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

Wed Apr 27 23:38:08 CEST 2022

I don't see the problem yet, Neil. For one thing, (-2+2i) can certainly be
written as (1+i)^3. And while there are certainly Gaussian integers that
can't be written as the products of positivish Gaussian primes, the
positive real integers always can be, and A353151 is a function of ordinary
positive integers, so I don't think such difficulties arise.
-- Allan

On Wed, Apr 27, 2022 at 4:41 PM Neil Fernandez <primeness at borve.org> wrote:

> Hi Allan,
>
> > 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.
>
> But -2+2i and -2-2i cannot be so expressed.
>
> -2+2i = i * (1+i)^2 * (1-i)
>
> Is your rule to count divisors from any of the following three sets:
>
> * products of positivish numbers (defined as having positive real part
> not smaller than their imaginary part, i.e. numbers in the 1st or 8th
> octants, inclusive of boundaries)
>
> * products of positivish numbers and a unit
>
> * 1 (but none of the other units, and in particular not -1) ?
>
> Neil
>
>
> In message <CADy-sGFFUBxQwW_BUz=NwavYO3hZV_0B0y3YbrL6jL3AU6GXaw at mail.gma
> il.com>, Allan Wechsler <acwacw at gmail.com> writes
> >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
> >>
> >> --
> >> Seqfan Mailing list - http://list.seqfan.eu/
> >>
> >
> >--
> >Seqfan Mailing list - http://list.seqfan.eu/
> >
>
> --
> Neil Fernandez
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>