greedy computation of pseudoperfect numbers?

[Joerg Arndt]

Thanks for the good pointer!

I was just too sure no odd n would turn up.

allowing divisor one, greedy gives these odd n<10000:
945, 1575, 2205, 2835, 3465, 4725, 5355, 5985, 6435, 6615, 6825, 7245,
7425, 7875, 8085, 8415, 8505, 8925, 9135, 9765,

forbidding divisor one, greedy gives:
1575, 2835, 3465, 6435, 6615, 7245, 7425, 7875, 8085, 8415, 8505,
8925, 9135, 9765,

 From the webpage:
"the only odd obundant numbers for which the greedy
 algorithm doesn't succeed are" (n<40000)
4095 5775 9555 12915 21945 22275 24255 27405 29925 35805 38745

(note all divisible by 5.)
It is not said whether these are pseudoperfect.

All in all an interesting but apparently hard problem.


* Joshua Zucker <joshua.zucker at gmail.com> [Apr 08. 2008 14:36]:
> On Mon, Apr 7, 2008 at 9:14 PM, Joerg Arndt <arndt at jjj.de> wrote:
> > For both sorts of pseudoperfect numbers, allowing divisor one A005835,
> >  and forbidding divisor one A136446, a greedy method _seems_ to always
> >  work for deciding whther a given n in in the sequence:
> 
> I think if you scroll down at
>   http://www.ics.uci.edu/~eppstein/numth/egypt/odd-one.html
> 
> you'll find that it only USUALLY works, not always.
> 
> >  Moreover, (disallowing the divisors one,) can any odd number n be in
> >  sequence A136446?  This is of interest because all gcd computations in
> >  a Rabin-like irreducibility test for polynomials over finite fields
> >  (have to write it up) can be avoided for all numbers _not_ in A136446.
> 
> In the same source it looks like there's a conjecture that there is an
> odd number in the sequence but that nobody had crunched enough numbers
> to find it.
> 
> --Joshua Zucker