

A166332


Number of primes in (n^(3/2)*(log(n))^(1/2)..(n+1)^(3/2)*(log(n+1))^(1/2)] semiopen intervals, n >= 1.


3



1, 2, 1, 2, 2, 1, 2, 1, 3, 1, 2, 3, 2, 1, 3, 3, 1, 3, 2, 3, 3, 2, 2, 2, 4, 2, 3, 4, 1, 4, 1, 4, 2, 4, 2, 3, 4, 4, 2, 4, 3, 1, 3, 4, 4, 4, 4, 3, 3, 3, 4, 3, 3, 3, 5, 4, 4, 2, 3, 3, 5, 3, 5, 5, 4, 4, 2, 3, 4, 5, 3, 5, 5, 2, 3, 2, 5, 5, 6, 3, 4, 5, 6, 3, 4, 4, 4, 4, 5, 2, 5, 5, 3, 3, 6, 5, 3, 6, 6, 3, 3, 4, 5, 5, 5
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OFFSET

1,2


COMMENTS

Number of primes in (n*(n*log(n))^(1/2)..(n+1)*((n+1)*log(n+1))^(1/2)] 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.
a(n) = pi((n+1)^(3/2)*(log(n+1))^(1/2))  pi(n^(3/2)*(log(n))^(1/2)) since the intervals are semiopen properly.
The nth interval length is: ~ (1/2)*(n+1/2)^(1/2)*(3*(log(n+1/2))^(1/2)+(log(n+1/2))^(1/2)) ~ (3/2)*n^(1/2)*(log(n))^(1/2) as n goes to infinity.
The nth interval prime density is: ~ 2/(3*log(n+1/2)+log(log(n+1/2))) ~ 2/(3*log(n)) as n goes to infinity.
The expected number of primes for nth interval is: ~ (n+1/2)^(1/2)*(3*(log(n+1/2))^(1/2)+(log(n+1/2))^(1/2))/ (3*log(n+1/2)+log(log(n+1/2))) ~ n^(1/2)/(log(n))^(1/2) as n goes to infinity.
Using Excel 2003, for n in [1..1123], I obtain a(n) >= 1 (at least one prime per interval).
CAUTION: I will submit the bfile, but since Excel 2003 is limited to 15digit precision, the rounding might assign to the wrong interval a prime which is extremely close to the limit of 2 successive intervals. The bfile NEEDS TO BE VERIFIED using interval arithmetic! (SEE NEXT)
CAUTION (ADDENDA): for n in [1..1123], the minimum ratio of... ABS(n^(3/2)*(log(n))^(1/2)ROUND(n^(3/2)*(log(n))^(1/2)))/(n^(3/2)*(log(n))^(1/2)) that I got is 5.04999E09 which is well above 1E15 (15digit limit of Excel 2003), so no interval ended too close to an integral value and every prime has then been assigned to its proper interval. My bfile should then be reliable.
If it can be proved that each interval always contains at least one prime, this would constitute shorter intervals than A143898(n) as n gets large.
The sequence A166363 gives even shorter intervals that seem to always contain at least one prime (for n > 1)!


LINKS

Daniel Forgues, Table of n, a(n) for n = 1..1123


CROSSREFS

Cf. A143898, A134034, A143935 (for primes between successive n^K, for different K).
Cf. A144140 (showing that for n^K, K=3/2, some intervals fails to contain primes).
Cf. A166363 (for primes in even shorter intervals).
Cf. A014085 (for primes between successive squares).
Cf. A000720.
Sequence in context: A014675 A308186 A107362 * A022303 A113189 A143098
Adjacent sequences: A166329 A166330 A166331 * A166333 A166334 A166335


KEYWORD

nonn


AUTHOR

Daniel Forgues, Oct 12 2009


EXTENSIONS

Corrected and edited by Daniel Forgues, Oct 14 2009
Edited by Daniel Forgues, Oct 20 2009


STATUS

approved



