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

[M. F. Hasler]

On Sat, Oct 3, 2020 at 2:13 AM Peter Munn <techsubs at pearceneptune.co.uk>
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.

1) The first property (any nonzero integer divides or is a multiple of some
term)
applies to many many sequences (primes, composites, k-almost primes,
(primitive) abundant / deficient numbers, ....).
So many that I think it could be interesting to consider the contrary
(infinite sets of numbers that do NOT have this property).
That (opposite property) would apply, e.g., to sequences made of numbers
having only (or not having) specific prime factors,
like "powers of p", or p-smooth numbers, or numbers having only prime
factors congruent to x mod y, etc.

2) The second property holds here *by definition* (of
*primitive* nondeficient numbers :<=>
all divisors are deficient  <=>  has no nondeficient divisor  =>  no term
divides another term)

In any set of numbers with fixed bigomega (as "k-almost primes", k >= 1)
it's also obvious that no element can divide (or be a multiple) of any
other element.

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 --
where I consider just the first part, "any positive integer is divisor or
multiple of some term",
given that the second part is true by definition of "primitive".)

Any nondeficient number N has divisors in this set: just arbitrarily remove
prime factors until you cannot remove any more without making the result
deficient.
The result is obviously nondeficient (by construction) and primitive (if it
had a nondeficient proper divisor, you could have removed more factors
without making it deficient).

Conversely, if the number N is deficient then by multiplying it with some
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 is
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, cf. e.g.
the formula log(n+1) < prod(1+1/p) * sum(1/k²) in this
Proof_that_the_series_exhibits_log-log_growth
<https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes#Proof_that_the_series_exhibits_log-log_growth>
after "Combining....".)

and does the property have an established name?
>

I'm not aware of a special name for this property but I don't think that's
needed or appropriate,
because for both of the properties (which are combined here), there are
completely different reasons why it is so:

- not having a divisor in the same set : for sequences defined as "
*primitive* xxx numbers" it is so by definition of "primitive",
for other sequences it will be impossible for totally different reasons ;
e.g. of numbers with fixed bigomega = k
 it is obviously impossible, also by definition: if they must be composed
of k prime factors, you obviously can't take away any of them.

- any positive integer having a divisor or multiple in the set: here (and
for abundant numbers, primitive or not, and for deficient numbers)
 it is really specific to the construction of the numbers in the set,
but in other cases (e.g. primes ; composites ; multiples of x ; ....) it is
so for quite different reasons :
 for primes because all numbers are made of prime factors ;
 for composites because by definition if you multiply a number > 1 by
something > 1 it's composite,
 for "multiples of x" it's similarly true by definition
 etc.

- Maximilian