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.

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, 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, 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

Offset: 1

Author

Benoit Cloitre, Nov 07 2004

Keywords

• nonn

Comments

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

The "n-th nonsquare integer" in the definition is A005117(n + 1). - Michael B. Porter, Jun 06 2018

Links

• A.H.M. Smeets, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A005117, A013647, A013943.

Programs

Python

from math import isqrt
from sympy.ntheory.continued_fraction import continued_fraction_periodic
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