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A063896
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2^Fibonacci(n) - 1.
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10
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0, 1, 1, 3, 7, 31, 255, 8191, 2097151, 17179869183, 36028797018963967, 618970019642690137449562111, 22300745198530623141535718272648361505980415
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OFFSET
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0,4
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COMMENTS
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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
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
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LINKS
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Table of n, a(n) for n=0..12.
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FORMULA
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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)-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
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MATHEMATICA
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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} ]
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]
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CROSSREFS
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Cf. A000045 and A000301.
Cf. A000045, A061107.
See A131293 for a base-10 analog with Fib(n) 1's.
Sequence in context: A073917 A030521 A105767 * A156895 A074047 A121810
Adjacent sequences: A063893 A063894 A063895 * A063897 A063898 A063899
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KEYWORD
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nonn
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AUTHOR
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Robert G. Wilson v, Aug 29 2001
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STATUS
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approved
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