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a(n) = 2^(binomial(2*n+2,3)/4).
4

%I #20 Oct 18 2024 21:43:08

%S 1,2,32,16384,1073741824,36028797018963968,

%T 2475880078570760549798248448,

%U 1393796574908163946345982392040522594123776,25711008708143844408671393477458601640355247900524685364822016

%N a(n) = 2^(binomial(2*n+2,3)/4).

%C Hankel transform of A185994.

%H G. C. Greubel, <a href="/A185995/b185995.txt">Table of n, a(n) for n = 0..20</a>

%F a(n) = Product_{k=0..n} (2^(2k+1))^(n-k).

%F a(n) = 2^A000330(n).

%F a(n) = 2^(1/6*n*(2*n^2+3*n+1)). - _Alexander R. Povolotsky_, Feb 13 2011

%t Table[2^(1/6*n*(2*n^2 + 3*n + 1)), {n, 0, 25}] (* _G. C. Greubel_, Feb 20 2017 *)

%t Table[2^(Binomial[2n+2,3]/4),{n,0,10}] (* _Harvey P. Dale_, Sep 13 2024 *)

%o (PARI) for(n=0,15, print1(2^(1/6*n*(2*n^2 + 3*n + 1)), ", ")) \\ _G. C. Greubel_, Feb 20 2017

%K nonn,easy

%O 0,2

%A _Paul Barry_, Feb 08 2011