a(n) = n*a(n-1) + 1, a(0) = 1.
a(n) = n! * Sum_{k=0..n} 1/k! = n! * (e - Sum_{k>=n+1} 1/k!). -
Michael Somos, Mar 26 1999
a(0) = 1; for n > 0, a(n) = floor(e*n!).
E.g.f.: exp(x)/(1-x).
Integral representation as n-th moment of a nonnegative function on a positive half-axis: a(n) = e*Integral_{x>=0} x^n*e^(-x)*Heaviside(x-1) dx. -
Karol A. Penson, Oct 01 2001
Formula, in Mathematica notation: Special values of Laguerre polynomials, a(n)=(-1)^n*n!*LaguerreL[n, -1-n, 1], n=1, 2, ... . This relation cannot be checked by Maple, as it appears that Maple does not incorporate Laguerre polynomials with second index equal to negative integers. It does check with Mathematica. -
Karol A. Penson and Pawel Blasiak ( blasiak(AT)lptl.jussieu.fr), Feb 13 2004
G.f.: Sum_{k>=0} k!*x^k/(1-x)^(k+1). a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*k^(n-k)*(k+1)^k. -
Vladeta Jovovic, Aug 18 2002
a(n) = e*Gamma(n+1,1) where Gamma(z,t) = Integral_{x>=t} e^(-x)*x^(z-1) dx is incomplete gamma function. -
Michael Somos, Jul 01 2004
a(n) = 1 + n + n*(n-1) + n*(n-1)*(n-2) + ... + n!. -
Jonathan Sondow, Aug 18 2006
a(n) = Sum_{k=0..n} binomial(n,k) * k!; interpretation: for all k-subsets (sum), choose a subset (binomial(n,k)), and permutation of subset (k!). -
Joerg Arndt, Dec 09 2012
a(n) = Sum_{k=0..n}
A094816(n, k), n>=0 (row sums of Poisson-Charlier coefficient matrix). -
N. J. A. Sloane, Nov 10 2007
Act on 1/(1-x) with 1/(1-xDx) = Sum_{j>=0} (xDx)^j = Sum_{j>=0} x^j*D^j*x^j = Sum_{j>=0} j!*x^j*L(j,-:xD:,0) where Lag(n,x,0) are the Laguerre polynomials of order 0, D the derivative w.r.t. x and (:xD:)^j = x^j*D^j. Truncating the operator series at the j = n term gives an o.g.f. for a(0) through a(n) consistent with Jovovic's.
These results and those of Penson and Blasiak, Arnold, Bottomley and Deleham are related by the operator
A094587 (the reverse of
A008279), which is the umbral equivalent of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j)*j!*x^(n-j) = Sum_{j=0..n} (n!/j!)*x^j. Umbral substitution of b(.) for x and then letting b(n)=1 for all n connects the results. See
A132013 (the inverse of
A094587) for a connection between these operations and 1/(1-xDx).
(End)
a(n) = n!*e - 1/(n + 1/(n+1 + 2/(n+2 + 3/(n+3 + ...)))).
Asymptotic result (Ramanujan): n!*e - a(n) ~ 1/n - 1/n^3 + 1/n^4 + 2/n^5 - 9/n^6 + ..., where the sequence [1,0,-1,1,2,-9,...] = [(-1)^k*
A000587(k)], for k>=1.
a(n) is a difference divisibility sequence, that is, the difference a(n) - a(m) is divisible by n - m for all n and m (provided n is not equal to m). For fixed k, define the derived sequence a_k(n) = (a(n+k)-a(k))/n, n = 1,2,3,... . Then a_k(n) is also a difference divisibility sequence.
For example, the derived sequence a_0(n) is just a(n-1). The set of integer sequences satisfying the difference divisibility property forms a ring with term-wise operations of addition and multiplication.
Recurrence relations: a(0) = 1, a(n) = (n-1)*(a(n-1) + a(n-2)) + 2, for n >= 1. a(0) = 1, a(1) = 2, D-finite with recurrence: a(n) = (n+1)*a(n-1) - (n-1)*a(n-2) for n >= 2. The sequence b(n) := n! satisfies the latter recurrence with the initial conditions b(0) = 1, b(1) = 1. This leads to the finite continued fraction expansion a(n)/n! = 1/(1-1/(2-1/(3-2/(4-...-(n-1)/(n+1))))), n >= 2.
Limit_{n->oo} a(n)/n! = e = 1/(1-1/(2-1/(3-2/(4-...-n/((n+2)-...))))). This is the particular case m = 0 of the general result m!/e - d_m = (-1)^(m+1) *(1/(m+2 -1/(m+3 -2/(m+4 -3/(m+5 -...))))), where d_m denotes the m-th derangement number
A000166(m).
