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A141822 Maximum term in the continued fraction of A141821(n)/n. 7
2, 2, 3, 2, 5, 2, 2, 3, 3, 2, 2, 2, 3, 3, 3, 2, 2, 2, 4, 2, 3, 3, 3, 3, 2, 2, 4, 2, 2, 2, 3, 3, 2, 3, 3, 3, 4, 3, 3, 2, 4, 2, 2, 2, 2, 2, 3, 2, 2, 3, 3, 3, 5, 2, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 3, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 2, 4, 3, 3, 3, 3, 3, 4, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET
2,1
COMMENTS
Consider the continued fraction [0;c1,c2,...,cm] of k/n, with k<n and gcd(k,n)=1. Let f(k,n) be the maximum of the ci. Then a(n) is the minimum value of f(k,n).
Zaremba conjectured that a(n)<=5, a bound that is attained for n in A195901. It appears that n=150 may be the largest integer with a(n)=5, while n=6234 may be the largest integer with a(n)=4.
LINKS
Robin Visser, Table of n, a(n) for n = 2..10000 (terms n = 2..2000 from T. D. Noe).
MATHEMATICA
Table[c=ContinuedFraction[Select[Range[n-1], GCD[ #, n]==1&]/n]; Min[Max/@c], {n, 150}]
PROG
(PARI) vecmax(v)=my(mx=v[1]); for(i=2, #v, mx=max(mx, v[i])); mx
a(n)=vecmin([vecmax(contfrac(k/n))|k<-[1..n], gcd(k, n)==1]) \\ Charles R Greathouse IV, Jul 18 2014
CROSSREFS
See A141821 for the least value of k for each n.
See A141832, A141833, A141823, and A195901 for the integers n>1 such that a(n) = 2, 3, 4, and 5, respectively.
Cf. A006839 (where cm is constrained to be 1).
Sequence in context: A102247 A054249 A160273 * A033099 A330833 A337331
KEYWORD
nonn
AUTHOR
T. D. Noe, Jul 08 2008
EXTENSIONS
Edited by Max Alekseyev, Sep 25 2011
STATUS
approved

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Last modified March 19 03:33 EDT 2024. Contains 370952 sequences. (Running on oeis4.)