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a(n) = Sum_{k=0..n} binomial(4*n,4*k).
20

%I #43 Nov 10 2024 20:40:25

%S 1,2,72,992,16512,261632,4196352,67100672,1073774592,17179738112,

%T 274878431232,4398044413952,70368752566272,1125899873288192,

%U 18014398643699712,288230375614840832,4611686020574871552,73786976286248271872,1180591620751771041792,18889465931341141901312

%N a(n) = Sum_{k=0..n} binomial(4*n,4*k).

%C Also the cogrowth sequence of the 16-element group C4 X C4 = <S,T | S^4, T^4, [S,T]>. - _Sean A. Irvine_, Nov 09 2024

%H Seiichi Manyama, <a href="/A070775/b070775.txt">Table of n, a(n) for n = 0..830</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (12,64).

%F a(n) = (1/2)*(-4)^n + (1/4)*16^n for n > 0.

%F Let b(n) = a(n) - 2^(4n)/4 then b(n+1) = 4*b(n) - _Benoit Cloitre_, May 27 2004

%F G.f.: (1 - 10*x - 16*x^2)/((1-16*x)*(1+4*x)). - _Seiichi Manyama_, Mar 15 2019

%F G.f.: ((cos(x) + cosh(x))/2)^2 = Sum_{n >= 0} a(n)*x(4*n)/(4*n)!. - _Peter Bala_, Jun 20 2022

%p a := n -> if n = 0 then 1 else 4^(n - 1)*(2*(-1)^n + 4^n) fi:

%p seq(a(n), n = 0..19); # _Peter Luschny_, Jul 02 2022

%t Table[Sum[Binomial[4n,4k],{k,0,n}],{n,0,30}] (* or *) Join[{1}, LinearRecurrence[{12,64},{2,72},30]] (* _Harvey P. Dale_, Apr 24 2011 *)

%o (PARI) a(n)=sum(k=0,n,binomial(4*n,4*k))

%o (PARI) N=66; x='x+O('x^N); Vec((1-10*x-16*x^2)/((1-16*x)*(1+4*x))) \\ _Seiichi Manyama_, Mar 15 2019

%Y Sum_{k=0..n} binomial(b*n,b*k): A000079 (b=1), A081294 (b=2), A007613 (b=3), this sequence (b=4), A070782 (b=5), A070967 (b=6), A094211 (b=7), A070832 (b=8), A094213 (b=9), A070833 (b=10).

%K easy,nonn

%O 0,2

%A Sebastian Gutierrez and Sarah Kolitz (skolitz(AT)mit.edu), May 15 2002