A160520 a(n) is the number of distinct values in the periodic part of the continued fraction of the square root of n-th nonsquare positive integer.
1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 2, 1, 2, 4, 2, 3, 4, 3, 2, 1, 2, 3, 3, 2, 4, 2, 3, 3, 2, 1, 2, 2, 2, 2, 2, 4, 3, 3, 5, 3, 2, 1, 2, 4, 3, 4, 2, 2, 3, 2, 4, 3, 5, 3, 2, 1, 2, 5, 2, 4, 3, 4, 2, 3, 2, 2, 5, 4, 3, 3, 2, 1, 2, 2
Offset: 1
Keywords
Examples
The 10th nonsquare positive integer is 13; sqrt(13) = cf[3;(1,1,1,1,6)], which has period 5, and 2 distinct values in its periodic part.
Programs
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Mathematica
Length@Union@Last@ContinuedFraction@Sqrt@#&/@Select[Range@100,!IntegerQ@Sqrt@#&] (* Giorgos Kalogeropoulos, May 05 2022 *)
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Python
import math def findperiod( d ) : if len(d) == 0 : return 0 for p in range( 1, len(d) - 1 ) : isPeriod = True for i in range(0, p) : for j in range(i + p, len(d), p ) : if not d[i] == d[j] : isPeriod = False break if not isPeriod : break if isPeriod : return len( set( d[:p] ) ) return -1 def nextv( a, b, n, less = True ) : # print a, b, a[1]*a[1], n * a[0] * a[0] d = -1 while (a[1]*a[1] < n * a[0] * a[0]) == less : d += 1 a = ( a[0] + b[0], a[1] + b[1] ) a = ( a[0] - b[0], a[1] - b[1] ) return d, a, b def generated( n ) : maxperiod = 100 s = int( math.sqrt( n ) ) if s * s == n : return [] a = (1, 0) b = (0, 1) ds = [] for i in range( maxperiod ): d, b, a = nextv( a, b, n ) ds.append( d ) d, b, a = nextv( a, b, n, less = False ) ds.append( d ) return ds[1:] maxn = 1000 ns = [x for x in range( maxn ) if not int( math.sqrt( x ) ) ** 2 == x ] v = list(map( findperiod, map( generated, ns ) )) if v.count( -1 ) == 0 : print(v) -
Python
from math import isqrt from sympy.ntheory.continued_fraction import continued_fraction_periodic def A160520(n): return len(set(continued_fraction_periodic(0,1,n+(k:=isqrt(n))+int(n>=k*(k+1)+1))[-1])) # Chai Wah Wu, Jul 20 2024