[seqfan] Re: A Gaussian-integer analog of the sum-of-divisors function
Allan Wechsler
acwacw at gmail.com
Wed Apr 27 18:04:54 CEST 2022
Ten thousand thanks to David Corneth for the almost ten thousand new
elements.
Now I suppose the Usual Multiperfect Suspects (Michel Marcus, David
Corneth, I, some others?) will go trawling for more analog-multiperfect
numbers. The consecutive ones so far are 1, 5, 10, and 52, and I have
spotted some bigger ones, like 8528 (2^4 * 13 * 41, also of order 5), and,
if I haven't made an embarrassing arithmetic error, the somewhat croggling
order-4 example 52000 = 2^5 * 5^3 * 13.
On Wed, Apr 27, 2022 at 10:57 AM Allan Wechsler <acwacw at gmail.com> wrote:
> 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/
>>
>
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