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 A166363 Number of primes in the half-open 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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) 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, Cramer-Granville 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 half-open 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: A072528 A227083 A327571 * 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

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Last modified January 29 16:58 EST 2020. Contains 331347 sequences. (Running on oeis4.)