A275974 Partial sums of the Jeffreys binary sequence A275973.

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

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

Alois P. Heinz, Table of n, a(n) for n = 1..16385 (first 2100 terms from Stanislav Sykora)

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.

Maple

b:= n-> (p-> `if`(2^p=n, (-1)^p, 0))(ilog2(n)):
g:= proc(n) g(n):= `if`(n=1, 1, g(n-1)-b(n-1)) end:
a:= proc(n) a(n):= `if`(n<1, 0, a(n-1)+g(n)) end:
seq(a(n), n=1..68);  # Alois P. Heinz, Feb 18 2024

Prog

(PARI) JeffreysSequence(nmax) = {
  my(a=vector(nmax), n=0, p=1); a[n++]=1;
  while(n<nmax,
    for(k=2^(p-1)+1, 2^p, a[n++]=0; if(n==nmax, break));
    if(n<nmax, for(k=2^p+1, 2^(p+1), a[n++]=1; if(n==nmax, break)));
    p+=2; );
  return(a); }
a = JeffreysSequence(2100);
for(n=2, #a, a[n]+=a[n-1]); a

Crossrefs

Cf. A275973.

Sequence in context: A327041 A277267 A246262 * A052288 A280455 A284725

Adjacent sequences: A275971 A275972 A275973 * A275975 A275976 A275977

Keyword

nonn

Author

Stanislav Sykora, Aug 15 2016

Status

approved