A212541 Let p_n=prime(n), n>=1. Then a(n) is the maximal prime p which differs from p_n, for which the intervals (p/2,p_n/2), (p,p_n], if pp_n, contain the same number of primes, and a(n)=0, if no such prime p exists.

0, 11, 11, 11, 7, 17, 13, 29, 29, 23, 41, 41, 37, 47, 43, 59, 53, 67, 61, 0, 97, 97, 97, 97, 89, 0, 107, 103, 127, 149, 109, 149, 149, 151, 137, 139, 167, 167, 163, 179, 173, 0, 227, 229, 229, 233, 229, 227, 223, 211, 199, 0, 0, 263, 263, 257, 0, 281, 281

Offset: 1

Author

Vladimir Shevelev and Peter J. C. Moses, May 20 2012

Keywords

• nonn

Comments

a(n)A104272).

a(n)=0 if and only if p_n is a peculiar prime, i.e., simultaneously Ramanujan and Labos (A080359) prime (see sequence A164554).

Examples

			Let n=4, p_n=7. Since 7 is not Ramanujan prime, then a(4) = A104272(4-pi(3.5)) = A104272(2) = 11.
		

Links

• V. Shevelev, Ramanujan and Labos primes, their generalizations, and classifications of primes, J. Integer Seq. 15 (2012) Article 12.5.4

Crossrefs

Cf. A212493, A104272, A080359, A164554.

Formula

If p_n is not a Ramanujan prime, then a(n) = A104272(n-pi(p_n/2)).