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%I #18 Nov 06 2023 11:08:03
%S 1,2,3,2,5,5,3,7,3,8,5,5,11,4,7,12,13,7,9,8,17,7,7,7,19,19,23,12,11,
%T 12,25,10,13,27,11,10,9,14,11,29,11,31,31,19,17,34,37,18,19,40,41,14,
%U 17,21,15,16,17,18,47,17,23,46,45,46,25,49,49,50,29,26,19,27,31,29,55,34,61
%N Least number k < n and coprime to n such that the largest term of the continued fraction of k/n is as small as possible.
%C See A141822 for the value of the largest term in the continued fraction of a(n)/n. Zaremba conjectured that the largest value is 5.
%D R. K. Guy, Unsolved problems in number theory, F20.
%D S. K. Zaremba, ed., "Applications of number theory to numerical analysis," Proceedings of the Symposium at the Centre for Research in Mathematics, University of Montreal, Academic Press, New York, London (1972).
%H Robin Visser, <a href="/A141821/b141821.txt">Table of n, a(n) for n = 2..10000</a> (terms n = 2..2000 from T. D. Noe).
%H T. W. Cusick, <a href="https://doi.org/10.1090/S0025-5718-1993-1189517-7">Zaremba's conjecture and sums of the divisor function</a>, Math. Comp. 61 (1993), 171-176.
%H Takao Komatsu, <a href="http://www.anubih.ba/Journals/vol-1,no-1,y05/03revkomatsu.pdf">On a Zaremba's conjecture for powers</a>, Sarajevo J. Math. 1 (2005), 9-13.
%e For n=7, the six continued fractions for k/7 are (0, 7), (0, 3, 2), (0, 2, 3), (0, 1, 1, 3), (0, 1, 2, 2) and (0, 1, 6). It is easy to see that the fifth one, for 5/7, has the smallest maximum term, 2. Hence a(7)=5.
%t Table[k=Select[Range[n-1], GCD[ #,n]==1&]; c=ContinuedFraction[k/n]; mx=Max/@c; mn=Min[mx]; k[[Position[mx,mn,1,1][[1,1]]]], {n,2,100}]
%K nonn
%O 2,2
%A _T. D. Noe_, Jul 08 2008