A166363 Number of primes in the half-open interval (n*(log(n))^2..(n+1)*(log(n+1))^2].

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

Offset: 1

Author

Daniel Forgues, Oct 12 2009

Keywords

• nonn

Comments

The open-closed half-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 ~ (log(n+1/2))^2 + 2*log(n+1/2) ~ (log(n))^2 as n goes to infinity.

The n-th 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 n-th 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 Cramer-Granville Conjecture tell us that the intervals of length (log(n))^2 are of very critical length (the constant M > 1 of the Cramer-Granville 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)

The similar sequence A345755 relies on intervals that are slightly more than twice as wide as those in the present sequence. A345755 does not include zero entries for n <= 2772, suggesting that the lengths of prime gaps may be bracketed by the two sequences. We conjecture that prime gaps may be larger than log(p)^2, but are not larger than log_2(p)^2. - Hal M. Switkay, Aug 29 2023

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Prime Gaps.
• Eric Weisstein's World of Mathematics, Cramer-Granville Conjecture.
• Eric Weisstein's World of Mathematics, Shanks Conjecture (and Wolf Conjecture.)

Crossrefs

Cf. A166332, A000720, A111943, A143898, A134034, A143935, A144140 (primes between successive n^K, for different K), A014085 (primes between successive squares).

Cf. A182315, A345755.

Programs

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

Formula

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

Extensions

Edited by Daniel Forgues, Oct 18 2009 and Nov 01 2009

Edited by Charles R Greathouse IV, May 13 2010