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A105151
Greatest numerator among the n! ratios equal to the continued fractions which have the permutations of (1,2,3,...,n) for terms.
3
1, 3, 11, 48, 253, 1576, 11331, 92467, 845064, 8554195, 95032146, 1149773923, 15050556403, 211951761735, 3195468293093, 51354400809456, 876431092504915, 15830294577832786, 301703171661686235, 6050766978392127541, 127383588868883838996, 2808790552014917701633
OFFSET
1,2
LINKS
FORMULA
a(n) ~ c * n!, where c = 2.61781739197011756877854274... . - Vaclav Kotesovec, Aug 25 2014
From Mark van Hoeij, Jul 08 2024: (Start)
Conjecture: a(n) = n*a(n-1) + (n-1)*a(n-3) + a(n-4).
Consequently: a(n) = round( sqrt(2/Pi) * (cosh(1)*BesselI(0,1)+ sinh(1)*BesselI(1,1)) * BesselK(n/2+1/2,1) * BesselK(n/2+1,1) ). (End)
Consequently: c = cosh(1)*BesselI(0,1)+ sinh(1)*BesselI(1,1). - Vaclav Kotesovec, Jul 09 2024
EXAMPLE
a(4) = 48 because the continued fractions [4;2,1,3] (= 48/11) and [3;1,2,4] (= 48/13) have the greatest numerators among continued fraction which each have a permutation of (1,2,3,4) for terms.
MAPLE
r:= proc(l) local j; infinity; for j to nops(l) do l[j] +1/% od end: gl:= proc(n) local i, l; l:=[]; for i to n do l:= `if` (irem (i, 2)=0, [l[], i], [i, l[]]) od; l end: a:= n-> numer (r (gl (n))): seq (a(n), n=1..30); # Alois P. Heinz, Nov 18 2009
MATHEMATICA
Table[Max@ Map[Numerator@ FromContinuedFraction@ # &, Permutations@ Range@ n], {n, 10}] (* Michael De Vlieger, Sep 25 2017 *)
CROSSREFS
Sequence in context: A372623 A186374 A187249 * A111680 A095822 A025539
KEYWORD
nonn
AUTHOR
Leroy Quet, Apr 10 2005
EXTENSIONS
More terms from Vladeta Jovovic and David W. Wilson, Apr 12 2005
Further terms from Alois P. Heinz, Nov 18 2009
STATUS
approved