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Sum[Sum[C(2k,k),{k,1,m}],{m,1,n}], where C(2k,k)=(2k)!/(k!)^2=A000984[k].
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%I #13 Nov 21 2013 12:48:59

%S 2,10,38,136,486,1760,6466,24042,90238,341190,1297574,4958114,

%T 19019254,73196994,282492254,1092867904,4236849774,16455966944,

%U 64020347914,249431257704,973100041934,3800867789884,14862066265434,58170868424084

%N Sum[Sum[C(2k,k),{k,1,m}],{m,1,n}], where C(2k,k)=(2k)!/(k!)^2=A000984[k].

%C a(2(p-1)) is divisible by p^2 for p=7,13,19,31,37,43,61,67.. A002476 Primes of form 6n + 1.

%H Vincenzo Librandi, <a href="/A120278/b120278.txt">Table of n, a(n) for n = 1..200</a>

%F a(n) = Sum[Sum[(2k)!/(k!)^2,{k,1,m}],{m,1,n}].

%F a(n) = 2 * Sum[ A079309[k], {k,1,n} ] = Sum[ A066796[k], {k,1,n} ]. - _Alexander Adamchuk_, Sep 01 2006

%F G.f.: x*(1/Sqrt[1-4*x]-1)/(x(x-1)^2) [From Harvey P. Dale, May 24 2011]

%F Recurrence: n*a(n) = 2*(3*n-1)*a(n-1) - (9*n-4)*a(n-2) + 2*(2*n-1)*a(n-3). - _Vaclav Kotesovec_, Oct 19 2012

%F a(n) ~ 2^(2*n+4)/(9*sqrt(Pi*n)). - _Vaclav Kotesovec_, Oct 19 2012

%t Table[Sum[Sum[(2k)!/(k!)^2,{k,1,m}],{m,1,n}],{n,1,50}]

%t CoefficientList[Series[(1/Sqrt[1-4 x]-1)/((x-1)^2 x),{x,0,50}],x] (* _Harvey P. Dale_, May 24 2011 *)

%Y Cf. A000984, A066796, A002476.

%Y Cf. A066796, A079309.

%K nonn

%O 1,1

%A _Alexander Adamchuk_, Jul 04 2006