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

Neil Fernandez primeness at borve.org
Wed Apr 27 22:40:51 CEST 2022


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



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