Concerning A023199

[Matthew Conroy]

It is worth noting that A023199 is a subsequence of A004394,
and that A004394 does not have this property.

For a number n to be in A023199 (or A004394),
sigma(n)/n must be 'large' while n is 'small'.
Consider the function sigma(p^k)/p^k for
p a prime and k a positive integer.
This is a decreasing function of p and
an increasing function of k.  Using this,
you should be able to show that if n is in A023199, then
n's prime factorization looks like
    (p1^k1)(p2^k2)...(pm^km)
where pi is the i-th prime,  ki>0 for all 1<i<m,
and k1>=k2>=...>=km.

To get larger elements of A023199, you (generally) increase
the exponents and/or add new prime factors, keeping the
non-increasing feature of the exponents.  The result
is that each element is (generally?) a multiple of
previous elements.

Notice that this is not true of A004394.  Each new element
of A023199 represents such a large increase in sigma(n)/n
that this should only be possible by increasing most (perhaps
all) of the prime factor exponents, and adding prime factors.
On the other hand, with A004394, small gains in sigma(n)/n
are possible by small changes in the exponents, some increasing
while others decrease (for instance, 24 is followed by 36
in A004394).  Note that this is where things get interesting:
while sigma(p^k)/p^k is an increasing function of k, n also
increases as k increases, and we're trying to keep n small.

I hope this helps.  Let me know if more details are desired.


cheers,

Matthew Conroy




David Wilson wrote:

> Santi Sparado has asked me if each element of A023199 divides the next, 
> and if so why.
> 
> Can anyone help here?
>