A063896 - OEIS

A063896

a(n) = 2^Fibonacci(n) - 1.

0, 1, 1, 3, 7, 31, 255, 8191, 2097151, 17179869183, 36028797018963967, 618970019642690137449562111, 22300745198530623141535718272648361505980415

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OFFSET

0,4

COMMENTS

The recurrence can also be written a(n)+1 = (a(n-1)+1)*(a(n-2)+1) or log_p(a(n)+1) = log_p(a(n-1)+1) + log_p(a(n-2)+1), respectively. Setting a(1)=p-1 for any natural p>1, it follows that log_p(a(n)+1)=Fibonacci(n). Hence any other sequence p^Fibonacci(n)-1 could also serve as a valid solution to that recurrence, depending only on the value of the term a(1). - Hieronymus Fischer, Jun 27 2007

Written in binary, a(n) contains Fibonacci(n) 1's. Thus the sequence converted to base-2 is A007088(a(n)) = 0, 1, 1, 11, 111, 11111, 11111111, ... . - Hieronymus Fischer, Jun 27 2007

In general, if b(n) is defined recursively by b(0) = p, b(1) = q, b(n) = b(n-1)*b(n-2) + b(n-1) + b(n-2) for n >= 2 then b(n) = p^Fibonacci(n-1) * q^Fibonacci(n) - 1. - Rahul Goswami, Apr 15 2020

a(n) is also the numerator of the continued fraction [2^F(0), 2^F(1), 2^F(2), 2^F(3), ..., 2^F(n-2)] for n>0. For the denominator, see A005203. - Chinmay Dandekar and Greg Dresden, Sep 19 2020

LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..18

M. Tamba and Y. S. Valaulikar, A nonlinear extension of Fibonacci sequence, Turkish Journal of Analysis and Number Theory, Vol. 4, No. 4, 2016.

FORMULA

The solution to the recurrence a(0) = 0; a(1) = 1; a(n) = a(n-1)*a(n-2) + a(n-1) + a(n-2).

a(n) = A000301(n) - 1. - R. J. Mathar, Apr 26 2007

a(n) = a(n-2)*2^ceiling(log_2(a(n-1))) + a(n-1) for n>1. - Hieronymus Fischer, Jun 27 2007

a(n) = A000225(A000045(n)). - Alois P. Heinz, Mar 19 2020

MAPLE

a:= n-> 2^(<<0|1>, <1|1>>^n)[1, 2]-1:

seq(a(n), n=0..15); # Alois P. Heinz, Aug 12 2017

MATHEMATICA