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A111166 Let p < q be consecutive primes; p is in the sequence if p/(q-p) is a record. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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]

This sequence corresponds with A001359 for all terms less than 10^100. - Charles R Greathouse IV, May 14 2012

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..10000

EXAMPLE

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

MATHEMATICA

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 *)

CROSSREFS

Cf. A001359.

Sequence in context: A156850 A156611 A143509 * A064337 A076873 A089440

Adjacent sequences:  A111163 A111164 A111165 * A111167 A111168 A111169

KEYWORD

easy,nonn

AUTHOR

Bernardo Boncompagni, Oct 21 2005

STATUS

approved

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Last modified June 7 01:26 EDT 2020. Contains 334836 sequences. (Running on oeis4.)