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Let f(n) = Sum_{j>=1} j^n/binomial(2*j,j) = r_n*Pi*sqrt(3)/3^{t_n} + s_n/3; sequence gives r_n.
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%I #60 May 15 2020 02:50:19

%S 2,2,10,74,238,938,13130,23594,1298462,26637166,201403930,5005052234,

%T 135226271914,1315508114654,13747435592810,153590068548062,

%U 202980764290906,69141791857625242,2766595825017102650,38897014541363246798,1724835471991750464238,80219728936311383557694

%N Let f(n) = Sum_{j>=1} j^n/binomial(2*j,j) = r_n*Pi*sqrt(3)/3^{t_n} + s_n/3; sequence gives r_n.

%H Petros Hadjicostas, <a href="/A181334/b181334.txt">Table of n, a(n) for n = 0..101</a>

%H F. J. Dyson, N. E. Frankel and M. L. Glasser, <a href="http://arxiv.org/abs/1009.4274">Lehmer's Interesting Series</a>, arXiv:1009.4274 [math-ph], 2010-2011.

%H F. J. Dyson, N. E. Frankel and M. L. Glasser, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.120.02.116">Lehmer's interesting series</a>, Amer. Math. Monthly, 120 (2013), 116-130.

%H D. H. Lehmer, <a href="https://www.jstor.org/stable/2322496">Interesting series involving the central binomial coefficient</a>, Amer. Math. Monthly, 92(7) (1985), 449-457.

%F a(n) = numerator(2*Sum_{m=1..n+1} Sum_{p=0..m-1} (-1)^p * (m!/((p+1)*3^(m+2))) * Stirling2(n+1,m) * binomial(2*p,p) * binomial(m-1,p))). [It follows from Theorem 1 in Dyson et al. (2010-2011, 2013).] - _Petros Hadjicostas_, May 15 2020

%p LehmerSer := n -> 2*add(add((-1)^p*(m!/((p+1)*3^(m+2)))*Stirling2(n+1,m)

%p *binomial(2*p, p)*binomial(m-1, p), p=0..m-1), m=1..n+1):

%p a := n -> numer(LehmerSer(n)): seq(a(n), n=0..21);

%p # (after _Petros Hadjicostas_) _Peter Luschny_, May 15 2020

%t f[n_] := Sum[j^n/Binomial[2*j, j], {j, 1, Infinity}];

%t a[n_] := Expand[ FunctionExpand[ f[n] ] ][[2, 1]] // Numerator;

%t Table[a[n], {n, 0, 21}] (* _Jean-François Alcover_, Nov 24 2017 *)

%o (PARI) a(n)=numerator(2*sum(m=1, n+1, sum(p=0, m-1, (-1)^p*(m!/((p+1)*3^(m+2)))*stirling(n+1,m,2)*binomial(2*p,p)*binomial(m-1,p)))) \\ _Petros Hadjicostas_, May 15 2020

%Y Cf. A098830 (s_n), A185585 (t_n), A181374, A180875, A014307.

%K nonn

%O 0,1

%A _N. J. A. Sloane_, Feb 09 2011, following a suggestion from _Herb Conn_

%E a(11)-a(21) from _Nathaniel Johnston_, Apr 07 2011