Let P(n,X) = (X+1)*(X+2)*(X+3)*...*(X+n); then a(n) is the coefficient of X; or a(n) = P'(n,0). -
Benoit Cloitre, May 09 2002
Sum_{k > 0} a(k) * x^k/ k!^2 = exp(x) *(Sum_{k>0} (-1)^(k+1) * x^k / (k * k!)). -
Michael Somos, Mar 24 2004; corrected by
Warren D. Smith, Feb 12 2006
a(n) is the coefficient of x^(n+2) in (-log(1-x))^2, multiplied by (n+2)!/2.
a(n) = n! * Sum_{i=1..n} 1/i = n! * H(n), where H(n) =
A001008(n)/
A002805(n) is the n-th harmonic number.
a(n) ~ 2^(1/2)*Pi^(1/2)*log(n)*n^(1/2)*e^-n*n^n. - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
E.g.f.: log(1 - x) / (x-1). (= (log(1 - x))^2 / 2 if offset 1). -
Michael Somos, Feb 05 2004
D-finite with recurrence: a(n) = a(n-1) * (2*n - 1) - a(n-2) * (n - 1)^2, if n > 1. -
Michael Somos, Mar 24 2004
a(n) = n!*Sum_{k=1..n} (-1)^(k+1)*binomial(n, k)/k. -
Vladeta Jovovic, Jan 29 2005
p^2 divides a(p-1) for prime p > 3. a(n) = (Sum_{i=1..n} 1/i) / Product_{i=1..n} 1/i. -
Alexander Adamchuk, Jul 11 2006
a(n) = 3*
A001710(n) + 2*
A001711(n-3) for n > 2; e.g., 11 = 3*3 + 2*1, 50 = 3*12 + 2*7, 274 = 3*60 + 2*47, ... -
Gary Detlefs, May 24 2010
a(n+1) = Sum_{i=1..floor((n-1)/2)} n!/((n-i)*i) + Sum_{i=ceiling(n/2)..floor(n/2)} n!/(2*(n-i)*i). -
Shanzhen Gao, Sep 14 2010
a(n) = (a(n-1)*(n^2 - 2*n + 1) + (n + 1)!)/(n - 1) for n > 2.
It appears that, with the exception of n = 2, (a(n+1)^2 - a(n)^2) mod n^2 = 0 if n is composite and 4*n if n is prime.
It appears that, with the exception of n = 2, (a(n+1)^3 - a(n)^2) mod n = 0 if n is composite and n - 2 if n is prime.
It appears that, with the exception of n = 2, (a(n)^2 + a(n+1)^2) mod n = 0 if n is composite and = 2 if n is prime. (End)
a(n) = Integral_{x=0..oo} (x^n - n!)*log(x)*exp(-x) dx. -
Groux Roland, Mar 28 2011
a(n) = 3*n!/2 + 2*(n-2)!*Sum_{k=0..n-3} binomial(k+2,2)/(n-2-k) for n >= 2. -
Gary Detlefs, Sep 02 2011
a(n)/(n-1)! = ml(n) = n*ml(n-1)/(n-1) + 1 for n > 1, where ml(n) is the average number of random draws from an n-set with replacement until the total set has been observed. G.f. of ml: x*(1 - log(1 - x))/(1 - x)^2. -
Paul Weisenhorn, Nov 18 2011
a(n) = det(|S(i+2, j+1)|, 1 <= i,j <= n-2), where S(n,k) are Stirling numbers of the second kind. -
Mircea Merca, Apr 06 2013
0 = a(n) * (a(n+4) - 6*a(n+3) + 7*a(n+2) - a(n+1)) - a(n+1) * (4*a(n+3) - 6*a(n+2) + a(n+1)) + 3*a(n+2)^2 unless n=0. -
Michael Somos, Nov 28 2013
For a simple way to calculate the sequence, multiply n! by the integral from 0 to 1 of (1 - x^n)/(1 - x) dx. -
Rahul Jha, Feb 18 2015
Inverse binomial transform of
A073596.
a(n) ~ sqrt(2*Pi*n) * n^n * (log(n) + gamma)/exp(n), where gamma is the Euler-Mascheroni constant
A001620. (End)
a(n) = ((-1)^(n+1)/2*(n+1))*Sum_{k=1..n} k*Bernoulli(k-1)*Stirling1(n,k). -
Vladimir Kruchinin, Nov 20 2016
a(n) = (n)! * (digamma(n+1) + gamma), where gamma is the Euler-Mascheroni constant
A001620. -
Pedro Caceres, Mar 10 2018
Gamma'(x) = a(x-1) - (x-1)!*gamma, where Gamma'(x) is the derivative of the gamma function at positive integers and gamma is the Euler-Mascheroni constant. E.g.:
Gamma'(1) = -gamma, Gamma'(2) = 1-gamma, Gamma'(3) = 3-2*gamma,
Gamma'(22) = 186244810780170240000 - 51090942171709440000*gamma. (End)
The following are all conjectural:
E.g.f.: for nonzero m, (1/m)*Sum_{n >= 1} (-1)^(n+1)*(1/n)*binomial(m*n,n)* x^n/(1 - x)^(m*n+1) = x + 3*x^2/2! + 11*x^3/3! + 50*x^4/4! + ....
For nonzero m, a(n) = (1/m)*n!*Sum_{k = 1..n} (-1)^(k+1)*(1/k)*binomial(m*k,k)* binomial(n+(m-1)*k,n-k).
a(n)^2 = (1/2)*n!^2*Sum_{k = 1..n} (-1)^(k+1)*(1/k^2)*binomial(n,k)* binomial(n+k,k). (End)
a(n) = -Sum_{k=0..n} k!*
A021009(n, k+1).
a(n) = Sum_{k=0..n} k!*
A094587(n, k+1). (End)
a(n) = n! * 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n - 1)^2/((2*n - 1)))))). -
Peter Bala, Mar 16 2024
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