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A111166
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Let p < q be consecutive primes; p is in the sequence if p/(q-p) is a record.
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1
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2, 5, 11, 17, 29, 41, 59, 71, 101, 107, 137, 149, 179, 191, 197, 227, 239, 269, 281, 311, 347, 419, 431, 461, 521, 569, 599, 617, 641, 659, 809, 821, 827, 857, 881, 1019, 1031, 1049, 1061, 1091, 1151, 1229, 1277, 1289, 1301, 1319, 1427, 1451, 1481, 1487, 1607
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OFFSET
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1,1
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COMMENTS
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Conjecture: Except for first term, the sequence coincides with A001359. This is true for all primes < 2*10^7.
Conjecture: Except for first term, the sequence coincides with A001359. This is true for all primes < 7*10^16. Let n >= 2 be an integer, N +- 1 and M +- 1 two consecutive twin pairs where M>n*N. Finding a counterexample is the same as finding two consecutive primes P1 and P2 with n*N<P1<M and P2-P1 <= n. However, such gaps are unknown even for n=2.
The smallest prime(n) such that prime(n+1)/prime(n) is decreasing. [Thomas Ordowski, May 13 2012]
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LINKS
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EXAMPLE
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a(0)=2 and the record is 2/(3-2)=2; a(1)<>3 because 3/(5-3)=1.5; a(1)=5 because 5/(7-5)=2.5
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MATHEMATICA
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rmax = 0; p = 2; seq = {}; Do[q = NextPrime[p]; r = p/(q-p); If[r > rmax, rmax = r; AppendTo[seq, p]]; p = q, {100}]; seq (* Amiram Eldar, Dec 24 2019 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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