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A010186
Continued fraction for sqrt(125).
5
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
0,1
COMMENTS
Periodic with period [5,1,1,5,22] of length 5 (after the initial term). - M. F. Hasler, Sep 09 2011
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
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 *)
PROG
(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
CROSSREFS
Cf. A172074.
Sequence in context: A141240 A298892 A038318 * A097531 A131029 A033331
KEYWORD
nonn,cofr,easy,nice
STATUS
approved