%I #34 Jan 30 2020 21:29:14
%S 1,3,17,163,2241,39971,874513,22652547,677742593,22996109251,
%T 872449527441,36595485309923,1681600030358977,84005018253431523,
%U 4532832802360066961,262732854317051785411,16280199853832658463233,1073958487530496802770307
%N E.g.f.: exp(x)/sqrt(1-4*x).
%D R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see page 191.
%H Vincenzo Librandi, <a href="/A052143/b052143.txt">Table of n, a(n) for n = 0..200</a>
%H Vladimir Kruchinin, D. V. Kruchinin, <a href="http://arxiv.org/abs/1103.2582">Composita and their properties</a>, arXiv:1103.2582 [math.CO], 2011-2013.
%F a(n) = n!*Sum_{k=0..n} A000984(k)/(n-k)!. - _Vladimir Kruchinin_, Sep 10 2010
%F a(n) = Sum_{k=0..n} binomial(n,k)*(2*k)!/k!. - _Vladimir Kruchinin_, Sep 10 2010
%F a(n) ~ sqrt(2)*4^n*n^n/exp(n-1/4). - _Vaclav Kotesovec_, Jun 27 2013
%F D-finite with recurrence: a(n) - (4*n-1)*a(n-1) + 4*(n-1)*a(n-2) = 0. - _R. J. Mathar_, Sep 27 2013
%F a(n) = U(1/2, n+3/2, 1/4)/2 where U denotes the Kummer U function. - _Peter Luschny_, Nov 26 2014
%F From _Peter Bala_, Nov 21 2017: (Start)
%F a(n+k) = a(n) (mod k) for all n and k. It follows that the sequence a(n) taken modulo k is periodic with the exact period dividing k. For example, modulo 10 the sequence becomes 1, 3, 7, 3, 1, 1, 3, 7, 3, 1, ... with exact period 5.
%F The e.g.f. A(x) = 1/sqrt(1 - 4*x)*exp(x) satisfies the differential equation (1 - 4*x)A' - (3 - 4*x)*A = 0 with A(0) = 1. Mathar's recurrence above follows from this. (End)
%p A052143 := n -> KummerU(1/2, n+3/2, 1/4)/2:
%p seq(simplify(A052143(n)), n=0..17); # _Peter Luschny_, Dec 18 2017
%t CoefficientList[Series[E^x/Sqrt[1-4*x], {x, 0, 20}], x]* Range[0, 20]! (* _Vaclav Kotesovec_, Jun 27 2013 *)
%o (Maxima) makelist(sum(binomial(n,k)*binomial(2*k,k)*(k)!,k,0,n),n,0,12); /* _Emanuele Munarini_, Dec 17 2017 */
%o (PARI) x='x+O('x^99); Vec(serlaplace(exp(x)/sqrt(1-4*x))) \\ _Altug Alkan_, Dec 17 2017
%Y Cf. A000984.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Jan 23 2000