A217689 - OEIS

%I #36 May 21 2022 08:30:19

%S 2,3,4,6,8,12,16,19,23,24,31,32,38,43,46,48,59,61,62,64,73,76,83,86,

%T 92,96,103,107,109,113,118,122,124,128,146,151,152,163,166,172,179,

%U 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

%N 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.

%C 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.

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

%H Charles R Greathouse IV, <a href="/A217689/b217689.txt">Table of n, a(n) for n = 1..10000</a>

%F 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).

%e 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.

%e Further, for n=7, k+1=6: a(6)=12. Since prime(8)=19<2*a(6)=24, then a(8)=19.

%t v = Prime[Range[100]]; v[[3]] = 4; k = 1;

%t 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]]]];

%t v (* _Jean-François Alcover_, May 21 2022, after _Charles R Greathouse IV_ *)

%o (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

%Y Cf. A217671.

%K nonn,easy

%O 1,1

%A _Vladimir Shevelev_, Oct 11 2012