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A167713 a(n) = 16^n * Sum_{k=0..n} binomial(2*k, k) / 16^k.

%I A167713 #23 Aug 06 2025 01:02:33
%S A167713 1,18,294,4724,75654,1210716,19372380,309961512,4959397062,
%T A167713 79350401612,1269606610548,20313706474200,325019306291356,
%U A167713 5200308911062296,83204942617113336,1331279082028930896,21300465313063974726,340807445011357201836,5452919120190790364676
%N A167713 a(n) = 16^n * Sum_{k=0..n} binomial(2*k, k) / 16^k.
%C A167713 p^2 divides a((p-3)/2) for prime p of the form p = 6k + 1 (A002476).
%C A167713 The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), a(n) = A167713 (B=16).
%H A167713 Vincenzo Librandi, <a href="/A167713/b167713.txt">Table of n, a(n) for n = 0..200</a>
%H A167713 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CentralBinomialCoefficient.html">Central Binomial Coefficient</a>.
%H A167713 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BinomialSums.html">Binomial Sums</a>.
%F A167713 a(n) = 16^n * Sum_{k=0..n} ((2k)!/(k!)^2) / 16^k.
%F A167713 a(n) = 16^n * Sum_{k=0..n} binomial(2k,k) / 16^k.
%F A167713 G.f.: 1/((1-16*x)*sqrt(1-4*x)). - _R. J. Mathar_, Nov 13 2009
%F A167713 From _Vaclav Kotesovec_, Oct 20 2012: (Start)
%F A167713 Recurrence: n*a(n) = 2*(10*n-1)*a(n-1) - 32*(2*n-1)*a(n-2).
%F A167713 a(n) ~ 2^(4*n+1)/sqrt(3). (End)
%p A167713 A167713 := proc(n) coeftayl( 1/(1-16*x)/sqrt(1-4*x),x=0,n) ; end proc: seq(A167713(n),n=0..40) ; # _R. J. Mathar_, Nov 13 2009
%t A167713 Table[ 16^n * Sum[ (2k)!/(k!)^2 / 16^k, {k,0,n} ], {n,0,20} ]
%t A167713 CoefficientList[Series[1 / ((1 - 16 x) Sqrt[1 - 4 x]), {x, 0, 20}], x] (* _Vincenzo Librandi_, May 27 2013 *)
%Y A167713 Cf. A000984, A002476, A066796, A006134, A082590, A132310, A002457, A144635.
%K A167713 nonn
%O A167713 0,2
%A A167713 _Alexander Adamchuk_, Nov 10 2009
%E A167713 Extended by _R. J. Mathar_, Nov 13 2009