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 A213809 Position of the maximum element in the simple continued fraction of Fibonacci(n+1)^5/Fibonacci(n)^5. 1
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,5 COMMENTS 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. LINKS FORMULA 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. EXAMPLE 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. MAPLE 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 CROSSREFS Sequence in context: A114227 A187469 A112593 * A204854 A113215 A105591 Adjacent sequences: A213806 A213807 A213808 * A213810 A213811 A213812 KEYWORD nonn AUTHOR Art DuPre, Jun 20 2012 STATUS approved

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Last modified February 4 20:58 EST 2023. Contains 360082 sequences. (Running on oeis4.)