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 A275974 Partial sums of the Jeffreys binary sequence A275973. 2
 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, 11, 11, 11, 11, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 43, 43, 43, 43 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS 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)}. LINKS Stanislav Sykora, Table of n, a(n) for n = 1..2100 Hsien-Kuei Hwang, S. Janson, T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint 2016. Hsien-Kuei Hwang, S. Janson, T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585. FORMULA 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] For any n, d(n) = a(n)/n > 1/3. liminf_{n->infinity} d(n) = 1/3 and limsup_{n->infinity} d(n) = 2/3. PROG (PARI) JeffreysSequence(nmax) = {   my(a=vector(nmax), n=0, p=1); a[n++]=1;   while(n

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Last modified July 12 21:31 EDT 2020. Contains 335669 sequences. (Running on oeis4.)