A214926 Difference A214925(n) - A214924(n), prime count between Ramanujan primes bounding maximal gap primes.

4, 4, 4, 3, 5, 3, 8, 7, 5, 7, 7, 3, 10, 6, 8, 24, 19, 6, 24, 25, 16, 8, 30, 17, 12, 13, 12, 11

Offset: 1

Author

John W. Nicholson, Aug 06 2012

Keywords

• nonn

Comments

Conjecture: For every n > 0, a(n) > 1.

Let rho(m) = A179196(m), for any n, let m be an integer such that p_(rho(m)) <= p_n and p_(n+1) <= p_(rho(m+1)), then rho(m) <= n < n + 1 <= rho(m + 1), therefore A001223(n) = p_(n+1) - p_n <= p_rho(m+1) - p_rho(m) = A182873(m). For all rho(m) = A179196(m), A001223(rho(m)) < A165959(m). (Comment copied from A001223). John W. Nicholson, Nov 17 2013

Examples

			a(4) = pi(A214757(4)) - pi(A214756(4)) = 10 - 7 = 3
		

Crossrefs

Cf. A214924, A214925, A214756, A214757.

Cf. A000101, A104272, A001223, A002386.

Formula

a(n) = pi(A214757(n)) - pi(A214756(n)).

a(n) = rho(A214757(n)) - rho(A214756(n)).

Extensions

Extension to a(28) added by John W. Nicholson, Nov 11 2013