A166737 Number of primes in (n^2*log(n)..(n+1)^2*log(n+1)] semi-open intervals, n >= 1.

1, 3, 4, 4, 6, 6, 8, 8, 10, 11, 10, 13, 13, 14, 16, 14, 17, 20, 18, 21, 21, 22, 21, 24, 22, 30, 22, 31, 28, 25, 34, 32, 32, 33, 33, 34, 36, 38, 41, 35, 41, 40, 41, 45, 41, 41, 48, 49, 48, 49, 48, 48, 48, 54, 56, 54, 51, 56, 56, 61, 62, 57, 60, 62, 63, 59, 65, 66, 64, 65, 77, 67

Offset: 1

Author

Daniel Forgues, Oct 21 2009

Keywords

• nonn

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

Crossrefs

Cf. A166712 (for intervals containing an asymptotic average of one prime.)

Cf. A014085 (for primes between successive squares.)

Cf. A166332, A166363.

Cf. A000720.

Formula

a(n) = pi((n+1)^2*log(n+1)) - pi(n^2*log(n)) since the intervals are semi-open properly.

Extensions

Corrected and edited by Daniel Forgues, Oct 23 2009