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
KEYWORD
nonn
AUTHOR
Stanislav Sykora, Aug 15 2016
STATUS
approved