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%I #20 Sep 08 2022 08:44:59
%S 1,6,44,352,2940,25194,219604,1937520,17250012,154663960,1394538288,
%T 12631852688,114858935204,1047772373340,9584557428600,87885886492320,
%U 807564936805020,7434289153896264,68551275793965328,633038816547052800
%N a(n) = (5n+1)*C(4n,n)/(3n+1).
%H G. C. Greubel, <a href="/A052204/b052204.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: (2*g-1)*g/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - _Mark van Hoeij_, Nov 11 2011
%F Conjecture: 6*n*(3*n-1)*(3*n+1)*a(n) + (-809*n^3 + 1444*n^2 - 1505*n + 582)*a(n-1) + 88*(4*n-5)*(4*n-7)*(2*n-3)*a(n-2) = 0. - _R. J. Mathar_, Sep 29 2012
%F a(n) ~ 5*2^(8*n+1/2)*3^(-3*n-3/2)/sqrt(Pi*n). - _Ilya Gutkovskiy_, Aug 10 2016
%p A052204:=n->(5*n+1)*binomial(4*n,n)/(3*n+1): seq(A052204(n), n=0..20); # _Wesley Ivan Hurt_, Aug 10 2016
%t Table[(5 n + 1) Binomial[4 n, n]/(3 n + 1), {n, 0, 20}] (* _Wesley Ivan Hurt_, Aug 10 2016 *)
%o (Magma) [(5*n+1)*Binomial(4*n,n)/(3*n+1) : n in [0..20]]; // _Wesley Ivan Hurt_, Aug 10 2016
%o (PARI) for(n=0,25, print1((5*n+1)*binomial(4*n,n)/(3*n+1), ", ")) \\ _G. C. Greubel_, Feb 16 2017
%Y Cf. A002293, A051945.
%K easy,nonn
%O 0,2
%A _Barry E. Williams_, Jan 28 2000
%E More terms from _James A. Sellers_, Jan 31 2000
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