A330556 a(n) = (number of primes p <= 2*n+1 with Delta(p) == 2 mod 4) - (number of primes p <= 2*n+1 with Delta(p) == 0 mod 4), where Delta(p) = nextprime(p) - p.

0, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 4, 4, 4, 3, 3, 4, 3, 3, 4, 4, 4, 5, 5, 5, 6, 7, 7, 7, 6, 6, 7, 8, 8, 8, 7, 7, 8, 8, 8, 7, 7, 7, 7, 6, 6, 7, 6, 6, 7, 6, 6, 7, 7, 7, 7, 7, 7, 7, 6, 6, 7, 7, 7, 8, 9, 9, 9, 9, 9, 10, 11, 11, 11, 12, 12, 12, 11, 11, 12, 12, 12, 13, 13, 13

Offset: 0

Author

N. J. A. Sloane, Dec 29 2019

Keywords

• nonn

Comments

a(n) = A330557(n) - A330558(n).

Since Delta(prime(n)) grows roughly like log n, this probably changes sign infinitely often. When is the next time a(n) is zero, or the first time a(n) < 0 (if these values exist)?

Examples

			n=5, 2*n+1=11: there are three primes <= 11 with Delta(p) == 2 mod 4, namely 3,5,11; and one with Delta(p) == 0 mod 4, namely 7; so a(5) = 3-1 = 2.
		

Links

• N. J. A. Sloane, Table of n, a(n) for n = 0..99999
• StackExchange, Asymptotic Distribution of Prime Gaps in Residue Classes.

Crossrefs

Sequences related to the differences between successive primes: A001223 (Delta(p)), A028334, A080378, A104120, A330556-A330561.