A076736 - OEIS

A076736

Let u(1) = u(2) = u(3) = 2, u(n) = (1 + u(n-1)*u(n-2))/u(n-3); then a(n) is the denominator of u(n).

1, 1, 1, 2, 1, 4, 2, 8, 4, 16, 8, 32, 16, 64, 32, 128, 64, 256, 128, 512, 256, 1024, 512, 2048, 1024, 4096, 2048, 8192, 4096, 16384, 8192, 32768, 16384, 65536, 32768, 131072, 65536, 262144, 131072, 524288, 262144, 1048576, 524288, 2097152

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,4

COMMENTS

The sequence 1,4,2,8,4,... has g.f. (1+4x)/(1-2x^2) and a(n)=(2^(n/2)(1+2*sqrt(2) + (1-2*sqrt(2))(-1)^n)/2. - Paul Barry, Apr 26 2004

The sequence 2,1,4,2,8,... has a(n) = 2^(n/2)(1+2*sqrt(2)-(1-2*sqrt(2))(-1)^n)/(2*sqrt(2)) and is essentially the pair-reversal of A016116. - Paul Barry, Apr 26 2004

LINKS

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (0,2).

FORMULA

For n > 4, a(n) = 2^A028242(n-4).

From Colin Barker, Oct 14 2014: (Start)

For n > 5, a(n) = 2*a(n-2).

G.f.: x*(x-1)*(x^3+x^2+2*x+1) / (2*x^2-1). (End)

MATHEMATICA

LinearRecurrence[{0, 2}, {1, 1, 1, 2, 1}, 50] (* Harvey P. Dale, Aug 25 2015 *)

CROSSREFS

Cf. A005246, A076737.