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%I #30 Jan 29 2021 04:48:08
%S 5,25,625,390625,152587890625,23283064365386962890625,
%T 542101086242752217003726400434970855712890625,
%U 293873587705571876992184134305561419454666389193021880377187926569604314863681793212890625
%N a(n) = 5^(2^n).
%C Also the hypotenuse of primitive Pythagorean triangles obtained by repeated application of basic formula c(n)=p(n)^2+q(n)^2 starting p(0)=2, q(0)=1, see A100686, A098122. Example: a(2)=25 since starting (2,1) gives Pythagorean triple (3,4,5) using (3,4) as new generators gives triple (7,24,25) hypotenuse 25=a(2). - _Carmine Suriano_, Feb 04 2011
%F a(n) = A165423(n+3).
%F a(n+1) = a(n)^2 with a(0)=5.
%F a(n-1) = (Im((2+i)^(2^n))^2 + Re((2+i)^(2^n))^2)^(1/2). - _Carmine Suriano_, Feb 04 2011
%F Sum_{n>=0} 1/a(n) = A078886. - _Amiram Eldar_, Nov 09 2020
%F Product_{n>=0} (1 + 1/a(n)) = 5/4. - _Amiram Eldar_, Jan 29 2021
%o (PARI) a(n) = 5^(2^n); \\ _Michel Marcus_, Jan 26 2016
%Y Cf. A001146, A011764, A078886, A098122, A100686, A165423.
%Y Cf. A185457, A120905, A139011. - _Carmine Suriano_, Feb 04 2011
%K nonn
%O 0,1
%A _Vincenzo Librandi_, Apr 21 2010
%E Offset corrected by _R. J. Mathar_, Jun 18 2010
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