For sequences satisfying the more general recurrence a(n) = (n+1+r)*a(n-1) - (n-1)*a(n-2), which yield series acceleration formulas for e/r! that involve the Poisson-Charlier polynomials c_r(-n;-1), refer to
A001339 (r=1),
A082030 (r=2),
A095000 (r=3) and
A095177 (r=4).
For the corresponding results for the constants log(2), zeta(2) and zeta(3) refer to
A142992,
A108625 and
A143007 respectively.
(End)
G.f. satisfies: A(x) = 1/(1-x)^2 + x^2*A'(x)/(1-x). -
Paul D. Hanna, Sep 03 2008
G.f.: 1/(1-2*x-x^2/(1-4*x-4*x^2/(1-6*x-9*x^2/(1-8*x-16*x^2/(1-10*x-25*x^2/(1-... (continued fraction);
G.f.: 1/(1-x-x/(1-x/(1-x-2*x/(1-2*x/(1-x-3*x/(1-3*x/(1-x-4*x/(1-4*x/(1-x-5*x/(1-5*x/(1-... (continued fraction).
(End)
O.g.f.: Sum_{n>=0} (n+2)^n*x^n/(1 + (n+1)*x)^(n+1). -
Paul D. Hanna, Sep 19 2011
G.f. hypergeom([1,k],[],x/(1-x))/(1-x), for k=1,2,...,9 is the generating function for
A000522,
A001339,
A082030,
A095000,
A095177,
A096307,
A096341,
A095722, and
A095740. -
Mark van Hoeij, Nov 07 2011
G.f.: 1/U(0) where U(k) = 1 - x - x*(k+1)/(1 - x*(k+1)/U(k+1)); (continued fraction). -
Sergei N. Gladkovskii, Oct 14 2012
E.g.f.: 1/U(0) where U(k) = 1 - x/(1 - 1/(1 + (k+1)/U(k+1))); (continued fraction). -
Sergei N. Gladkovskii, Nov 16 2012
G.f.: 1/(1-x)/Q(0), where Q(k) = 1 - x/(1-x)*(k+1)/(1 - x/(1-x)*(k+1)/Q(k+1)); (continued fraction). -
Sergei N. Gladkovskii, May 19 2013
G.f.: 2/(1-x)/G(0), where G(k) = 1 + 1/(1 - x*(2*k+2)/(x*(2*k+3) - 1 + x*(2*k+2)/G(k+1))); (continued fraction). -
Sergei N. Gladkovskii, May 31 2013
G.f.: (B(x)+ 1)/(2-2*x) = Q(0)/(2-2*x), where B(x) be g.f.
A006183, Q(k) = 1 + 1/(1 - x*(k+1)/(x*(k+1) + (1-x)/Q(k+1) )); (continued fraction). -
Sergei N. Gladkovskii, Aug 08 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(k+1) - x^2*(k+1)^2/Q(k+1); (continued fraction). -
Sergei N. Gladkovskii, Sep 30 2013
E.g.f.: e^x/(1-x) = (1 - 12*x/(Q(0) + 6*x - 3*x^2))/(1-x), where Q(k) = 2*(4*k+1)*(32*k^2 + 16*k + x^2 - 6) - x^4*(4*k-1)*(4*k+7)/Q(k+1); (continued fraction). -
Sergei N. Gladkovskii, Nov 18 2013
G.f.: conjecture: T(0)/(1-2*x), where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 - (1 - 2*x*(k+1))*(1 - 2*x*(k+2))/T(k+1) ); (continued fraction). -
Sergei N. Gladkovskii, Nov 18 2013
0 = a(n)*(+a(n+1) - 3*a(n+2) + a(n+3)) + a(n+1)*(+a(n+1) - a(n+3)) + a(n+2)*(+a(n+2)) for all n>=0. -
Michael Somos, Jul 04 2014
a(n) = F(n), where the function F(x) := Integral_{0..infinity} e^(-u)*(1 + u)^x du smoothly interpolates this sequence to all real values of x. Note that F(-1) = G and for n = 2,3,... we have F(-n) = (-1)^n/(n-1)! *(
A058006(n-2) - G ), where G = 0.5963473623... denotes Gompertz's constant - see
A073003.
a(n) = n!*e - e*( Sum_{k >= 0} (-1)^k/((n + k + 1)*k!) ).
(End)
a(n) = (e/(2*Pi))*Integral_{x=-oo..oo} (n+1+i*x)!/(1+i*x) dx. -
David Ulgenes, Apr 18 2023
Sum_{i=0..n} (-1)^(n-i) * binomial(n, i) * a(i) = n!. -
Werner Schulte, Apr 03 2024
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