A296618 - OEIS

%I #25 Jun 23 2023 15:55:51

%S 1,1,9,89,1265,22929,506809,13220521,397585761,13543386785,

%T 515418398441,21673889807481,998003450868049,49942515803293489,

%U 2698849517019693465,156631203355106962889,9716434375682706344129,641592631434102757993281

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

%H Peter Bala, <a href="/A047974/a047974_1.pdf">Integer sequences that become periodic on reduction modulo k for all k</a>

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

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

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

%F 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!.

%F Conjectures:

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

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

%F a(n) ~ 2^(2*n + 1/2) * n^n / exp(n + 1/4). - _Vaclav Kotesovec_, Dec 17 2017

%F From _Peter Bala_, Jun 19 2023: (Start)

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

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

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

%p A296618 := n -> (-1)^n*(I/2)*KummerU(1/2, n+3/2, -1/4):

%p seq(simplify(A296618(n)), n=0..17); # _Peter Luschny_, Dec 18 2017

%t Table[Sum[Binomial[n,k]Binomial[2k,k]k! (-1)^(n-k),{k,0,n}],{n,0,18}]

%t CoefficientList[Series[Exp[-x]/Sqrt[1-4x], {x,0,18}], x] Range[0,18]!

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

%o (PARI) x='x+O('x^99); Vec(serlaplace(exp(-x)/sqrt(1-4*x))) \\ _Altug Alkan_, Dec 17 2017

%Y Cf. A052143.

%K nonn

%O 0,3

%A _Emanuele Munarini_, Dec 17 2017