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a(n) = 2^Fibonacci(n) + 1.
0

%I #27 Aug 12 2017 07:43:28

%S 2,3,3,5,9,33,257,8193,2097153,17179869185,36028797018963969,

%T 618970019642690137449562113,

%U 22300745198530623141535718272648361505980417,13803492693581127574869511724554050904902217944340773110325048447598593

%N a(n) = 2^Fibonacci(n) + 1.

%F a(n+2) = a(n+1)*a(n) - a(n) - a(n+1) + 2, a(0)=2, a(1)=3.

%F Binet type formula: log_2(a(n)-1) = (1/sqrt(5)) * (r^n - s^n), where r and s are the roots of x^2-x-1. (this is true by definition).

%F a(n) = A000301(n) + 1 = A063896(n) + 2. - _Alois P. Heinz_, Aug 12 2017

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

%p seq(a(n), n=0..15); # _Alois P. Heinz_, Aug 12 2017

%t Table[2^Fibonacci[n] + 1, {n, 0, 13}] (* _T. D. Noe_, Sep 07 2013 *)

%Y Cf. A000045, A000301, A063896.

%K nonn

%O 0,1

%A _Yeshwant Shivrai Valaulikar_ and M. Tamba, Sep 04 2013