A010186 Continued fraction for sqrt(125).

11, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5, 22, 5, 1, 1, 5

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• cofr
• easy
• nice

Comments

Periodic with period [5,1,1,5,22] of length 5 (after the initial term). - M. F. Hasler, Sep 09 2011

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..300
• G. Xiao, 'Contfrac' tool on WIMS.
• Index entries for continued fractions for constants
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1).

Crossrefs

Cf. A172074.

Programs

Mathematica

ContinuedFraction[Sqrt[125], 300] (* Vladimir Joseph Stephan Orlovsky, Mar 12 2011 *)
LinearRecurrence[{0,0,0,0,1},{11,5,1,1,5,22},80] (* or *) PadRight[ {11},80,{22,5,1,1,5}] (* Harvey P. Dale, Oct 05 2016 *)

PARI

default(realprecision,199); contfrac(sqrt(125))  \\ _M. F. Hasler, Sep 09 2011

PARI

a(n)=[22, 5, 1, 1, 5][n%5+1]-11*!n  \\ M. F. Hasler, Sep 09 2011

Python

from sympy import sqrt
from sympy.ntheory.continued_fraction import continued_fraction_iterator
def aupton(nn):
    gen = continued_fraction_iterator(sqrt(125))
    return [next(gen) for i in range(nn+1)]
print(aupton(74)) # Michael S. Branicky, Nov 05 2021

Python

# second version based on linear recurrence
def a(n): return 11 if n == 0 else [5, 1, 1, 5, 22][(n-1)%5]
print([a(n) for n in range(75)]) # Michael S. Branicky, Nov 05 2021

Formula

G.f.: (11+5*x+x^2+x^3+5*x^4+11*x^5)/((1-x)*(1+x+x^2+x^3+x^4)). - Bruno Berselli, Sep 10 2011

a(n) = a(n-5) for n >= 6. - Wesley Ivan Hurt, Mar 01 2023