A010187 Continued fraction for sqrt(126).

11, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4, 22, 4, 2, 4

Offset

0,1

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..100

G. Xiao, Contfrac

Index entries for continued fractions for constants

Index entries for periodic sequences.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Mathematica

ContinuedFraction[Sqrt[126], 300] (* Vladimir Joseph Stephan Orlovsky, Mar 12 2011 *)

Prog

(Python)
from sympy import sqrt
from sympy.ntheory.continued_fraction import continued_fraction_iterator
def aupton(nn):
    gen = continued_fraction_iterator(sqrt(126))
    return [next(gen) for i in range(nn+1)]
print(aupton(71)) # Michael S. Branicky, Nov 04 2021
(Python) # second version based on linear recurrence
def a(n): return 11 if n == 0 else [4, 2, 4, 22][(n-1)%4]
print([a(n) for n in range(72)]) # Michael S. Branicky, Nov 04 2021
(PARI) a(n)=if(n, [22, 4, 2, 4][n%4+1], 11) \\ Charles R Greathouse IV, May 31 2026

Crossrefs

Sequence in context: A088073 A385331 A352162 * A381151 A372761 A248175

Adjacent sequences: A010184 A010185 A010186 * A010188 A010189 A010190

Keyword

nonn,cofr,easy

Author

N. J. A. Sloane

Status

approved