cp's OEIS Frontend

A099725 a(n) is the number of 1's in the period of the continued fraction of the square root of the n-th nonsquare integer.

%I A099725 #30 Jul 20 2024 15:45:21
%S A099725 0,1,0,0,3,1,0,0,0,4,2,1,0,0,2,0,4,2,2,1,0,0,0,2,0,4,3,2,2,1,0,0,0,0,
%T A099725 0,0,6,6,2,6,2,1,0,0,2,2,2,0,0,4,6,2,2,4,2,1,0,0,4,0,2,2,2,0,4,4,3,6,
%U A099725 2,2,2,1,0,0,0,2,6,0,3,0,0,5,4,6,8,2,2,8,2,1,0,0,6,0,0,4,2,4,4,0,4,4,6,2,7
%N A099725 a(n) is the number of 1's in the period of the continued fraction of the square root of the n-th nonsquare integer.
%C A099725 For sufficiently large period lengths, the fraction of 1's in the repeating part tends to log(4/3)/log(2) = 0.415... as from the Gauss-Kuzmin distribution, i.e., a(n) tends to 0.415...*A013943(n) for sufficiently large A013943(n). - _A.H.M. Smeets_, Jun 02 2018
%C A099725 The "n-th nonsquare integer" in the definition is A005117(n + 1). - _Michael B. Porter_, Jun 06 2018
%H A099725 A.H.M. Smeets, <a href="/A099725/b099725.txt">Table of n, a(n) for n = 1..10000</a>
%o A099725 (Python)
%o A099725 from math import isqrt
%o A099725 from sympy.ntheory.continued_fraction import continued_fraction_periodic
%o A099725 def A099725(n): return (continued_fraction_periodic(0,1,n+(k:=isqrt(n))+int(n>=k*(k+1)+1))[-1]).count(1) # _Chai Wah Wu_, Jul 20 2024
%Y A099725 Cf. A005117, A013647, A013943.
%K A099725 nonn
%O A099725 1,5
%A A099725 _Benoit Cloitre_, Nov 07 2004