

A166737


Number of primes in (n^2*log(n)..(n+1)^2*log(n+1)] semiopen intervals, n >= 1.


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
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OFFSET

1,2


COMMENTS

Number of primes in (n*(n*log(n))..(n+1)*((n+1)*log(n+1))] semiopen intervals, n >= 1.
The semiopen intervals form a partition of the real line for x > 0, thus each prime appears in a unique interval.
The nth interval length is:
(n+1/2)*[2*log(n+1/2)+1]
2*n*log(n) as n goes to infinity
The nth 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 nth 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 nth 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 semiopen properly.


CROSSREFS

Cf. A166712 (for intervals containing an asymptotic average of one prime.)
Cf. A014085 (for primes between successive squares.)
Cf. A166332, A166363.
Cf. A000720.
Sequence in context: A229022 A014683 A213222 * A088847 A120613 A277193
Adjacent sequences: A166734 A166735 A166736 * A166738 A166739 A166740


KEYWORD

nonn


AUTHOR

Daniel Forgues, Oct 21 2009


EXTENSIONS

Corrected and edited by Daniel Forgues, Oct 23 2009


STATUS

approved



