E.g.f.: exp(exp(x) - 1).
Recurrence: a(n+1) = Sum_{k=0..n} a(k)*binomial(n, k).
a(n) = Sum_{k=0..n} Stirling2(n, k).
G.f.: (Sum_{k>=0} 1/((1-k*x)*k!))/exp(1) = hypergeom([-1/x], [(x-1)/x], 1)/exp(1) = ((1-2*x)+LaguerreL(1/x, (x-1)/x, 1)+x*LaguerreL(1/x, (2*x-1)/x, 1))*Pi/(x^2*sin(Pi*(2*x-1)/x)), where LaguerreL(mu, nu, z) = (gamma(mu+nu+1)/(gamma(mu+1)*gamma(nu+1)))* hypergeom([-mu], [nu+1], z) is the Laguerre function, the analytic extension of the Laguerre polynomials, for mu not equal to a nonnegative integer. This generating function has an infinite number of poles accumulating in the neighborhood of x=0. -
Karol A. Penson, Mar 25 2002
a(n) = exp(-1)*Sum_{k >= 0} k^n/k! [Dobinski]. -
Benoit Cloitre, May 19 2002
a(n) is asymptotic to n!*(2 Pi r^2 exp(r))^(-1/2) exp(exp(r)-1) / r^n, where r is the positive root of r exp(r) = n. See, e.g., the Odlyzko reference.
a(n) is asymptotic to b^n*exp(b-n-1/2)*sqrt(b/(b+n)) where b satisfies b*log(b) = n - 1/2 (see Graham, Knuth and Patashnik, Concrete Mathematics, 2nd ed., p. 493). -
Benoit Cloitre, Oct 23 2002, corrected by
Vaclav Kotesovec, Jan 06 2013
Lovasz (Combinatorial Problems and Exercises, North-Holland, 1993, Section 1.14, Problem 9) gives another asymptotic formula, quoted by Mezo and Baricz. -
N. J. A. Sloane, Mar 26 2015
G.f.: Sum_{k>=0} x^k/(Product_{j=1..k} (1-j*x)) (see Klazar for a proof). -
Ralf Stephan, Apr 18 2004
For n>0, a(n) = Aitken(n-1, n-1) [i.e., a(n-1, n-1) of Aitken's array (
A011971)]. -
Gerald McGarvey, Jun 26 2004
a(n) = Sum_{k=1..n} (1/k!)*(Sum_{i=1..k} (-1)^(k-i)*binomial(k, i)*i^n + 0^n). -
Paul Barry, Apr 18 2005
a(n) = (2*n!/(Pi*e))*Im( Integral_{x=0..Pi} e^(e^(e^(ix))) sin(nx) dx ) where Im denotes imaginary part [Cesaro]. -
David Callan, Sep 03 2005
O.g.f.: 1/(1-x-x^2/(1-2*x-2*x^2/(1-3*x-3*x^2/(.../(1-n*x-n*x^2/(...)))))) (continued fraction due to Ph. Flajolet). -
Paul D. Hanna, Jan 17 2006
Representation of Bell numbers B(n), n=1,2,..., as special values of hypergeometric function of type (n-1)F(n-1), in Maple notation: B(n)=exp(-1)*hypergeom([2,2,...,2],[1,1,...,1],1), n=1,2,..., i.e., having n-1 parameters all equal to 2 in the numerator, having n-1 parameters all equal to 1 in the denominator and the value of the argument equal to 1.
Examples:
B(1)=exp(-1)*hypergeom([],[],1)=1
B(2)=exp(-1)*hypergeom([2],[1],1)=2
B(3)=exp(-1)*hypergeom([2,2],[1,1],1)=5
B(4)=exp(-1)*hypergeom([2,2,2],[1,1,1],1)=15
B(5)=exp(-1)*hypergeom([2,2,2,2],[1,1,1,1],1)=52
(Warning: this formula is correct but its application by a computer may not yield exact results, especially with a large number of parameters.)
