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

1,5

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

Prog

(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

Crossrefs

Cf. A005117, A013647, A013943.

Sequence in context: A285631 A316836 A058612 * A285118 A128208 A154721

Adjacent sequences: A099722 A099723 A099724 * A099726 A099727 A099728

Keyword

nonn

Author

Benoit Cloitre, Nov 07 2004

Status

approved