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A063896 a(n) = 2^Fibonacci(n) - 1. 13

%I

%S 0,1,1,3,7,31,255,8191,2097151,17179869183,36028797018963967,

%T 618970019642690137449562111,

%U 22300745198530623141535718272648361505980415

%N a(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, depending only 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 (* _Vladimir Joseph Stephan Orlovsky_, Sep 13 2009 *)

%t 2^Fibonacci[Range[0,15]]-1 (* _Harvey P. Dale_, May 20 2014 *)

%t RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == (a[n - 1] + 1)*(a[n - 2] + 1) - 1}, a[n], {n, 0, 12}] (* _Ray Chandler_, Jul 30 2015 *)

%o (PARI) a(n) = 2^fibonacci(n) - 1 \\ _Charles R Greathouse IV_, Oct 03 2016

%Y Cf. A000045, A000301, 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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Last modified December 7 03:42 EST 2016. Contains 278840 sequences.