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Search: id:A166737
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| A166737 |
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Number of primes in (n^2*log(n)..(n+1)^2*log(n+1)] semi-open intervals, n >= 1. |
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1
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| 1, 3, 4, 4, 6, 6, 8, 8, 10, 11, 10, 13, 13, 14, 16, 14, 17, 20, 18, 21, 21, 22, 21, 24, 22, 30, 22, 31, 28, 25, 34, 32, 32, 33, 33, 34, 36, 38, 41, 35, 41, 40, 41, 45, 41, 41, 48, 49, 48, 49, 48, 48, 48, 54, 56, 54, 51, 56, 56, 61, 62, 57, 60, 62, 63, 59, 65, 66, 64, 65, 77, 67
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Number of primes in (n*(n*log(n))..(n+1)*((n+1)*log(n+1))] 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.
The n_th interval length is:
~ (n+1/2)*[2*log(n+1/2)+1]
~ 2*n*log(n) as n goes to infinity
The n_th interval prime density is:
~ 1/[2*log(n+1/2)+log(log(n+1/2))]
~ 1/(2*log(n)) as n goes to infinity
The expected number of primes for n_th interval is:
~ (n+1/2)*[2*log(n+1/2)+1]/[2*log(n+1/2)+log(log(n+1/2))]
~ n as n goes to infinity
The actual number of primes for n_th interval seems to be (from graph): a(n) = n + O(n^(1/2))
The partial sums of this sequence give:
pi((n+1)^2*log(n+1)) = Sum_{i=1}^n {a(i)} ~ Sum_{i=1}^n {i} = t_n = n*(n+1)/2
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LINKS
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Daniel Forgues, Table of n, a(n) for n=1..141
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FORMULA
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a(n) = pi((n+1)^2*log(n+1)) - pi(n^2*log(n)) since the intervals are semi-open properly.
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CROSSREFS
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Cf. A166712 (for intervals containing an asymptotic average of one prime.)
Cf. A014085 (for primes between succesive squares.)
Cf. A166332, A166363.
Cf. A000720.
Sequence in context: A061117 A111234 A014683 this_sequence A088847 A120613 A164326
Adjacent sequences: A166734 A166735 A166736 this_sequence A166738 A166739 A166740
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KEYWORD
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nonn
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AUTHOR
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Daniel Forgues (squid(AT)zensearch.com), Oct 21 2009
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EXTENSIONS
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Corrected and edited by Daniel Forgues (squid(AT)zensearch.com), Oct 23 2009
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