|
%I #14 May 03 2015 23:51:51
%S 1,2,3,4,5,7,9,10,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,
%T 79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,
%U 173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263
%N Average of divisors of n (sigma(n)/d(n)) is greater than average of divisors for all k < n.
%C Contribution from _Charles R Greathouse IV_, Apr 13 2010: (Start)
%C For composite n to be a member of the sequence, it must be in a prime gap of length > (p+2)/3 or the average of the divisors of p will be at least as great.
%C An unpublished paper of Kyle D. Balliet proves that, for n >= 3, there is a prime in [3n, 4n].
%C If p > 5 is a prime congruent to 2 mod 3, then there is a prime in [p+1, 4(p+1)/3] and hence the associated gap is at most (p+1)/3.
%C If p > 7 is a prime congruent to 1 mod 3, then there is a prime in [p+2, 4(p+2)/3] and hence the associated gap is at most (p+2)/3 as desired.
%C Thus if n > 11 is a member of this sequence, n is prime. As a result, a(n) = prime(n-4) for n >= 9.
%C (End)
%H Harry J. Smith, <a href="/A066050/b066050.txt">Table of n, a(n) for n = 1..1000</a>
%e 4 is included because sigma(4)/d(4) = 7/3 is greater than sigma(k)/d(k) for k = 1, 2, or 3.
%o (PARI) n=x=0; for (m=1, 10^9, a=sigma(m)/numdiv(m); if (a > x, x=a; write("b066050.txt", n++, " ", m); if (n==1000, return))) \\ _Harry J. Smith_, Nov 08 2009
%o (PARI) a(n)=if(n<9,[1, 2, 3, 4, 5, 7, 9, 10][n],prime(n-4)) \\ _Charles R Greathouse IV_, Aug 26 2011
%K nonn,easy
%O 1,2
%A _Leroy Quet_, Dec 29 2001
%E OFFSET changed from 0 to 1 by _Harry J. Smith_, Nov 08 2009
|
