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%I #10 Jan 14 2021 21:15:27
%S 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,
%T 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,
%U 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
%N Number of primes in (n^(3/2)*(log(n))^(1/2)..(n+1)^(3/2)*(log(n+1))^(1/2)] semi-open intervals, n >= 1.
%C Number of primes in (n*(n*log(n))^(1/2)..(n+1)*((n+1)*log(n+1))^(1/2)] semi-open intervals, n >= 1.
%C The semi-open intervals form a partition of the real line for x > 0, thus each prime appears in a unique interval.
%C 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.
%C 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.
%C 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.
%C 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.
%C Using Excel 2003, for n in [1..1123], I obtain a(n) >= 1 (at least one prime per interval).
%C 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)
%C 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.
%C 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.
%C The sequence A166363 gives even shorter intervals that seem to always contain at least one prime (for n > 1)!
%H Daniel Forgues, <a href="/A166332/b166332.txt">Table of n, a(n) for n = 1..1123</a>
%Y Cf. A143898, A134034, A143935 (for primes between successive n^K, for different K).
%Y Cf. A144140 (showing that for n^K, K=3/2, some intervals fails to contain primes).
%Y Cf. A166363 (for primes in even shorter intervals).
%Y Cf. A014085 (for primes between successive squares).
%Y Cf. A000720.
%K nonn
%O 1,2
%A _Daniel Forgues_, Oct 12 2009
%E Corrected and edited by _Daniel Forgues_, Oct 14 2009
%E Edited by _Daniel Forgues_, Oct 20 2009
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