A143414 Apéry-like numbers for the constant 1/e: a(n) = (1/(n-1)!)*Sum_{k = 0..n-1} binomial(n-1,k)*(2*n-k)!.

0, 2, 30, 492, 9620, 222630, 5989242, 184139480, 6377545512, 245868202890, 10446648201110, 485126443539012, 24449173476952380, 1329144227959100462, 77535552689576436210, 4831278674685354629040, 320262424087652686405712

Offset: 0

Author

Peter Bala, Aug 14 2008

Keywords

• easy
• nonn

Comments

This sequence satisfies the recursion (n-1)^2*a(n) - n^2*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1), which leads to a rapidly converging series for the constant 1/e: 1/e = 1/2 - 2 * Sum_{n >= 2} (-1)^n * n^2/(a(n)*a(n-1)).

Notice the striking resemblance to the theory of the Apéry numbers A(n) = A005258(n), which satisfy a similar recurrence relation n^2*A(n) - (n-1)^2*A(n-2) = (11*n^2-11*n+3)*A(n-1) and which appear in the series acceleration formula zeta(2) = 5*Sum_{n>=1} 1/(n^2*A(n)*A(n-1)). Compare with A143413 and A143415.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..365
• A. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3). An informal report, Math. Intelligencer 1 (1978/79), no. 4, 195-203.

Crossrefs

Cf. A143413, A143415.

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:
A143414 := n -> `if`(n=0, 0, ((2*n)!/(n-1)!)*hypergeom([1-n], [-2*n], 1)):
seq(simplify(A143414(n)), n = 0..16); # 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} binomial(n-1,k)*(2*n-k)!.

Recurrence relation: a(0) = 0, a(1) = 2, (n-1)^2*a(n) - n^2*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1), n >= 2.

Let b(n) denote the solution to this recurrence with initial conditions b(0) = -1, b(1) = 1. Then b(n) = A143413(n) = (1/(n-1)!)*Sum_{k = 0..n+1} (-1)^k*binomial(n+1,k)*(2*n-k)!.

The rational number b(n)/a(n) is equal to the Padé approximation to exp(x) of degree (n+1,n-1) evaluated at x = -1 and b(n)/a(n) -> 1/e very rapidly. For example, |b(100)/a(100) - 1/e| is approximately 2.177 * 10^(-437).

The identity a(n)*b(n-1) - a(n-1)*b(n) = (-1)^n *2*n^2 leads to rapidly converging series for the constants 1/e and e: 1/e = 1/2 - 2*Sum_{n >= 2} (-1)^n * n^2/(a(n)*a(n-1)) = 1/2 - 2*(2^2/(2*30) - 3^2/(30*492) + 4^2/(492*9620) - ...); e = 2 * Sum_{n >= 1} (-1)^n * n^2/(b(n)*b(n-1)) = 2*(1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...).

a(n) = (BesselK(n-1/2,1/2)-(1-2*n)*BesselK(n+1/2,1/2)) * exp(1/2)/(2*Pi^(1/2)). - Mark van Hoeij, Nov 12 2009

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

a(n) ~ 2^(2*n + 1/2) * n^(n+1) / exp(n - 1/2). - Vaclav Kotesovec, Jul 11 2021