

A166363


Number of primes in the halfopen interval (n*(log(n))^2..(n+1)*(log(n+1))^2].


5



0, 2, 2, 1, 3, 1, 2, 3, 2, 2, 3, 2, 2, 4, 1, 2, 3, 3, 3, 3, 2, 2, 5, 2, 3, 4, 1, 3, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 1, 3, 3, 5, 3, 4, 4, 3, 3, 3, 4, 3, 3, 4, 4, 4, 2, 3, 4, 3, 3, 4, 5, 3, 5, 4, 2, 3, 3, 6, 2, 4, 5, 3, 2, 2, 3, 6, 3, 6, 3, 4, 4, 6, 3, 4, 3, 4, 4, 4, 2, 3, 6, 3, 3, 2, 6, 5, 2, 6, 3, 5, 3, 2, 5, 4, 4
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OFFSET

1,2


COMMENTS

The openclosed halfopen intervals form a partition of the real line for x > 0, thus each prime appears in a unique interval.
The nth interval length is ~ (log(n+1/2))^2 + 2*log(n+1/2) ~ (log(n))^2 as n goes to infinity.
The nth interval prime density is ~ 1/[log(n+1/2)+2*log(log(n+1/2))] ~ 1/log(n) as n goes to infinity.
The expected number of primes in the nth interval is ~ [(log(n+1/2))^2 + 2*log(n+1/2)] / [log(n+1/2)+2*log(log(n+1/2))] ~ log(n) as n goes to infinity.
For n = 1 there is no prime.
If it can be proved that each interval always contains at least one prime, this would constitute even shorter intervals than A166332(n), let alone A143898(n), as n gets large.
The Shanks Conjecture and the CramerGranville Conjecture tell us that the intervals of length (log(n))^2 are of very critical length (the constant M > 1 of the CramerGranville Conjecture definitely matters!). There seems to be some risk that one such interval does not contain a prime.
The Wolf Conjecture (which agrees better with numerical evidence) seems more in favor of each interval's containing at least one prime.
From Charles R Greathouse IV, May 13 2010: (Start)
Not all intervals > 1 contain primes!
a(n) = 0 for n = 1, 4977, 17512, 147127, 76082969 (and no others up to 10^8).
Higher values include 731197850, 2961721173, 2103052050563, 188781483769833, 1183136231564246 but this list is not exhaustive.
The intervals have length (log n)^2 + 2*log n + o(1). In the Cramer model, the probability that a given integer in the interval would be prime is approximately 1/(log n + 2*log log n). Tedious calculation gives the probability that a(n) = 0 in the Cramer model as 3C(log n)^2/n * (1 + o(1)) with C = exp(5/2)/3. Thus under that model we would expect to find roughly C*(log N)^3 numbers n up to N with a(n) = 0. In fact, the numbers are not that common since the probabilities are not independent.
(End)


LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, PrimeGaps.
Eric Weisstein's World of Mathematics, CramerGranville Conjecture.
Eric Weisstein's World of Mathematics, Shanks Conjecture (and Wolf Conjecture.)


FORMULA

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


PROG

(PARI) a(n)=sum(k=ceil(n*log(n)^2), floor((n+1)*log(n+1)^2), isprime(k)) \\ Charles R Greathouse IV, Aug 21 2015


CROSSREFS

Cf. A166332, A000720, A111943, A143898, A134034, A143935, A144140 (primes between successive n^K, for different K), A014085 (primes between successive squares).
Cf. A182315.
Sequence in context: A320040 A072528 A227083 * A117470 A070786 A255542
Adjacent sequences: A166360 A166361 A166362 * A166364 A166365 A166366


KEYWORD

nonn


AUTHOR

Daniel Forgues, Oct 12 2009


EXTENSIONS

Edited by Daniel Forgues, Oct 18 2009 and Nov 01 2009
Edited by Charles R Greathouse IV, May 13 2010


STATUS

approved



