This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A089636 #24 Jul 14 2019 21:59:29 %S A089636 1,3,5,9,23,39,79,158,281,741,1145,2297,4495,10111,20223,40446,80983, %T A089636 162009,323369,646271,1216723,2592211,5184422,9733109,20739329, %U A089636 41467565,81559503,163649289,311481083,662667007,1325334527,2628708543,5308871023,9627863373 %N A089636 Least k, 1 <= k <= 2^n, such that the continued fraction for 2^n/k contains the maximum number of elements. %C A089636 From _Jon E. Schoenfield_, Nov 05 2014: (Start) %C A089636 Several terms are close to 2^n / phi, where phi = (1 + sqrt(5))/2 = 1.6180339... (see A001622); e.g., 2^22/a(22) = 4194304/2592211 = 1.6180411... . %C A089636 When a ratio r of two integers is expressed as a continued fraction, it cannot have a relatively large number of elements (i.e., relative to other fractions whose numerators and denominators are very roughly the same size as the numerator and denominator of r, respectively) if any of the elements of the continued fraction are large, so the elements of the continued fractions for 2^n / a(n) tend to consist only of small numbers, mostly ones; e.g., 131072/80983 = cf[1; 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 3] (20 elements, consisting of 17 ones, 2 threes, and 1 five). %C A089636 For n = 1..34, the maximum number of elements is 1, 2, 4, 4, 5, 7, 9, 9, 10, 12, 12, 14, 15, 17, 19, 19, 20, 22, 22, 24, 26, 27, 27, 29, 30, 32, 33, 34, 35, 37, 39, 39, 41, 41. %C A089636 If a(n) is even, then a(n) = 2*a(n-1), so 2^n/a(n) reduces to 2^(n-1)/a(n-1), and the maximum number of elements is the same at n as it is at n-1. Up to n=34, a(n) is even only at n = 8, 16, and 23. (End) %F A089636 a(n) = A084242(2^n). %e A089636 From _Jon E. Schoenfield_, Nov 05 2014: (Start) %e A089636 The continued fractions for 2^3/k for k = 1..2^3 are %e A089636 8/1 = 8 (1 element) %e A089636 8/2 = 4 (1 element) %e A089636 8/3 = 2 + 1/(1 + 1/2) = cf[2;1,2] (3 elements) %e A089636 8/4 = 2 (1 element) %e A089636 8/5 = 1 + 1/(1 + 1/(1 + 1/2)) = cf[1;1,1,2] (4 elements) %e A089636 8/6 = 4/3 = 1 + 1/3 = cf[1;3] (2 elements) %e A089636 8/7 = 1 + 1/7 = cf[1;7] (2 elements) %e A089636 8/8 = 1 (1 element) %e A089636 so the first (and, in this case, only) value of k at which the maximum number of elements (i.e., 4) occurs is k=5; thus, a(3)=5. (End) %Y A089636 Cf. A084242. %K A089636 nonn %O A089636 1,2 %A A089636 _Benoit Cloitre_, Jan 01 2004 %E A089636 More terms from _Jon E. Schoenfield_, Nov 05 2014