A217689 a(1)=2, a(2)=3, a(3)=4; for n>=4, a(n) is the largest number <= prime(n) such that no terms of the sequence are between a(n-1)/2 and a(n)/2.

2, 3, 4, 6, 8, 12, 16, 19, 23, 24, 31, 32, 38, 43, 46, 48, 59, 61, 62, 64, 73, 76, 83, 86, 92, 96, 103, 107, 109, 113, 118, 122, 124, 128, 146, 151, 152, 163, 166, 172, 179, 181, 184, 192, 197, 199, 206, 214, 218, 226, 233, 236, 241, 244, 248, 256, 269, 271, 277, 281, 283, 292, 302, 304, 313, 317, 326, 332, 344, 349, 353, 358

Offset

1,1

Comments

Every term has the form p*2^k, where p>=2 is prime and k>=0 (see A093641). For example, for a(3)=4, p=2, k=1. The sequence contains infinitely many primes and, therefore, limsup a(n)/(n*log(n))=1.

What is liminf a(n)/(n*log(n))?

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Formula

Let, for n>=3, a(k) <= a(n)/2 < a(k+1). Then a(n+1) = 2*a(k+1) if prime(n+1) > 2*a_(k+1), otherwise, a(n+1) = prime(n+1).

Example

For n=6, a(4)=6<a(5)=8, i.e., k+1=5 and a(k+1)=8. Since prime(7)=17>2*a(5)=16, then a(7)=2*a(6)=16.
Further, for n=7, k+1=6: a(6)=12. Since prime(8)=19<2*a(6)=24, then a(8)=19.

Mathematica

v = Prime[Range[100]]; v[[3]] = 4; k = 1;
For[n = 4, n <= Length[v], n++, While[v[[k+1]] <= v[[n-1]]/2, k++]; v[[n]] = Min[2*v[[k+1]], v[[n]]]];
v (* Jean-François Alcover, May 21 2022, after Charles R Greathouse IV *)

Prog

(PARI) v=primes(100); v[3]=4; k=1; for(n=4, #v, while(v[k+1]<=v[n-1]/2, k++); v[n]=min(2*v[k+1], v[n])); v \\ Charles R Greathouse IV, Oct 11 2012

Crossrefs

Cf. A217671.

Sequence in context: A092824 A355578 A084094 * A018718 A079647 A261205

Adjacent sequences: A217686 A217687 A217688 * A217690 A217691 A217692

Keyword

nonn,easy

Author

Vladimir Shevelev, Oct 11 2012

Status

approved