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A010172 Continued fraction for sqrt(106). 3
10, 3, 2, 1, 1, 1, 1, 2, 3, 20, 3, 2, 1, 1, 1, 1, 2, 3, 20, 3, 2, 1, 1, 1, 1, 2, 3, 20, 3, 2, 1, 1, 1, 1, 2, 3, 20, 3, 2, 1, 1, 1, 1, 2, 3, 20, 3, 2, 1, 1, 1, 1, 2, 3, 20, 3, 2, 1, 1, 1, 1, 2, 3, 20, 3, 2, 1, 1, 1, 1, 2, 3, 20, 3, 2, 1, 1, 1, 1, 2, 3, 20, 3, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..999

G. Xiao, Contfrac

Index entries for continued fractions for constants

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

FORMULA

a(n) = (1/162)*(-289*(n mod 9) - ((n+1) mod 9) - ((n+2) mod 9) + 17*((n+3) mod 9) + 17*((n+4) mod 9) + 17*((n+5) mod 9) + 35*((n+6) mod 9) + 35*((n+7) mod 9) + 323*((n+8) mod 9)) - 10*(C(2*n,n) mod 2), with n >= 0. - Paolo P. Lava, Jul 24 2009

MATHEMATICA

ContinuedFraction[Sqrt[106], 300] (* Vladimir Joseph Stephan Orlovsky, Mar 11 2011 *)

PadRight[{10}, 120, {20, 3, 2, 1, 1, 1, 1, 2, 3}] (* Harvey P. Dale, Aug 19 2021 *)

PROG

(Python)

from sympy import sqrt

from sympy.ntheory.continued_fraction import continued_fraction_iterator

def aupton(terms):

gen = continued_fraction_iterator(sqrt(106))

return [next(gen) for i in range(terms)]

print(aupton(82)) # Michael S. Branicky, Oct 03 2021

CROSSREFS

Sequence in context: A159005 A144859 A280519 * A224365 A087869 A167764

Adjacent sequences: A010169 A010170 A010171 * A010173 A010174 A010175

KEYWORD

nonn,cofr

AUTHOR

N. J. A. Sloane

STATUS

approved

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Last modified March 28 03:48 EDT 2023. Contains 361577 sequences. (Running on oeis4.)