

A066050


Average of divisors of n (sigma(n)/d(n)) is greater than average of divisors for all k < n.


1



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, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263
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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An unpublished paper of Kyle D. Balliet proves that, for n 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(n4) 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(n4)) \\ Charles R Greathouse IV, Aug 26 2011


CROSSREFS

Sequence in context: A325367 A160718 A122090 * A004050 A123538 A092999
Adjacent sequences: A066047 A066048 A066049 * A066051 A066052 A066053


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



