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a(1) = 1, a(2) = 6, a(n) = product of the previous terms for n >= 3.
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%I #8 Oct 19 2018 03:30:29

%S 1,6,6,36,1296,1679616,2821109907456,7958661109946400884391936,

%T 63340286662973277706162286946811886609896461828096

%N a(1) = 1, a(2) = 6, a(n) = product of the previous terms for n >= 3.

%H G. C. Greubel, <a href="/A165424/b165424.txt">Table of n, a(n) for n = 1..13</a>

%F a(1) = 1, a(2) = 6, a(n) = Product_{i=1..n-1} a(i), n >= 3.

%F a(1) = 1, a(2) = 6, a(n) = A000400(2^(n-3)) = 6^(2^(n-3)), n >= 3.

%F a(1) = 1, a(2) = 6, a(3) = 6, a(n) = (a(n-1))^2, n >= 4.

%t a[1]:= 1; a[2]:= 6; a[n_]:= Product[a[j], {j,1,n-1}]; Table[a[n],{n,1, 12}] (* _G. C. Greubel_, Oct 19 2018 *)

%o (PARI) {a(n) = if(n==1, 1, if(n==2, 6, prod(j=1,n-1, a(j))))};

%o for(n=1,10, print1(a(n), ", ")) \\ _G. C. Greubel_, Oct 19 2018

%Y See A173501 for another version. - _N. J. A. Sloane_, Feb 23 2010

%K nonn

%O 1,2

%A _Jaroslav Krizek_, Sep 17 2009