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%I #28 Jan 26 2021 10:17:48
%S 161,1,4,11,1,1,3,6,1,13,8,1,6,1,4,1,1,2,1,1,1,1,13,2,1,3,8,1,2,19,1,
%T 54,1,19,2,1,8,3,1,2,13,1,1,1,1,2,1,1,4,1,6,1,8,13,1,6,3,1,1,11,4,1,
%U 222
%N Continued fraction expansion of sqrt(12500)+50 = 100*phi, where phi=(sqrt(5)+1)/2 is the golden ratio.
%C The 62 trailing terms are repeated infinitely.
%C This is just one of an infinite set of continued fractions, related to the golden ratio, and more specifically to the square root of 125, 12500, 1250000...
%C Taking phi*10^k, one can look at sqrt(125) + 5, sqrt(12500) + 50 (this sequence, sqrt(1250000) + 500, etc.
%C This is not an efficient way to calculate phi - _Franklin T. Adams-Watters_, Sep 10 2011.
%C Periodic with a period of length 62, starting right after the initial term. Moreover, the sequence is symmetric when any 54 or 222 is taken as central value (cf. formula). - _M. F. Hasler_, Sep 09 2011
%F a(31*k - n) = a(31*k + n), for all n < 31k, k > 0. - _M. F. Hasler_, Sep 09 2011
%t ContinuedFraction[N[Sqrt[12500], 50000], 63]
%t ContinuedFraction[100*GoldenRatio,100] (* _Harvey P. Dale_, Dec 30 2018 *)
%o (PARI) default(realprecision, 199); contfrac((sqrt(5)+1)/.02) \\ _M. F. Hasler_, Sep 09 2011
%o (PARI) a(n)=[222-61*!n, 1, 4, 11, 1, 1, 3, 6, 1, 13, 8, 1, 6, 1, 4, 1, 1, 2, 1, 1, 1, 1, 13, 2, 1, 3, 8, 1, 2, 19, 1, 54][32-abs(n%62-31)] \\ _M. F. Hasler_, Sep 09 2011
%Y Cf. A001622, A010186.
%K cofr,nonn,nice
%O 0,1
%A _Shane Findley_, Jan 25 2010
%E Clarified the definition, following an observation by _Franklin T. Adams-Watters_. _M. F. Hasler_, Sep 09 2011
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