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A165423 a(1) = 1, a(2) = 5, a(n) = product of the previous terms for n >= 3.

%I A165423 #10 Oct 19 2018 03:32:58
%S A165423 1,5,5,25,625,390625,152587890625,23283064365386962890625,
%T A165423 542101086242752217003726400434970855712890625
%N A165423 a(1) = 1, a(2) = 5, a(n) = product of the previous terms for n >= 3.
%H A165423 Vincenzo Librandi, <a href="/A165423/b165423.txt">Table of n, a(n) for n = 1..13</a>
%F A165423 a(1) = 1, a(2) = 5, a(n) = Product_{i=1..n-1} a(i), n >= 3.
%F A165423 a(1) = 1, a(2) = 5, a(n) = A000351(2^(n-3)) = 5^(2^(n-3)), n >= 3.
%F A165423 a(1) = 1, a(2) = 5, a(3) = 5, a(n) = (a(n-1))^2, n >= 4.
%t A165423 a[1]:= 1; a[2]:= 5; a[n_]:= Product[a[j], {j,1,n-1}]; Table[a[n],{n,1, 12}] (* _G. C. Greubel_, Oct 19 2018 *)
%o A165423 (PARI) {a(n) = if(n==1, 1, if(n==2, 5, prod(j=1,n-1, a(j))))};
%o A165423 for(n=1,10, print1(a(n), ", ")) \\ _G. C. Greubel_, Oct 19 2018
%K A165423 nonn
%O A165423 1,2
%A A165423 _Jaroslav Krizek_, Sep 17 2009