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 A010171 Continued fraction for sqrt(103). 2
 10, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20 (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 linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,1). FORMULA a(n) = (1/264)*(-291*(n mod 12) - 93*((n+1) mod 12) + 39*((n+2) mod 12) - 5*((n+3) mod 12) + 17*((n+4) mod 12) + 193*((n+5) mod 12) - 159*((n+6) mod 12) + 17*((n+7) mod 12) + 39*((n+8) mod 12) - 5*((n+9) mod 12) + 127*((n+10) mod 12) + 325*((n+11) mod 12)) - 10*(C(2*n,n) mod 2), with n >= 0. - Paolo P. Lava, Jul 24 2009 MATHEMATICA ContinuedFraction[Sqrt[103], 300] (* Vladimir Joseph Stephan Orlovsky, Mar 10 2011 *) PROG (Python) from sympy import sqrt from sympy.ntheory.continued_fraction import continued_fraction_iterator def aupton(terms): gen = continued_fraction_iterator(sqrt(103)) return [next(gen) for i in range(terms)] print(aupton(85)) # Michael S. Branicky, Oct 06 2021 CROSSREFS Cf. A187768 (sqrt(103)). Sequence in context: A276348 A281400 A283726 * A006518 A094175 A193952 Adjacent sequences: A010168 A010169 A010170 * A010172 A010173 A010174 KEYWORD nonn,cofr AUTHOR STATUS approved

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