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%I #21 May 18 2020 19:56:14
%S 1,1,1,2,4,100,100,405700,12160319136100,820172805359644669394378100,
%T 833851539054293743258980868256234722984430395524650100,
%U 4967797610707807275548011269672520283591314213501804232311981072548746611502282121553381887584145907359338100
%N a(n) = denominator of r(n): r(1)=1, r(n+1) = [b(1,n);b(2,n),...,b(n,n)], a continued fraction of rational terms, where {b(k,n)} is the permutation of the first n terms of {r(k)} such that r(n+1) is maximized.
%H Alois P. Heinz, <a href="/A128773/b128773.txt">Table of n, a(n) for n = 1..15</a>
%e The first 5 terms of {r(k)} are: 1,1,2,5/2,13/4. The continued fraction, whose terms are the permutation of the first 5 terms of {r(k)} which leads to the largest r(6), is [13/4;1,5/2,1,2] = 401/100.
%p Ltoc := proc(L) numtheory[nthconver](L,nops(L)-1) ; end: r := proc(n) option remember ; local m,rL,rp,L ; if n = 1 then 1; else m := -1 ; rL := [seq(procname(i),i=1..n-1)] ; rp := combinat[permute](rL) ; for L in rp do m := max(m, Ltoc(L)) ; od: m ; fi; end: A128773 := proc(n) denom(r(n)) ; end: for n from 1 do printf("%d,\n", A128773(n)) ; od: # _R. J. Mathar_, Jul 30 2009
%p tor:= proc(l) local j; infinity; for j from nops(l) to 1 by -1 do l[j]+1/% od end: sh:= proc(l) local ll, h, s, m; ll:= []; h:= nops(l); s:= 1; m:= h; while s<=h do ll:= [ll[], l[m]]; if m=h then h:= h-1; m:= s else s:= s+1; m:= h fi od; ll end: r:= proc(n) option remember; local j; `if`(n=1, 1, tor(sh(sort([seq(r(j), j=1..n-1)])))) end: a:= n-> denom(r(n)): seq(a(n), n=1..12); # _Alois P. Heinz_, Aug 04 2009
%t r[1] = r[2] = 1;
%t r[n_] := r[n] = FromContinuedFraction /@ Permutations[Array[r, n - 1]] // Max;
%t Table[a[n] = Denominator[r[n]]; Print[n, " ", a[n]]; a[n], {n, 1, 10}] (* _Jean-François Alcover_, May 18 2020 *)
%Y Cf. A128772, A128774, A128775.
%K frac,nonn
%O 1,4
%A _Leroy Quet_, Mar 27 2007
%E 3 more terms from _R. J. Mathar_, Jul 30 2009
%E a(10)-a(12) from _Alois P. Heinz_, Aug 04 2009
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