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a(n) = Sum_{k=0..n} binomial(2n+1, 2k)*3^(n-k).
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%I #24 Mar 24 2023 18:08:45

%S 1,6,44,328,2448,18272,136384,1017984,7598336,56714752,423324672,

%T 3159738368,23584608256,176037912576,1313964867584,9807567290368,

%U 73204678852608,546407161659392,4078438577864704,30441879976280064

%N a(n) = Sum_{k=0..n} binomial(2n+1, 2k)*3^(n-k).

%C In general, Sum_{k=0..n} binomial(2n+1,2k)*r^(n-k) has g.f. (1-(r-1)x)/(1-2(r+1)+(r-1)^2x^2) and a(n) = ((sqrt(r)-1)^(2n+1) + (sqrt(r)+1)^(2n+1))/(2*sqrt(r)).

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-4).

%F G.f.: (1-2x)/(1-8x+4x^2);

%F a(n) = 8*a(n-1) - 4*a(n-2);

%F a(n) = sqrt(3)*(sqrt(3)-1)^(2n+1)/6 + sqrt(3)*(sqrt(3)+1)^(2n+1)/6.

%F a(n) = 2^n*A079935(n). - _R. J. Mathar_, Sep 20 2012

%F a(n) = 2^(2*n+1)*Sum_{k >= n} binomial(2*k,2*n)*(1/3)^(k+1). Cf. A099156. - _Peter Bala_, Nov 29 2021

%F 3*a(n)^2 = A107903(n)^2 + 2^(2*n+1). - _Philippe Deléham_, Mar 21 2023

%t LinearRecurrence[{8,-4},{1,6},20] (* _Harvey P. Dale_, Sep 28 2021 *)

%Y Bisection of A080879 or A002605.

%Y Cf. A066443, A079935, A099156, A102592, A107903.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Jan 22 2005