

A303112


Primes p such that (rq)/(qp) = 2 or 1/2, and p < q < r are three consecutive primes.


1



2, 5, 7, 11, 13, 17, 37, 41, 67, 89, 97, 101, 103, 107, 191, 193, 223, 227, 277, 307, 311, 347, 389, 397, 449, 457, 461, 479, 487, 491, 503, 613, 641, 739, 757, 761, 821, 823, 853, 857, 877, 881, 907, 929, 991, 1087, 1091, 1231, 1277, 1297, 1301, 1423, 1427, 1439, 1447, 1453
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OFFSET

1,1


COMMENTS

Conjecture: The two most frequent ratios between consecutive prime gaps are 2 and 1/2, and both ratios occur with about the same frequency.


LINKS



FORMULA

Conjecture: lim_{n>inf} n/primepi(a(n)) > k > 0 for some k.


EXAMPLE

The first three consecutive primes are 2, 3 and 5, and (53)/(32)=2, so the first term is a(1)=2, that is, the first prime of (2,3,5).
The next three consecutive primes are 3, 5 and 7, and (75)/(53)=1, so the first prime of (3,5,7) is not in the list.
The next three consecutive primes are 5, 7 and 11, and (117)/(75)=2, so the second term is a(2)=5, that is, the first prime of (5,7,11).
The prime 13 is also in the list because (1917)/(1713)=1/2.


MATHEMATICA

b={};
Do[If[Abs[Log[2, (Prime[j+2]Prime[j+1])/(Prime[j+1]Prime[j])]]==1, AppendTo[b, Prime[j]]], {j, 1, 200}];
Print@b
Select[Partition[Prime[Range[250]], 3, 1], (#[[3]]#[[2]])/(#[[2]]#[[1]]) == 2(#[[3]]#[[2]])/(#[[2]]#[[1]])==1/2&][[All, 1]] (* Harvey P. Dale, Mar 14 2022 *)


PROG

(PARI) isok(p) = my(q = nextprime(p+1), r = nextprime(q+1), f = (rq)/(qp)); (f == 2)  (f == 1/2);
forprime(p=2, 1000, if (isok(p), print1(p, ", "))); \\ Michel Marcus, Apr 23 2018


CROSSREFS

Cf. A257762 (indices of primes with above ratio = 2).


KEYWORD

nonn


AUTHOR



STATUS

approved



