A176594 - OEIS

A176594

a(n) = 5^(2^n).

5, 25, 625, 390625, 152587890625, 23283064365386962890625, 542101086242752217003726400434970855712890625, 293873587705571876992184134305561419454666389193021880377187926569604314863681793212890625

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OFFSET

0,1

COMMENTS

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

LINKS

Table of n, a(n) for n=0..7.

FORMULA

a(n) = A165423(n+3).

a(n+1) = a(n)^2 with a(0)=5.

a(n-1) = (Im((2+i)^(2^n))^2 + Re((2+i)^(2^n))^2)^(1/2). - Carmine Suriano, Feb 04 2011

Sum_{n>=0} 1/a(n) = A078886. - Amiram Eldar, Nov 09 2020

Product_{n>=0} (1 + 1/a(n)) = 5/4. - Amiram Eldar, Jan 29 2021

PROG

(PARI) a(n) = 5^(2^n); \\ Michel Marcus, Jan 26 2016

CROSSREFS

Cf. A001146, A011764, A078886, A098122, A100686, A165423.