A345755 a(n) is the number of primes p satisfying n*(log_2(n))^2 < p <= (n+1)*(log_2(n+1))^2.

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

Offset: 1

Author

Hal M. Switkay, Jun 27 2021

Keywords

• nonn

Comments

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

Examples

			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.
		

Links

• Hal M. Switkay, Table of n, a(n) for n = 1..2772
• Wikipedia, Cramér's conjecture

Crossrefs

Cf. A014085, A143935, A166363.

Programs

Mathematica

Differences @ Table[PrimePi[n*Log2[n]^2], {n, 1, 100}] (* Amiram Eldar, Jun 27 2021 *)

PARI

f(n) = n*(log(n)/log(2))^2;
a(n) = primepi(f(n+1)) - primepi(f(n)); \\ Michel Marcus, Jun 30 2021