(End)
a(n+1) = 1 + Sum_{k=0..n-1} Sum_{i=0..k} binomial(k,i)*(2^(k-i))*a(i). -
Yalcin Aktar, Feb 27 2007
a(n) = [1,0,0,...,0] T^(n-1) [1,1,1,...,1]', where T is the n X n matrix with main diagonal {1,2,3,...,n}, 1's on the diagonal immediately above and 0's elsewhere. [Meier]
a(n) = ((2*n!)/(Pi * e)) * ImaginaryPart(Integral[from 0 to Pi](e^e^e^(i*theta))*sin(n*theta) dtheta). -
Jonathan Vos Post, Aug 27 2007
a(n) = T(n,1) = Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * Lag(n,-1,j-n) = Sum_{j=0..n} [ E(n,j)/n! ] * [ n!*Lag(n,-1, j-n) ] where T(n,x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m. Note that E(n,j)/n! = E(n,j) / (Sum_{k=0..n} E(n,k)).
The Eulerian numbers count the permutation ascents and the expression [n!*Lag(n,-1, j-n)] is
A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to n!*a(n) = Sum_{j=0..n} E(n,j) * [n!*Lag(n,-1, j-n)].
(End)
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 for Bell numbers B_n we have B_n=1/e*f_n(1). -
Milan Janjic, May 30 2008
a(n) = (n-1)! Sum_{k=1..n} a(n-k)/((n-k)! (k-1)!) where a(0)=1. -
Thomas Wieder, Sep 09 2008
a(n+k) = Sum_{m=0..n} Stirling2(n,m) Sum_{r=0..k} binomial(k,r) m^r a(k-r). - David Pasino (davepasino(AT)yahoo.com), Jan 25 2009. (Umbrally, this may be written as a(n+k) = Sum_{m=0..n} Stirling2(n,m) (a+m)^k. -
N. J. A. Sloane, Feb 07 2009)
Sum_{k=1..n-1} a(n)*binomial(n,k) = Sum_{j=1..n}(j+1)*Stirling2(n,j+1). - [Zhao] -
R. J. Mathar, Jun 24 2024
a(n) = Sum_{k_1=0..n+1} Sum_{k_2=0..n}...Sum_{k_i=0..n-i}...Sum_{k_n=0..1}
delta(k_1,k_2,...,k_i,...,k_n)
where delta(k_1,k_2,...,k_i,...,k_n) = 0 if any k_i > k_(i+1) and k_(i+1) <> 0
and delta(k_1,k_2,...,k_i,...,k_n) = 1 otherwise.
(End)
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)=det(A). -
Milan Janjic, Jul 08 2010
G.f.: 1 / (1 - x / (1 - 1*x / (1 - x / (1 - 2*x / (1 - x / (1 - 3*x / ... )))))). -
Michael Somos, May 12 2012
a(n+1) = Sum_{m=0..n} Stirling2(n, m)*(m+1), n >= 0. Compare with the third formula for a(n) above. Here Stirling2 =
A048993. -
Wolfdieter Lang, Feb 03 2015
G.f.: (-1)^(1/x)*((-1/x)!/e + (!(-1-1/x))/x) where z! and !z are factorial and subfactorial generalized to complex arguments. -
Vladimir Reshetnikov, Apr 24 2013
The following formulas were proposed during the period Dec 2011 - Oct 2013 by
Sergei N. Gladkovskii: (Start)
E.g.f.: exp(exp(x)-1) = 1 + x/(G(0)-x); G(k) = (k+1)*Bell(k) + x*Bell(k+1) - x*(k+1)*Bell(k)*Bell(k+2)/G(k+1) (continued fraction).
G.f.: W(x) = (1-1/(G(0)+1))/exp(1); G(k) = x*k^2 + (3*x-1)*k - 2 + x - (k+1)*(x*k+x-1)^2/G(k+1); (continued fraction Euler's kind, 1-step).
G.f.: W(x) = (1 + G(0)/(x^2-3*x+2))/exp(1); G(k) = 1 - (x*k+x-1)/( ((k+1)!) - (((k+1)!)^2)*(1-x-k*x+(k+1)!)/( ((k+1)!)*(1-x-k*x+(k+1)!) - (x*k+2*x-1)*(1-2*x-k*x+(k+2)!)/G(k+1))); (continued fraction).
G.f.: A(x) = 1/(1 - x/(1-x/(1 + x/G(0)))); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1); (continued fraction, 1-step).
