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a(n) = 2^L(n+1) * 3^L(n)/12, where L(n) is the n-th Lucas number (A000032(n)).
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%I #9 Sep 08 2022 08:45:48

%S 2,36,864,373248,3869835264,17332899271409664,

%T 804905577934332296851095552,

%U 167416167663978753511691999938432197602574336

%N a(n) = 2^L(n+1) * 3^L(n)/12, where L(n) is the n-th Lucas number (A000032(n)).

%C For m>1, A166469(A002110(m)*a(n)) = L(m+n).

%C A166469(a(n)) = L(n+2) - 2 = A014739(n).

%H G. C. Greubel, <a href="/A166473/b166473.txt">Table of n, a(n) for n = 1..14</a>

%F a(n) = A166471(n)/12.

%F For n>1, a(n) = 12*a(n-1) * a(n-2).

%t Table[(2^LucasL[n+1] 3^LucasL[n])/12,{n,10}] (* _Harvey P. Dale_, Aug 17 2011 *)

%o (PARI) lucas(n) = fibonacci(n+1) + fibonacci(n-1);

%o vector(10, n, 2^(lucas(n+1)-2)*3^(lucas(n)-1) ) \\ _G. C. Greubel_, Jul 22 2019

%o (Magma) [2^(Lucas(n+1)-2)*3^(Lucas(n)-1): n in [1..10]]; // _G. C. Greubel_, Jul 22 2019

%o (Sage) [2^(lucas_number2(n+1,1,-1)-2)*3^(lucas_number2(n,1,-1)-1) for n in (1..10)] # _G. C. Greubel_, Jul 22 2019

%o (GAP) List([1..10], n-> 2^(Lucas(1,-1,n+1)[2]-2)*3^(Lucas(1,-1,n)[2]-1)); # _G. C. Greubel_, Jul 22 2019

%Y Subsequence of A003586, A025487.

%K nonn

%O 1,1

%A _Matthew Vandermast_, Nov 05 2009