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%I #25 Feb 18 2024 10:49:34
%S 1,1,2,3,3,3,3,3,4,5,6,7,8,9,10,11,11,11,11,11,11,11,11,11,11,11,11,
%T 11,11,11,11,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,
%U 30,31,32,33,34,35,36,37,38,39,40,41,42,43,43,43,43,43
%N Partial sums of the Jeffreys binary sequence A275973.
%C The ratio d(n) = a(n)/n, while obviously bounded, has no limit. Rather, it kind of 'oscillates', at an exponentially decreasing rate, between about 1/3 and 2/3. As mentioned by Jeffreys, the values of liminf and limsup of the set {d(n)} are 1/3 and 2/3, respectively. A proof of this fact by elementary means is relatively easy, for example, using the first formula below, but the following statement is a conjecture: Any real value c in the interval [1/3, 2/3] is an accumulation point of {d(n)}.
%H Alois P. Heinz, <a href="/A275974/b275974.txt">Table of n, a(n) for n = 1..16385</a> (first 2100 terms from Stanislav Sykora)
%H Hsien-Kuei Hwang, S. Janson, T.-H. Tsai, <a href="http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf">Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications</a>, Preprint 2016.
%H Hsien-Kuei Hwang, S. Janson, T.-H. Tsai, <a href="https://doi.org/10.1145/3127585">Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications</a>, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
%F For k>= 0 and 4^k <= m <= 2*4^k (i.e., m spanning a block of 0's in A275973), a(m) = 1 + 2*(1 + 4 + 4^2 + ... + 4^(k-1)) = 1/3 + (2/3)*4^k. [sequence reference corrected by _Peter Munn_, May 16 2019]
%F For any n, d(n) = a(n)/n > 1/3.
%F liminf_{n->infinity} d(n) = 1/3 and limsup_{n->infinity} d(n) = 2/3.
%p b:= n-> (p-> `if`(2^p=n, (-1)^p, 0))(ilog2(n)):
%p g:= proc(n) g(n):= `if`(n=1, 1, g(n-1)-b(n-1)) end:
%p a:= proc(n) a(n):= `if`(n<1, 0, a(n-1)+g(n)) end:
%p seq(a(n), n=1..68); # _Alois P. Heinz_, Feb 18 2024
%o (PARI) JeffreysSequence(nmax) = {
%o my(a=vector(nmax),n=0,p=1);a[n++]=1;
%o while(n<nmax,
%o for(k=2^(p-1)+1,2^p,a[n++]=0;if(n==nmax,break));
%o if(n<nmax,for(k=2^p+1,2^(p+1),a[n++]=1;if(n==nmax,break)));
%o p+=2;);
%o return(a);}
%o a = JeffreysSequence(2100);
%o for(n=2,#a,a[n]+=a[n-1]);a
%Y Cf. A275973.
%K nonn
%O 1,3
%A _Stanislav Sykora_, Aug 15 2016
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