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A166332
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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.
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3
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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, 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, 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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| 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 digits 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 digits limit of Excel 2003), so every interval did not end 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)!
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LINKS
| Daniel Forgues, Table of n, a(n) for n=1..1123
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CROSSREFS
| Cf. A143898, A134034, A143935 (for primes between successive n^K, for different K.)
Cf. A144140 (showing that for n^K, K=3/2, some intervals fails to contain primes.)
Cf. A166363 (for primes in even shorter intervals.)
Cf. A014085 (for primes between successive squares.)
Cf. A000720.
Sequence in context: A001468 A014675 A107362 * A022303 A113189 A143098
Adjacent sequences: A166329 A166330 A166331 * A166333 A166334 A166335
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KEYWORD
| nonn
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AUTHOR
| Daniel Forgues (squid(AT)zensearch.com), Oct 12 2009
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EXTENSIONS
| Corrected and edited by Daniel Forgues (squid(AT)zensearch.com), Oct 14 2009
Edited by Daniel Forgues (squid(AT)zensearch.com), Oct 20 2009
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