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A219162 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 3. 5
3, 47, 4870847, 562882766124611619513723647 (list; graph; refs; listen; history; text; internal format)
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(n-1)^4-4*a(n-1)^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

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Last modified January 17 18:02 EST 2020. Contains 330987 sequences. (Running on oeis4.)