G.f.: -1/U(0) where U(k) = x*k - 1 + x - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + x/U(0) where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + 1/(U(0) - x) where U(k) = 1 + x - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1 + x/(U(0)-x) where U(k) = 1 - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1/G(0) where G(k) = 1 - x/(1 - x*(2*k+1)/(1 - x/(1 - x*(2*k+2)/G(k+1) ))); (continued fraction).
G.f.: G(0)/(1+x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k-1) - x*(2*k+1)*(2*k+3)*(2*x*k-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+x-1)/G(k+1) )); (continued fraction).
G.f.: -(1+2*x) * Sum_{k >= 0} x^(2*k)*(4*x*k^2-2*k-2*x-1) / ((2*k+1) * (2*x*k-1)) * A(k) / B(k) where A(k) = Product_{p=0..k} (2*p+1), B(k) = Product_{p=0..k} (2*p-1) * (2*x*p-x-1) * (2*x*p-2*x-1).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: 1 + x*(S-1) where S = Sum_{k>=0} ( 1 + (1-x)/(1-x-x*k) )*(x/(1-x))^k/Product_{i=0..k-1} (1-x-x*i)/(1-x).
G.f.: (G(0) - 2)/(2*x-1) where G(k) = 2 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: -G(0) where G(k) = 1 - (x*k - 2)/(x*k - 1 - x*(x*k - 1)/(x + (x*k - 2)/G(k+1) )); (continued fraction).
G.f.: G(0) where G(k) = 2 - (2*x*k - 1)/(x*k - 1 - x*(x*k - 1)/(x + (2*x*k - 1)/G(k+1) )); (continued fraction).
G.f.: (G(0) - 1)/(1+x) where G(k) = 1 + 1/(1-k*x)/(1-x/(x+1/G(k+1) )); (continued fraction).
G.f.: 1/(x*(1-x)*G(0)) - 1/x where G(k) = 1 - x/(x - 1/(1 + 1/(x*k-1)/G(k+1) )); (continued fraction).
G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 + x/( x*k - 1 )/Q(k+1); (continued fraction).
G.f.: 1+x/Q(0), where Q(k) = 1 - x - x/(1 - x*(k+1)/Q(k+1)); (continued fraction).
G.f.: 1/(1-x*Q(0)), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))); (continued fraction).
G.f.: Q(0)/(1-x), where Q(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x*(k+1))*(1-x*(k+2))/Q(k+1) ); (continued fraction).
(End)
a(n) ~ exp(exp(W(n))-n-1)*n^n/W(n)^(n+1/2), where W(x) is the Lambert W-function. -
Vladimir Reshetnikov, Nov 01 2015
a(n) ~ n^n * exp(n/W(n)-1-n) / (sqrt(1+W(n)) * W(n)^n). -
Vaclav Kotesovec, Nov 13 2015
By taking logarithmic derivatives of the equivalent to Kotesovec's asymptotic for Bell polynomials at x=1, we obtain properties of the nth row of
A008277 as a statistical distribution (where W=W(n),X=W(n)+1)
a(n+1)/a(n) ~ n/W + W/(2*(W+1)^2) is 1 more than the expectation.
(2*a(n+1)+a(n+2))/a(n) - (a(n+1)/a(n))^2 - a(n+2)/a(n+1) ~ n/(W*X)+1/(2*X^2)-3/(2*X^3)+1/X^4 is 1 more than the variance.
(This is a complete asymptotic characterisation, since they converge to normal distributions; see Harper, 1967)
(End)
a(n) are the coefficients in the asymptotic expansion of -exp(-1)*(-1)^x*x*Gamma(-x,0,-1), where Gamma(a,z0,z1) is the generalized incomplete Gamma function. -
Vladimir Reshetnikov, Nov 12 2015
a(p^m) == m+1 (mod p) when p is prime and m >= 1 (see Lemma 3.1 in the Hurst/Schultz reference). -
Seiichi Manyama, Jun 01 2016
a(n) = Sum_{k=0..n} hypergeom([1, -k], [], 1)*Stirling2(n+1, k+1) = Sum_{k=0..n}
A182386(k)*Stirling2(n+1, k+1). -
Mélika Tebni, Jul 02 2022
a(n) is the n-th derivative of e^e^x divided by e at point x=0. -
Joan Ludevid, Nov 05 2024
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