

A273918


Numerator of z(n), where z(n) = z(n  1)^2 + 1/4 and z(0) = 1.


0




OFFSET

0,2


COMMENTS

a(8) is approximately 5.698 * 10^93.
The denominator of z(n) is 2^(2^n) for n > 0.
Given that the iteration of z(n) escapes to infinity, this shows that 1 is not in the Julia set for the function z^2 + 1/4. This is of course also true of 1.


LINKS

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


EXAMPLE

1^2 + 1/4 = 5/4, hence a(1) = 5.
(5/4)^2 + 1/4 = 25/16 + 4/16 = 29/16, hence a(2) = 29.


MATHEMATICA

Numerator[NestList[#^2 + 1/4 &, 1, 8]]


CROSSREFS

Cf. A015701, A020773.
Sequence in context: A263369 A072880 A112959 * A085553 A057208 A046842
Adjacent sequences: A273915 A273916 A273917 * A273919 A273920 A273921


KEYWORD

easy,nonn


AUTHOR

Alonso del Arte, Jun 04 2016


STATUS

approved



