Sum_{i=0..n} (-1)^i * i^n * binomial(n, i) = (-1)^n * n!. - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
Sum_{i=0..n} (-1)^i * (n-i)^n * binomial(n, i) = n!. - Peter C. Heinig (algorithms(AT)gmx.de), Apr 10 2007
The sequence trivially satisfies the recurrence a(n+1) = Sum_{k=0..n} binomial(n,k) * a(k)*a(n-k). -
Robert FERREOL, Dec 05 2009
D-finite with recurrence: a(n) = n*a(n-1), n >= 1. n! ~ sqrt(2*Pi) * n^(n+1/2) / e^n (Stirling's approximation).
a(0) = 1, a(n) = subs(x = 1, (d^n/dx^n)(1/(2-x))), n = 1, 2, ... -
Karol A. Penson, Nov 12 2001
a(n) = Sum_{k=0..n} (-1)^(n-k)*
A000522(k)*binomial(n, k) = Sum_{k=0..n} (-1)^(n-k)*(x+k)^n*binomial(n, k). -
Philippe Deléham, Jul 08 2004
a(n) = Sum_{i=1..n} ((-1)^(i-1) * sum of 1..n taken n - i at a time) - e.g., 4! = (1*2*3 + 1*2*4 + 1*3*4 + 2*3*4) - (1*2 + 1*3 + 1*4 + 2*3 + 2*4 + 3*4) + (1 + 2 + 3 + 4) - 1 = (6 + 8 + 12 + 24) - (2 + 3 + 4 + 6 + 8 + 12) + 10 - 1 = 50 - 35 + 10 - 1 = 24. -
Jon Perry, Nov 14 2005
a(n) = (n-1)*(a(n-1) + a(n-2)), n >= 2. - Matthew J. White, Feb 21 2006
1 / a(n) = determinant of matrix whose (i,j) entry is (i+j)!/(i!(j+1)!) for n > 0. This is a matrix with Catalan numbers on the diagonal. -
Alexander Adamchuk, Jul 04 2006
For n >= 2, a(n-2) = (-1)^n*Sum_{j=0..n-1} (j+1)*Stirling1(n,j+1). -
Milan Janjic, Dec 14 2008
G.f.: 1/(1-x-x^2/(1-3x-4x^2/(1-5x-9x^2/(1-7x-16x^2/(1-9x-25x^2... (continued fraction), hence Hankel transform is
A055209.
G.f. of (n+1)! is 1/(1-2x-2x^2/(1-4x-6x^2/(1-6x-12x^2/(1-8x-20x^2... (continued fraction), hence Hankel transform is
A059332. (End)
a(n) = Product_{p prime} p^(Sum_{k > 0} floor(n/p^k)) by Legendre's formula for the highest power of a prime dividing n!. -
Jonathan Sondow, Jul 24 2009
It appears that a(n) = (1/0!) + (1/1!)*n + (3/2!)*n*(n-1) + (11/3!)*n*(n-1)*(n-2) + ... + (b(n)/n!)*n*(n-1)*...*2*1, where a(n) = (n+1)! and b(n) =
A000255. -
Timothy Hopper, Aug 12 2009
a(n) = A_{n}(1) where A_{n}(x) are the Eulerian polynomials. -
Peter Luschny, Aug 03 2010
a(n) = n*(2*a(n-1) - (n-1)*a(n-2)), n > 1. -
Gary Detlefs, Sep 16 2010
1/a(n) = -Sum_{k=1..n+1} (-2)^k*(n+k+2)*a(k)/(a(2*k+1)*a(n+1-k)). -
Groux Roland, Dec 08 2010
a(n) = Product_{p prime, p <= n} p^(Sum_{i >= 1} floor(n/p^i)).
