A143413 Apéry-like numbers for the constant e: a(n) = 1/(n-1)!*Sum_{k = 0..n+1} (-1)^k*C(n+1,k)*(2*n-k)! for n >= 1.

-1, 1, 11, 181, 3539, 81901, 2203319, 67741129, 2346167879, 90449857081, 3843107102339, 178468044946621, 8994348275804891, 488964835817842021, 28523735794360301039, 1777328098986754744081, 117817961601577138782479, 8279178465722546926265329

Offset: 0

Author

Peter Bala, Aug 14 2008

Keywords

• easy
• sign

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 Napier's constant: e = 2 * Sum_{n >= 1} (-1)^n * n^2/(a(n)* a(n-1)).

Notice the striking parallels with 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)) = 5*[1/(1*3) + 1/(2^2*3*19) + 1/(3^2*19*147) + ...].

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. A005258, A060475, A086764, A143414, 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((-1)^k*binomial(n+1,k)*(2*n-k)!, k = 0..n+1):
seq(a(n), n = 1..19);
# Alternative
a := n -> `if`(n<2, 2*n-1, (2*n)!/(n-1)!*hypergeom([-n-1], [-2*n], -1)):
seq(simplify(a(n)), n=0..17); # Peter Luschny, Nov 14 2018

Mathematica

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

PARI

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

Formula

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

Apart from the initial term, this sequence is the second superdiagonal of the square array A060475; equivalently, the second subdiagonal of the square array A086764.

Recurrence relation: a(0) = -1, a(1) = 1, (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) = 0, b(1) = 2. Then b(n) = A143414(n) = 1/(n-1)!*sum {k = 0..n-1} C(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) -> e very rapidly.

For example, b(100)/a(100) - e is approximately 1.934 * 10^(-436). The identity b(n)*a(n-1) - b(n-1)*a(n) = (-1)^n *2*n^2 leads to rapidly converging series for e and 1/e: e = 2 * Sum_{n >= 1} (-1)^n * n^2/(a(n)*a(n-1)) = 2*[1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...]; 1/e = 1/2 - 2*Sum_{n >= 2} (-1)^n * n^2/(b(n)*b(n-1)) = 1/2 - 2*[2^2/(2*30) - 3^2/(30*492) + 4^2/(492*9620) - ...].

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

a(n) = ((2*n)!/(n-1)!)*hypergeom([-n-1], [-2*n], -1) for n >= 2. - Peter Luschny, Nov 14 2018

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