cp's OEIS Frontend

This value of K is conjectured to be the least possible such that there is at least one prime in the range n^K to (n+1)^K for n > 0. This value of K was found using exact interval arithmetic. For each n <= 110 and for each prime p in the range n to n^1.7, we computed an interval k(n,p) such that p is between n^k(n,p) and (n+1)^k(n,p). The intersection of all these intervals produces a list of intervals. The least value in those intervals is K, which is log(1151)/log(95). We computed 10^5 terms of this sequence to give us confidence that a(n) > 0 for all n.

More details about the algorithm: The n^1.7 limit was chosen because we were fairly certain that K would be less than 1.7. Let k(n) be the union of the intervals k(n,p) for p < n^1.7. Then k(n) is the set of exponents e such that the range n^e to (n+1)^e always contains a prime. Let k be the intersection of all the k(n) intervals for n up to N. Then k is the set of exponents e such that there is always a prime in the range n^e to (n+1)^e for n <= N. The number K is the least number in the set k. It appears that as N becomes larger, the set k converges. See A143935. - T. D. Noe, Sep 08 2008

The constant log(1151)/log(95) is A194362. - John W. Nicholson, Nov 25 2013

1151 counts as a qualifying prime towards both a(94)=1 and a(95)=3, in accordance with use of closed ranges. If prime p were counted only when n^K < p <= (n+1)^K, then term 95 would be 2. If prime p were counted only when n^K <= p < (n+1)^K, then term 94 would be 0. The conjecture in the author's comment implies K is the greatest real value such that for all e <= K there exists n > 0 with no prime p satisfying n^e <= p < (n+1)^e. - Peter Munn, Mar 02 2017

The author's description of the calculation of K implies that K is not an isolated qualifying value; equivalently that K is also the least real value for which there is a positive epsilon such that for all exponent e, K <= e <= K+epsilon and integer n > 0 there is a prime p satisfying n^e <= p <= (n+1)^e. This is a necessary precondition for my Mar 02 2017 deduction from the author's conjecture. - Peter Munn, Aug 21 2019

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

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

The semi-open 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 semi-open properly.

The n-th 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 n-th 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 n-th 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 b-file, but since Excel 2003 is limited to 15-digit precision, the rounding might assign to the wrong interval a prime which is extremely close to the limit of 2 successive intervals. The b-file 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 5.04999E-09 which is well above 1E-15 (15-digit 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 b-file 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)!