

A219162


Recurrence equation a(n+1) = a(n)^4  4*a(n)^2 + 2 with a(0) = 3.


5




OFFSET

0,1


COMMENTS

Bisection of A001566. Compare the following remarks with A001999.
The present sequence is the case x = 3 of the following general remarks. For other cases see A219163 (x = 4), A219164 (x = 5) and A219165 (x = 6).
Let x > 2 and let alpha := {x + sqrt(x^2  4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(4^n) + (1/alpha)^(4^n). Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^4 + 4*a(n)^2  2 with the initial condition a(0) = x.
We have the product expansion sqrt((x + 2)/(x  2)) = product {n = 0..inf} ((1 + 2/a(n))/(1  2/a(n)^2)).


LINKS

Table of n, a(n) for n=0..3.


FORMULA

Let alpha = 1/2*(3 + sqrt(5)) then a(n) = (alpha)^4^n + (1/alpha)^4^n.
a(n) = A001566(2*n) = A000032(2*4^n).
Product {n >= 0} ((1 + 2/a(n))/(1  2/a(n)^2)) = sqrt(5).


PROG

(PARI) a(n)={if(n==0, 3, a(n1)^44*a(n1)^2+2)} \\ Edward Jiang, Sep 11 2014


CROSSREFS

Cf. A000032, A001566, A001999, A002814, A145502, A219163, A219164, A219165.
Sequence in context: A239450 A187665 A088718 * A016548 A208059 A162333
Adjacent sequences: A219159 A219160 A219161 * A219163 A219164 A219165


KEYWORD

nonn,easy


AUTHOR

Peter Bala, Nov 13 2012


STATUS

approved



