A296618 Expansion of the e.g.f. exp(-x)/sqrt(1-4*x).

1, 1, 9, 89, 1265, 22929, 506809, 13220521, 397585761, 13543386785, 515418398441, 21673889807481, 998003450868049, 49942515803293489, 2698849517019693465, 156631203355106962889, 9716434375682706344129, 641592631434102757993281

Offset: 0

Author

Emanuele Munarini, Dec 17 2017

Keywords

• nonn

Links

• Peter Bala, Integer sequences that become periodic on reduction modulo k for all k

Crossrefs

Cf. A052143.

Programs

Maple

A296618 := n -> (-1)^n*(I/2)*KummerU(1/2, n+3/2, -1/4):
seq(simplify(A296618(n)), n=0..17); # Peter Luschny, Dec 18 2017

Mathematica

Table[Sum[Binomial[n,k]Binomial[2k,k]k! (-1)^(n-k),{k,0,n}],{n,0,18}]
CoefficientList[Series[Exp[-x]/Sqrt[1-4x], {x,0,18}], x] Range[0,18]!

Maxima

makelist(sum(binomial(n,k)*binomial(2*k,k)*k!*(-1)^(n-k),k,0,n),n,0,12);

PARI

x='x+O('x^99); Vec(serlaplace(exp(-x)/sqrt(1-4*x))) \\ Altug Alkan, Dec 17 2017

Formula

a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*k,k)*k!*(-1)^(n-k).

a(n) = (i/2)*(-1)^n*U(1/2,n+3/2,-1/4), where U denotes the Kummer U function.

D-finite with recurrence: a(n+2) - (4*n+5)*a(n+1) - 4*(n+1)*a(n) = 0.

Sum_{k=0..n} binomial(n,k)*a(k)*a(n-k) = Sum_{k=0..n} binomial(n,k)*(-1)^(n-k)*2^(n+k)*k!.

Conjectures:

1) a(n+1) == a(n) (mod n) for all n >= 1.

2) a(n+k) == (-1)^k*a(n) (mod k) for all n and k >= 1.

a(n) ~ 2^(2*n + 1/2) * n^n / exp(n + 1/4). - Vaclav Kotesovec, Dec 17 2017

From Peter Bala, Jun 19 2023: (Start)

a(n) == 1 (mod 4).

Conjecture 1) above follows immediately from the stated recurrence equation. In fact, a(n+1) == a(n) (mod 8*n) for n >= 1.

For a proof of Conjecture 2) see the Bala link (corollary to Theorem 1, p. 5). (End)