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A296618
Expansion of the e.g.f. exp(-x)/sqrt(1-4*x).
2
1, 1, 9, 89, 1265, 22929, 506809, 13220521, 397585761, 13543386785, 515418398441, 21673889807481, 998003450868049, 49942515803293489, 2698849517019693465, 156631203355106962889, 9716434375682706344129, 641592631434102757993281
OFFSET
0,3
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)
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]!
PROG
(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
CROSSREFS
Cf. A052143.
Sequence in context: A075507 A094935 A258388 * A230114 A187090 A078248
KEYWORD
nonn
AUTHOR
Emanuele Munarini, Dec 17 2017
STATUS
approved