A143415 Another sequence of Apery-like numbers for the constant 1/e: a(n) = 1/(n+1)!*Sum_{k = 0..n-1} C(n-1,k)*(2*n-k)!.

0, 1, 5, 41, 481, 7421, 142601, 3288205, 88577021, 2731868921, 94969529101, 3675200329841, 156725471006105, 7302990263511541, 369216917569411601, 20130327811188977621, 1177435382675193700021, 73546210385434763486705

Offset: 0

Author

Peter Bala, Aug 14 2008

Keywords

• easy
• nonn

Comments

This sequence is a modified version of A143414.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..366

Crossrefs

Cf. A143413, A143414.

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

Maple

a := n -> 1/(n+1)!*add (binomial(n-1,k)*(2*n-k)!,k = 0..n-1): seq(a(n),n = 0..19);
# Alternative:
A143415 := n -> `if`(n=0, 0, ((2*n)!/(n+1)!)*hypergeom([1-n], [-2*n], 1)):
seq(simplify(A143415(n)), n = 0..17); # Peter Luschny, May 14 2020

Mathematica

Table[(1/(n+1)!)*Sum[Binomial[n-1,k]*(2*n-k)!, {k,0,n-1}], {n,0,50}] (* G. C. Greubel, Oct 24 2017 *)

PARI

for(n=0,25, print1((1/(n+1)!)*sum(k=0,n-1, binomial(n-1,k)*(2*n-k)!), ", ")) \\ G. C. Greubel, Oct 24 2017

Formula

a(n) = 1/(n+1)!*sum {k = 0..n-1} C(n-1,k)*(2*n-k)!.

a(n) = 1/(n*(n+1))*A143414(n) for n > 0.

Recurrence relation: a(0) = 0, a(1) = 1, (n-1)*(n+1)*a(n) - (n-2)*n*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1) for n >= 2. 1/e = 1/2 - 2 * Sum_{n = 1..inf} (-1)^(n+1)/(n*(n+2)*a(n)*a(n+1)) = 1/2 - 2*[1/(3*1*5) - 1/(8*5*41) + 1/(15*41*481) - 1/(24*481*7421) + ...] .

Conjectural congruences: for r >= 0 and prime p, calculation suggests the congruences a(p^r*(p+1)) == a(p^r) (mod p^(r+1)) may hold.

a(n) = ((2*n)!/(n+1)!)*hypergeom([1-n], [-2*n], 1) for n > 0. - Peter Luschny, May 14 2020