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2, 71, 101, 181, 239, 241, 269, 349, 373, 409, 419, 433, 439, 491, 593, 599, 601, 607, 647, 653, 659, 823, 827, 857, 947, 1021, 1031, 1061, 1063, 1091, 1103, 1301, 1427, 1429, 1447, 1451, 1489, 1553, 1559, 1567, 1601, 1607, 1609, 1789, 1867, 1871, 1913, 1999, 2003
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OFFSET
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1,1
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COMMENTS
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For every n>=1, A104272(n) >= A080359(n), and the sequence shows where the inequality becomes an equality.
Let prime(m) < a(n)/2 < prime(m+1); then primes p<q exist such that p is in the interval (2*Prime(m), a(n))
and such that q is in the interval (a(n), 2*Prime(m+1)).
Let us call a prime p compatible with another prime q, if the intervals (p/2,q/2) and (p,q], if q>p, (or intervals (q/2,p/2) and (q,p], if q<p) contain the same number of primes. If p is compatible with no other prime, we call it a peculiar prime. The sequence lists the peculiar primes. [Vladimir Shevelev, Apr 25 2012]
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LINKS
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Table of n, a(n) for n=1..49.
V. Shevelev, On critical small intervals containing primes, arXiv:0908.2319 [math.NT] [From Vladimir Shevelev, Aug 20 2009]
V. Shevelev, Ramanujan and Labos primes, their generalizations, and classifications of primes, J. Integer Seq. 15 (2012) Article 12.5.4
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FORMULA
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All solutions of the equation A104272(x)=A080359(x) are x=pi(a(n))-pi(a(n)/2). - Vladimir Shevelev, Apr 25 2012
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EXAMPLE
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a(2)=71, such that 31<71/2<37, and we see that p=67 is in interval (62, 71) and q=73 is in interval (71, 74).
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CROSSREFS
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Cf. A104272, A080359, A164368, A164333, A164288, A164294, A212493, A212541.
Sequence in context: A132566 A151686 A201555 * A141908 A157368 A095935
Adjacent sequences: A164551 A164552 A164553 * A164555 A164556 A164557
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KEYWORD
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nonn
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AUTHOR
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Vladimir Shevelev, Aug 15 2009
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EXTENSIONS
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Terms beyond 659 from R. J. Mathar, Dec 17 2009
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STATUS
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approved
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