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A158013
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100*a(n)+13 and 100*a(n)+27 are consecutive primes, i.e. a prime gap 14
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1, 106, 133, 154, 184, 217, 220, 307, 334, 436, 454, 496, 505, 574, 580, 604, 616, 631, 805, 892, 1009, 1015, 1045, 1132, 1174, 1189, 1198, 1204, 1360, 1408, 1444, 1504, 1510, 1627, 1702, 1708, 1771, 1954, 1984, 2101, 2182, 2218, 2221, 2245, 2260, 2281
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| notes: 1) Necessarily a(n)=3k+1: a(n)=3k => 100*3k+27= 3*(100k+9), divisible by 3 a(n)=3k+2 => 100*(3k+2)+13=3*(100k+71), divisible by 3 2) It is conjectured that sequence is infinite 3) Each sequence 100*b(n)+13 and 100*c(n)+27 includes an infinite number of primes (because of DIRICHLET's theorem) 4) Analogic sequences for investigation of prime gaps are obvious and useful
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REFERENCES
| A. E. Ingham, On the difference between consecutive primes
N. G. Tchudakoff, On the difference between two neighboring prime numbers
R. K. Guy, Unsolved problems in number theory
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FORMULA
| p(k+1)=100*a(n)+27 and p(k)=100*a(n)+13 where p(k) is the k-th prime => prime gap p(k+1)-p(k)=14
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EXAMPLE
| 1) 113=P(30) und 127=P(31) => a(1)=1 2) 1613=P(255) and 1627=P(258) prime too but 1619=P(256), 1621=P(257) => 1613 and 1627 are not consecutive primes 3) next: 10613=P(1295), 10627 = P(1296) => a(2)=106
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MATHEMATICA
| fQ[n_] := PrimeQ[ Range[100 n + 13, 100 n + 27, 2]] == {True, False, False, False, False, False, False, True}; Select[ Range@ 2295, fQ@# &] [From Robert G. Wilson, v (rgwv(AT)rgwv.com), Mar 13 2009]
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CROSSREFS
| A157772 primes ending with "13" ordered in natural growing size
Sequence in context: A095645 A039554 A186458 * A163625 A070796 A045093
Adjacent sequences: A158010 A158011 A158012 * A158014 A158015 A158016
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KEYWORD
| nonn
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AUTHOR
| Ulrich Krug (leuchtfeuer37(AT)gmx.de), Mar 11 2009
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EXTENSIONS
| a(31)-a(46) from Robert G. Wilson, v (rgwv(AT)rgwv.com), Mar 13 2009
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