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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 comprised in the interval [1/3,2/3] is an accumulation point of {d(n)}. REFERENCES Hsien-Kuei Hwang, S Janson, TH 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; http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf. Also 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 LINKS Stanislav Sykora, Table of n, a(n) for n = 1..2100 FORMULA For k>=0 and 4^k <= m <= 2*4^k (i.e., m spanning a block of 0's in A275974), a(m) = 1+2*(1+4+4^2+...+4^(k-1)) = (1/3)+(2/3)*4^k. 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 December 18 21:01 EST 2018. Contains 318245 sequences. (Running on oeis4.)