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A071679 Least k such that the maximum number of elements among the continued fractions for k/1, k/2, k/3, k/4, ..., k/k equals n.

%I A071679 #38 Sep 08 2022 08:45:06
%S A071679 1,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,
%T A071679 17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269,
%U A071679 2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155
%N A071679 Least k such that the maximum number of elements among the continued fractions for k/1, k/2, k/3, k/4, ..., k/k equals n.
%H A071679 Seiichi Manyama, <a href="/A071679/b071679.txt">Table of n, a(n) for n = 1..4784</a>
%H A071679 Mohammad K. Azarian, <a href="http://www.m-hikari.com/ijcms/ijcms-2012/37-40-2012/azarianIJCMS37-40-2012.pdf">Fibonacci Identities as Binomial Sums</a>, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (See Corollary 1 (x)).
%H A071679 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).
%F A071679 Smallest k such that n = Max_{ i=1..k: Card[ contfrac(k/i) ] }.
%F A071679 a(1) = 1; for n>1 a(n) = F(n+2) where F(n)=A000045(n) are the Fibonacci numbers.
%F A071679 G.f.: (1+x)^2/(1-x-x^2); a(n) = 3*F(n+1) - F(n-1) - 0^n. - _Paul Barry_, Jul 26 2004
%F A071679 a(n) = Fibonacci(n+2) for n > 1. - _Charles R Greathouse IV_, Jan 17 2012
%t A071679 Join[{1}, LinearRecurrence[{1, 1}, {3, 5}, 50]] (* _Vincenzo Librandi_, Jul 12 2015 *)
%o A071679 (PARI) a(n)=if(n>1,fibonacci(n+2),1) \\ _Charles R Greathouse IV_, Jan 17 2012
%o A071679 (Magma) [1] cat [Fibonacci(n+2): n in [2..50]]; // _Vincenzo Librandi_, Jul 12 2015
%Y A071679 Cf. A000045, A324969.
%K A071679 easy,nonn
%O A071679 1,2
%A A071679 _Benoit Cloitre_, Jun 22 2002