A105151 - OEIS

%I #33 Jul 13 2024 02:50:41

%S 1,3,11,48,253,1576,11331,92467,845064,8554195,95032146,1149773923,

%T 15050556403,211951761735,3195468293093,51354400809456,

%U 876431092504915,15830294577832786,301703171661686235,6050766978392127541,127383588868883838996,2808790552014917701633

%N Greatest numerator among the n! ratios equal to the continued fractions which have the permutations of (1,2,3,...,n) for terms.

%H Alois P. Heinz, <a href="/A105151/b105151.txt">Table of n, a(n) for n = 1..175</a>

%F a(n) ~ c * n!, where c = 2.61781739197011756877854274... . - _Vaclav Kotesovec_, Aug 25 2014

%F From _Mark van Hoeij_, Jul 08 2024: (Start)

%F Conjecture: a(n) = n*a(n-1) + (n-1)*a(n-3) + a(n-4).

%F 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)

%F Consequently: c = cosh(1)*BesselI(0,1)+ sinh(1)*BesselI(1,1). - _Vaclav Kotesovec_, Jul 09 2024

%e 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.

%p 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

%t Table[Max@ Map[Numerator@ FromContinuedFraction@ # &, Permutations@ Range@ n], {n, 10}] (* _Michael De Vlieger_, Sep 25 2017 *)

%K nonn

%O 1,2

%A _Leroy Quet_, Apr 10 2005

%E More terms from _Vladeta Jovovic_ and _David W. Wilson_, Apr 12 2005

%E Further terms from _Alois P. Heinz_, Nov 18 2009