

A111166


Let p < q be consecutive primes; p is in the sequence if p/(qp) 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
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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 P2P1 <= 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/(32)=2; a(1)<>3 because 3/(53)=1.5; a(1)=5 because 5/(75)=2.5


MATHEMATICA

rmax = 0; p = 2; seq = {}; Do[q = NextPrime[p]; r = p/(qp); 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



