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A212493 Let p_n=prime(n), n>=1. Then a(n) is the least prime p which differs from p_n, for which the intervals (p/2,p_n/2), (p,p_n], if p<p_n, or the intervals (p_n/2,p/2), (p_n,p], if p>p_n, contain the same number of primes, and a(n)=0, if no such prime p exists. 5
0, 5, 3, 3, 3, 17, 13, 23, 19, 19, 37, 31, 31, 47, 43, 59, 53, 67, 61, 0, 79, 73, 73, 73, 73, 0, 107, 103, 127, 131, 109, 113, 113, 151, 113, 139, 163, 157, 157, 179, 173, 0, 223, 197, 193, 233, 193, 191, 191, 193, 199, 0, 0, 257, 251, 251, 0, 277, 271, 271 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

a(n)=0 if and only if p_n is a peculiar prime, i.e., simultaneously Ramanujan (A104272) and Labos (A080359) prime (see sequence A164554).

a(n)>p_n if and only if p_n is Labos prime but not Ramanujan prime.

LINKS

Table of n, a(n) for n=1..60.

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

J. Sondow, J. W. Nicholson, and T. D. Noe, Ramanujan Primes: Bounds, Runs, Twins, and Gaps, J. Integer Seq. 14 (2011) Article 11.6.2

FORMULA

If p_n is not a Labos prime, then a(n)=A080359(n-pi(p_n/2)).

EXAMPLE

Let n=5, p_5=11; p=2 is not suitable, since in (1,5.5) we have 3 primes, while in (2,11] there are 4 primes. Consider p=3. Now in intervals (1.5,5.5) and (3,11] we have the same number (3) of primes. Therefore, a(5)=3. The same value we can obtain by the formula. Since 11 is not a labos prime, then a(5)=A080359(5-pi(5.5))=A080359(2)=3.

CROSSREFS

Cf. A104272, A080359, A164554.

Sequence in context: A080134 A246728 A155685 * A011320 A208123 A090489

Adjacent sequences:  A212490 A212491 A212492 * A212494 A212495 A212496

KEYWORD

nonn

AUTHOR

Vladimir Shevelev and Peter J. C. Moses, May 18 2012

STATUS

approved

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Last modified December 10 00:24 EST 2016. Contains 278993 sequences.