The infinitary analog of this formula is: a(n) = Product_{q terms of
A050376 <= n} q^((n)_q), where (n)_q denotes the number of those numbers <= n for which q is an infinitary divisor (for the definition see comment in
A037445). (End)
The terms are the denominators of the expansion of sinh(x) + cosh(x). -
Arkadiusz Wesolowski, Feb 03 2012
G.f.: 1 / (1 - x / (1 - x / (1 - 2*x / (1 - 2*x / (1 - 3*x / (1 - 3*x / ... )))))). -
Michael Somos, May 12 2012
G.f. 1 + x/(G(0)-x) where G(k) = 1 - (k+1)*x/(1 - x*(k+2)/G(k+1)); (continued fraction, 2-step). -
Sergei N. Gladkovskii, Aug 14 2012
G.f.: W(1,1;-x)/(W(1,1;-x) - x*W(1,2;-x)), where W(a,b,x) = 1 - a*b*x/1! + a*(a+1)*b*(b+1)*x^2/2! - ... + a*(a+1)*...*(a+n-1)*b*(b+1)*...*(b+n-1)*x^n/n! + ...; see [A. N. Khovanskii, p. 141 (10.19)]. -
Sergei N. Gladkovskii, Aug 15 2012
G.f.: A(x) = 1 + x/(G(0) - x) where G(k) = 1 + (k+1)*x - x*(k+2)/G(k+1); (continued fraction).
Let B(x) be the g.f. for
A051296, then A(x) = 2 - 1/B(x). (End)
G.f.: 1 + x*(G(0) - 1)/(x-1) where G(k) = 1 - (2*k+1)/(1-x/(x - 1/(1 - (2*k+2)/(1-x/(x - 1/G(k+1) ))))); (continued fraction). -
Sergei N. Gladkovskii, Jan 15 2013
G.f.: 1 + x*(1 - G(0))/(sqrt(x)-x) where G(k) = 1 - (k+1)*sqrt(x)/(1-sqrt(x)/(sqrt(x)-1/G(k+1) )); (continued fraction). -
Sergei N. Gladkovskii, Jan 25 2013
G.f.: 1 + x/G(0) where G(k) = 1 - x*(k+2)/( 1 - x*(k+1)/G(k+1) ); (continued fraction). -
Sergei N. Gladkovskii, Mar 23 2013
a(n) = det(S(i+1, j), 1 <= i, j <=n ), where S(n,k) are Stirling numbers of the second kind. -
Mircea Merca, Apr 04 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(k+1)/(x*(k+1) + 1/G(k+1))); (continued fraction). -
Sergei N. Gladkovskii, May 24 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 1/(1 - 1/(2*x*(k+1)) + 1/G(k+1))); (continued fraction). -
Sergei N. Gladkovskii, May 29 2013
G.f.: G(0), where G(k) = 1 + x*(2*k+1)/(1 - x*(2*k+2)/(x*(2*k+2) + 1/G(k+1))); (continued fraction). -
Sergei N. Gladkovskii, Jun 07 2013
a(n) = P(n-1, floor(n/2)) * floor(n/2)! * (n - (n-2)*((n+1) mod 2)), where P(n, k) are the k-permutations of n objects, n > 0. -
Wesley Ivan Hurt, Jun 07 2013
G.f.: 1 + x/Q(0), m=+2, where Q(k) = 1 - 2*x*(2*k+1) - m*x^2*(k+1)*(2*k+1)/( 1 - 2*x*(2*k+2) - m*x^2*(k+1)*(2*k+3)/Q(k+1) ); (continued fraction). -
Sergei N. Gladkovskii, Sep 24 2013
a(n) = round(Sum_{k>=1} log(k)^n/k^2), for n>=1, which is related to the n-th derivative of the Riemann zeta function at x=2 as follows: round((-1)^n * zeta^(n)(2)). Also see
A073002. -
Richard R. Forberg, Dec 30 2014
a(n) ~ Sum_{j>=0} j^n/e^j, where e =
