604237119, 1071092025420867, 12008090971866207, 139014305916844739, 1659578039401022079, 20405708646650507075 (list; graph; refs; listen; history; text; internal format)
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.

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

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 *)

PROG

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

CROSSREFS

Cf. A000110, A137597, A222391, A129340.

Sequence in context: A062146 A216947 A090365 * A051296 A030832 A030865

Adjacent sequences:  A035006 A035007 A035008 * A035010 A035011 A035012

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified January 15 20:47 EST 2019. Contains 319184 sequences. (Running on oeis4.)