a(n-1) = Sum_{k=1..n} k*Stirling2(n, k) for n>=1.
E.g.f.: exp(exp(x) + 2*x - 1). First differences of Bell numbers (if offset 1). -
Michael Somos, Oct 09 2002
G.f.: Sum_{k>=0} (x^k/Product_{l=1..k} (1-(l+1)x)). -
Ralf Stephan, Apr 18 2004
a(n) = Sum_{i=0..n} 2^(n-i)*B(i)*binomial(n,i) where B(n) = Bell numbers
A000110(n). -
Fred Lunnon, Aug 04 2007 [Written umbrally, a(n) = (B+2)^n. -
N. J. A. Sloane, Feb 07 2009]
Representation as an infinite series: a(n-1) = Sum_{k>=2} (k^n*(k-1)/k!)/exp(1), n=1, 2, ... This is a Dobinski-type summation formula. -
Karol A. Penson, Mar 14 2002
Recurrence: a(n+1) = 1 + Sum_{j=1..n} (1+binomial(n, j))*a(j). -
Jon Perry, Apr 25 2005
Binomial transform of Bell numbers 1, 2, 5, 15, 52, 203, 877, 4140, ... (see
A000110).
Define f_1(x), f_2(x), ... such that f_1(x)=x*e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n=2,3,.... Then a(n-1) = e^(-1)*f_n(1). -
Milan Janjic, May 30 2008
Representation of numbers a(n), n=0,1..., as special values of hypergeometric function of type (n)F(n), in Maple notation: a(n)=exp(-1)*2^n*hypergeom([3,3...3],[2.2...2],1), n=0,1..., i.e., having n parameters all equal to 3 in the numerator, having n parameters all equal to 2 in the denominator and the value of the argument equal to 1. Examples: a(0)= 2^0*evalf(hypergeom([],[],1)/exp(1))=1 a(1)= 2^1*evalf(hypergeom([3],[2],1)/exp(1))=3 a(2)= 2^2*evalf(hypergeom([3,3],[2,2],1)/exp(1))=10 a(3)= 2^3*evalf(hypergeom([3,3,3],[2,2,2],1)/exp(1))=37 a(4)= 2^4*evalf(hypergeom([3,3,3,3],[2,2,2,2],1)/exp(1))=151 a(5)= 2^5*evalf(hypergeom([3,3,3,3,3],[2,2,2,2,2],1)/exp(1)) = 674. -
Karol A. Penson, Sep 28 2007
Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]=binomial(j-1,i-1), (i <= j), and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = (-1)^(n)charpoly(A,-2). -
Milan Janjic, Jul 08 2010
Continued fractions:
G.f.: 1/U(0) where U(k) = 1 - x*(k+3) - x^2*(k+1)/U(k+1).
G.f.: 1/(U(0)-x) where U(k) = 1 - x - x*(k+1)/(1 - x/U(k+1)).
G.f.: G(0)/(1-x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k+2*x-1) - x*(2*k+1)*(2*k+3)*(2*x*k+2*x-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+3*x-1)/G(k+1) )).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-2*x-k*x)/(1-x/(x-1/G(k+1) )).
G.f.: -G(0)/x where G(k) = 1 - 1/(1-k*x-x)/(1-x/(x-1/G(k+1) )).
G.f.: 1 - 2/x + (1/x-1)*S where S = sum(k>=0, ( 1 + (1-x)/(1-x-x*k) )*(x/(1-x))^k / prod(i=0..k-1, (1-x-x*i)/(1-x) ) ).
G.f.: (1-x)/x/(G(0)-x) - 1/x where G(k) = 1 - x*(k+1)/(1 - x/G(k+1) ).
G.f.: (1/G(0) - 1)/x^3 where G(k) = 1 - x/(x - 1/(1 + 1/(x*k-1)/G(k+1) )).
G.f.: 1/Q(0), where Q(k)= 1 - 2*x - x/(1 - x*(k+1)/Q(k+1)).
G.f.: G(0)/(1-3*x), where G(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1 - x*(k+3))*(1 - x*(k+4))/G(k+1) ). (End)
a(n) ~ exp(n/LambertW(n) + 3*LambertW(n)/2 - n - 1) * n^(n + 1/2) / LambertW(n)^(n+1). -
Vaclav Kotesovec, Jun 09 2020
a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=0..n-1} binomial(n-1,k) * a(k). -
Ilya Gutkovskiy, Jul 02 2020
a(n) ~ n^2 * Bell(n) / LambertW(n)^2 * (1 - LambertW(n)/n). -
Vaclav Kotesovec, Jul 28 2021
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