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%I
%S 0,1,1,3,7,31,255,8191,2097151,17179869183,36028797018963967,
%T 618970019642690137449562111,
%U 22300745198530623141535718272648361505980415
%N 2^Fibonacci(n) - 1.
%C 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)=Fib(n). Hence any other sequence p^Fib(n)-1 could also serve as a valid solution to that recurrence, only depending on the value of the term a(1). - _Hieronymus Fischer_, Jun 27 2007
%C Written in binary, a(n) contains Fib(n) 1's (Fib(n)=A000045(n)). Thus the sequence converted to base-2 is A007088(a(n))=0,1,1,11,111,11111,11111111,... - _Hieronymus Fischer_, Jun 27 2007
%F 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).
%F a(n)=A000301(n+1)-1. - _R. J. Mathar_, Apr 26 2007
%F a(n)=a(n-2)*2^ceiling(log_2(a(n-1)))+a(n-1) for n>1. - _Hieronymus Fischer_, Jun 27 2007
%t a[0, k_] = 0; a[1, k_] = 1; a[n_, k_] := (k - 1)*a[n - 1, k]*a[n - 2, k] + a[n - 1, k] + a[n - 2, k]; Table[ a[n, 2], {n, 0, 14} ]
%t a=0;b=1;lst={a,b};Do[c=a*b+a+b;AppendTo[lst,c];a=b;b=c,{n,3*3!}];lst [From _Vladimir Joseph Stephan Orlovsky_, Sep 13 2009]
%Y Cf. A000045 and A000301.
%Y Cf. A000045, A061107.
%Y See A131293 for a base-10 analog with Fib(n) 1's.
%K nonn
%O 0,4
%A _Robert G. Wilson v_, Aug 29 2001
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