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A213809
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Position of the maximum element in the simple continued fraction of Fibonacci(n+1)^5/Fibonacci(n)^5.
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1
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1, 1, 1, 1, 3, 3, 3, 3, 3, 5, 5, 3, 5, 5, 5, 5, 5, 5, 5, 7, 7, 5, 7, 7, 7, 7, 7, 7, 7, 9, 9, 7, 9, 9, 9, 9, 9, 9, 9, 11, 11, 9, 11, 11, 11, 11, 11, 11, 11, 13, 13, 11, 13, 13, 13, 13, 13, 13, 13, 15, 15, 13, 15, 15, 15, 15, 15, 15, 15, 17, 17, 15, 17, 17, 17, 17, 17, 17, 17, 19, 19, 17, 19, 19, 19, 19, 19, 19, 19, 21, 21, 19, 21, 21, 21, 21, 21, 21, 21, 23, 23, 21, 23, 23, 23, 23, 23, 23, 23
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OFFSET
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1,5
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COMMENTS
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The maximum elements themselves are in A113506.
The fractions F(n+1)^5/F(n)^5 are 1, 32, 243/32, 3125/243,... (see A056572). The continued fractions are [1], [32], [7,1,1,2,6], [12,1,6,6,1,4], [10,2,17,17,1,4]..., and for the first 4 of these the maximum element is the first, for the 5th of these the maximum element is the third.
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LINKS
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Table of n, a(n) for n=1..109.
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FORMULA
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a(10k+m)=3+2k if m=0,1,3,4,5,6,7,8,9, k>0.
a(10k+2)=1+2k, k>0.
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EXAMPLE
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The continued fraction of the fraction corresponding to [1,1,1,1,1,1,1,1,1,1,1,1,1]^5 is
[11,11,7,1,39282,2,5,11,11,1,11,11] and the maximum occurs at place 5, which according to the formula, should be 3+2k, and since 13=10k+3, k=1 and 3+2k=3+2=5.
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MAPLE
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A213809 := proc(n)
local c, a, i;
(combinat[fibonacci](n+1)/combinat[fibonacci](n))^5 ;
c := numtheory[cfrac](%, quotients) ;
a := 1 ;
for i from 2 to nops(c) do
if op(i, c) > op(a, c) then
a := i ;
end if;
end do:
a ;
end proc: # R. J. Mathar, Jul 06 2012
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CROSSREFS
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Sequence in context: A114227 A187469 A112593 * A204854 A113215 A105591
Adjacent sequences: A213806 A213807 A213808 * A213810 A213811 A213812
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KEYWORD
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nonn
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AUTHOR
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Art DuPre, Jun 20 2012
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STATUS
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approved
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