OFFSET
1,2
COMMENTS
Number of primes in (n*(n*log(n))..(n+1)*((n+1)*log(n+1))] semi-open intervals, n >= 1.
The semi-open intervals form a partition of the real line for x > 0, thus each prime appears in a unique interval.
The n-th interval length is:
(n+1/2)*[2*log(n+1/2)+1]
2*n*log(n) as n goes to infinity
The n-th interval prime density is:
1/[2*log(n+1/2)+log(log(n+1/2))]
1/(2*log(n)) as n goes to infinity
The expected number of primes for n-th interval is:
(n+1/2)*[2*log(n+1/2)+1]/[2*log(n+1/2)+log(log(n+1/2))]
n as n goes to infinity
The actual number of primes for n-th interval seems to be (from graph): a(n) = n + O(n^(1/2))
The partial sums of this sequence give:
pi((n+1)^2*log(n+1)) = Sum_{i=1}^n {a(i)} ~ Sum_{i=1}^n {i} = t_n = n*(n+1)/2
LINKS
Daniel Forgues, Table of n, a(n) for n=1..141
FORMULA
a(n) = pi((n+1)^2*log(n+1)) - pi(n^2*log(n)) since the intervals are semi-open properly.
CROSSREFS
KEYWORD
nonn
AUTHOR
Daniel Forgues, Oct 21 2009
EXTENSIONS
Corrected and edited by Daniel Forgues, Oct 23 2009
STATUS
approved