[seqfan] Re: A property of k-almost primes, but also A006039?

Peter Munn techsubs at pearceneptune.co.uk
Fri Oct 9 15:58:53 CEST 2020

On Thu, October 8, 2020 4:23 pm, M. F. Hasler wrote:
> On Sat, Oct 3, 2020 at 2:13 AM Peter Munn wrote:
>> For all k, the set of k-almost primes has the property
>> "all positive integers are either a divisor or multiple
>> of at least one member, but no member divides another member".
>> I've come to think A006039, the set of primitive nondeficient
>> numbers, also has this property,

> This is a composite property of two independent parts.

Satisfying one of the component properties is often trivial, and not of
great interest, as Maximilian explains. The interesting aspect (to me) is
that their effects are in a sense not independent: if I look for sets that
satisfy both, then the selection of sets that satisfy one of the
properties so constrains the process, that satisfying the other property
seems not at all trivial. I have not yet come up with examples unrelated
to those I mentioned, and knowing of some others would be enlightening.


>> and can see a likely route to prove it; but maybe I don't need
>> to, as this seems close to the mainstream and could be well-known.
>> So, does anyone here know?

> Yes that's certainly well known, although you might encounter it with
"abundant" rather than "nondeficient".
> (Making the set larger by adding the perfect numbers makes the statement
weaker --

It would be weaker, if the set was larger, as stated. However, of the two
primitive abundant number sequences in OEIS, the one that satisfies my
property (A091191) is not a subset of the primitive nondeficient numbers.
(For every perfect number, m, that is absent from A091191, there are an
infinite number of prime multiples of m that are extra.)


> Conversely, if the number N is deficient then by multiplying it with
> more prime factors you can make it abundant ; it is sufficient to choose
these additional factor(s) in a way to reach a nondeficient number which
> primitive, i.e., such that you just get abundancy >= 2, but < 2 whatever
factor you omit.
> (To be precise, any prime factor increases the abundancy by a factor 1 +
1/sum_{k=1..m+1} p^k if it's already there with multiplicity m >= 0
before adding it.
> Let x = max { sum_{k=1..m(p)} p^k ; p | N} so that 1+1/x is the smallest
such contribution of any prime factor in the original number N.
> Then you can simply add distinct prime factors p > max(x,P) until the
abundancy is >= 2 (since the infinite product of 1+1/p diverges

Thanks. That's exactly what I was looking for. It's a much better
elucidated, more general, process than the one I had put in the A337814


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