[seqfan] Re: A Gaussian-integer analog of the sum-of-divisors function

[Allan Wechsler]

Neil Fernandez: You are getting some different values because you are
adding a different set of divisors than I am; that is, you are selecting
different associates than I am. You are adding all the "positivish"
divisors. This is a perfectly well-defined operation, and maybe deserves
its own sequence, but it turns out that sequence is not multiplicative, so
I went looking for another version that *is* multiplicative.

What I am doing is adding divisors that are the products of
positivish Gaussian primes. This can be different, because the product of
two positivish numbers might not be positivish. My "canonical" divisors of
8 are 1, 1+i, 2i, -2+2i, 1-i, 2, 2+2i, 4i, -2i, 2-2i, 4, 4+4i, -2-2i, -4i,
4-4i, 8. Like your divisors, the ones with nonzero imaginary part appear in
conjugate pairs, so the imaginary parts cancel, and the real parts add to
25. These divisors are all of the form (1+i)^a * (1-i)^b, where a and b
range from 0 to 3, because the (canonicalized) Gaussian factorization of 8
is (1+i)^3 * (1-i)^3. I hope this clarifies where I am getting my 25.

It would be wonderful if you could adjust your program to see if you can
produce numbers that match mine. And, of course, I can *still *have made an
arithmetic error.

Maximilian Hasler: This is a good catch, and I may have to adjust my
verbiage to make things clearer. Essentially I am making a special case for
the factors of 2, 1+i and 1-i. I want to include both of them, since that
preserves the occurrence of divisors in conjugate pairs (as I think Neil
Fernandez was also pointing out in his second message), even though 1+i and
1-i are associates, and thus (according to another principle I claimed to
adhere to) one of them should be excluded.

I guess what I should say is that a "canonical" divisor is the product of
positivish Gaussian primes, period, and drop my claim of never including
two associated divisors. In fact in the case of 8, I include all the
associates of 2+2i, because they all occur as different products of powers
of 1+i and 1-i.

Things get even woolier with higher powers of 2, because then my procedures
include some divisors more than once! The simplest example occurs when
listing Gaussian divisors of 16. The natural, multiplicative way of making
this list includes -4 twice, once as (1+i)^4 and once as (1-i)^4. If I
punctiliously leave these duplicates out, I risk losing multiplicativity.

The best explanation of what I am doing is in the first paragraph of the
comments. I will have to change wording elsewhere where it conflicts with
that algorithm.

On Wed, Apr 27, 2022 at 8:22 AM Neil Fernandez <primeness at borve.org> wrote:

> Hi,
>
> If we allow only one divisor from {x+xi,x-xi} then we will get sums that
> are not real, e.g. for a(2).
>
> Neil
>
>
>
> In message <CABxCbJ1=Fv1dn=azG3AJYecpQhSngtOijC7C2=Y9=PxRVmDOWQ at mail.gma
> il.com>, M. F. Hasler <oeis at hasler.fr> writes
> >    On Wed, Apr 27, 2022, 07:30 Neil Fernandez <primeness at borve.org>
> >    wrote:
> >>       Hi Allan,
> >
> >>       'm not sure whether I'm applying your rules correctly, but I get
> >>        values for a(8) and a(10):
> >
> >
> >
> >>        in first octant or eighth (allowing both boundaries, not
> >>        8 itself):
> >>        1+i, 2, 2+2i, 4, 4+4i, 8, 1-i, 2-2i, 4-4i};
> >
> >
> >    Maybe I get something wrong, but the last three are related to
> >    earlier ones by a unit ( -i ) and so I would have thought they
> >    shouldn't be counted a second time.
> >
> >    - M.
> >
> >>        Allan Wechsler <acwacw at gmail.com> writes
> >
> >>       >I have a draft at A353151 for a sequence that is intended to be
> >>       an analog
> >>       >of A000205, the sum of the divisors of n.
> >>       >
> >>       >This endeavor is a little bit fraught because every Gaussian
> >>       divisor of n
> >>       >is one of a set of four "associate" divisors, which are related
> >>       by a
> >>       >factors of a Gaussian unit. When we add up the divisors, we
> >>       only want one
> >>       >of each associated set; which one shall we choose?
> >>       >
> >>       >My choice was to add up the divisors that are the products of
> >>       powers of
> >>       >"positivish" Gaussian primes. "Positivish" means that the real
> >>       part is
> >>       >positive, and the imaginary part doesn't exceed the real part.
>
> --
> Neil Fernandez
>
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