A001113. When substituting a generic variable for "e" this infinite sum is related to Eulerian polynomials. See
A008292. This approximation of n! is within 0.4% at n = 2. See
A255169. Accuracy, as a percentage, improves rapidly for larger n. -
Richard R. Forberg, Mar 07 2015
a(n) = Product_{k=1..n} (C(n+1, 2)-C(k, 2))/(2*k-1); see Masanori Ando link. -
Michel Marcus, Apr 17 2015
Sum_{n>=0} a(n)/(a(n + 1)*a(n + 2)) = Sum_{n>=0} 1/((n + 2)*(n + 1)^2*a(n)) = 2 - exp(1) - gamma + Ei(1) = 0.5996203229953..., where gamma =
A001620, Ei(1) =
A091725. -
Ilya Gutkovskiy, Nov 01 2016
a(2^n) = 2^(2^n - 1) * 1!! * 3!! * 7!! * ... * (2^n - 1)!!. For example, 16! = 2^15*(1*3)*(1*3*5*7)*(1*3*5*7*9*11*13*15) = 20922789888000. -
Peter Bala, Nov 01 2016
a(n) = sum(prod(B)), where the sum is over all subsets B of {1,2,...,n-1} and where prod(B) denotes the product of all the elements of set B. If B is a singleton set with element b, then we define prod(B)=b, and, if B is the empty set, we define prod(B) to be 1. For example, a(4)=(1*2*3)+(1*2)+(1*3)+(2*3)+(1)+(2)+(3)+1=24. -
Dennis P. Walsh, Oct 23 2017
Sum_{n >= 0} 1/(a(n)*(n+2)) = 1. - Multiplying the denominator by (n+2) in Jaume Oliver Lafont's entry above creates a telescoping sum. -
Fred Daniel Kline, Nov 08 2020
O.g.f.: Sum_{k >= 0} k!*x^k = Sum_{k >= 0} (k+y)^k*x^k/(1 + (k+y)*x)^(k+1) for arbitrary y. -
Peter Bala, Mar 21 2022
E.g.f.: 1/(1 + LambertW(-x*exp(-x))) = 1/(1-x), see
A258773. -(1/x)*substitute(z = x*exp(-x), z*(d/dz)LambertW(-z)) = 1/(1 - x). See
A075513. Proof: Use the compositional inverse (x*exp(-x))^[-1] = -LambertW(-z). See
A000169 or
A152917, and Richard P. Stanley: Enumerative Combinatorics, vol. 2, p. 37, eq. (5.52). -
Wolfdieter Lang, Oct 17 2022
1/a(n) = (e/(2*Pi*n)*Integral_{x=-oo..oo} cos(x-n*arctan(x))/(1+x^2)^(n/2) dx). Proof: take the real component of Laplace's integral for 1/Gamma(x).
a(n) = Integral_{x=0..1} e^(-t)*LerchPhi(1/e, -n, t) dt. Proof: use the relationship Gamma(x+1) = Sum_{n >= 0} Integral_{t=n..n+1} e^(-t)t^x dt = Sum_{n >= 0} Integral_{t=0..1} e^(-(t+n))(t+n)^x dt and interchange the order of summation and integration.
Conjecture: a(n) = 1/(2*Pi)*Integral_{x=-oo..oo}(n+i*x+1)!/(i*x+1)-(n+i*x-1)!/(i*x-1)dx. (End)
a(n) = e^(Integral_{x=1..n+1} Psi(x) dx) where Psi(x) is the digamma function. -
Andrea Pinos, Jan 10 2024
a(n) = Integral_{x=0..oo} e^(-x^(1/n)) dx, for n > 0. -
Ridouane Oudra, Apr 20 2024
O.g.f.: N(x) = hypergeometric([1,1], [], x) = LaplaceTransform(x/(1-x))/x, satisfying x^2*N'(x) + (x-1)*N(x) + 1 = 0, with N(0) = 1. -
Wolfdieter Lang, May 31 2025
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