OFFSET
1,2
COMMENTS
Contribution from Charles R Greathouse IV, Apr 13 2010: (Start)
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.
An unpublished paper of Kyle D. Balliet proves that, for n >= 3, there is a prime in [3n, 4n].
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.
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.
Thus if n > 11 is a member of this sequence, n is prime. As a result, a(n) = prime(n-4) for n >= 9.
(End)
LINKS
Harry J. Smith, Table of n, a(n) for n = 1..1000
EXAMPLE
4 is included because sigma(4)/d(4) = 7/3 is greater than sigma(k)/d(k) for k = 1, 2, or 3.
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Leroy Quet, Dec 29 2001
EXTENSIONS
OFFSET changed from 0 to 1 by Harry J. Smith, Nov 08 2009
STATUS
approved