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A345755
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a(n) is the number of primes p satisfying n*(log_2(n))^2 < p <= (n+1)*(log_2(n+1))^2.
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2
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1, 3, 2, 3, 3, 4, 4, 4, 5, 3, 4, 4, 6, 3, 5, 7, 3, 4, 6, 5, 5, 7, 5, 3, 6, 6, 7, 6, 4, 6, 5, 7, 5, 6, 5, 6, 7, 6, 8, 4, 6, 6, 9, 3, 5, 7, 9, 5, 7, 9, 4, 7, 7, 5, 7, 6, 5, 9, 7, 8, 3, 7, 8, 8, 8, 6, 4, 7, 6, 8, 10, 7, 8, 7, 6, 7, 6, 6, 6, 7, 7, 10, 4, 8, 9, 7
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OFFSET
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1,2
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COMMENTS
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Prime gaps appear to grow more slowly than any power function.
Cramér's conjecture states that prime gaps grow as follows: prime(n+1) - prime(n) = O(log(prime(n))^2).
Since prime(n) ~ n*log(n), we conjecture that a(n) > 0 for n > 0, and that the exponent 2 cannot be replaced by any smaller exponent.
Note: n*(log_2(n))^2 < n^(log(127)/log(16)) when n >= 267. Therefore the conjecture immediately above is stronger than the conjecture that A143935(n) > 0 when n > 0, which in turn is stronger than Legendre's conjecture.
This sequence relies on intervals that are slightly more than twice as wide as those in the similar sequence A166363. A comment at that sequence by Greathouse discovers zero values (representing prime-free intervals). In contrast, the present sequence 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
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LINKS
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EXAMPLE
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a(10) is the number of primes > 110.35 and <= 131.64. a(10) = 3, because the primes in this interval are 113, 127, and 131.
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MATHEMATICA
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Differences @ Table[PrimePi[n*Log2[n]^2], {n, 1, 100}] (* Amiram Eldar, Jun 27 2021 *)
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PROG
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(PARI) f(n) = n*(log(n)/log(2))^2;
a(n) = primepi(f(n+1)) - primepi(f(n)); \\ Michel Marcus, Jun 30 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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