A141822 Maximum term in the continued fraction of A141821(n)/n.

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

Offset: 2

Author

T. D. Noe, Jul 08 2008

Keywords

• nonn

Comments

Consider the continued fraction [0;c1,c2,...,cm] of k/n, with k

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).

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).

Programs

Mathematica

Table[c=ContinuedFraction[Select[Range[n-1],GCD[ #,n]==1&]/n]; Min[Max/@c], {n, 150}]

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

Extensions

Edited by Max Alekseyev, Sep 25 2011

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