One entry and just one entry

[Leo Hennig]

Hi Guys, 

I have the following question; what if I have just one entry?

I know that two entries are allowed, e.g. the sublime numbers...A081357

BUt, what for the following:

Numbers k such that sigma(k)+k is a perfect number ---

I know just one number that fulfills this criterium and it is 10 (our base! :) ).  

I'd be curious though to find another. 

Greets, LAH 

[Sean A. Irvine]

No, that would not be accepted.

What might be acceptable, is adding a comment to A000396 that the only known such number is 10.

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[Leo Hennig]

I think it should be, since it is a good starting point for others to search for other numbers. Is not this a good thing?

In fact I think it is a good challenge to find another - for everyone. 

[Marc LeBrun]

Alas, no, it's not a good ideal to start flooding the OEIS with singleton "sequences" (nor adding comments about magical sporadic properties of individual values).

If you think this is an interesting fact worthy of further exploration, you should write it up and post it somewhere -- perhaps either as a short note (eg attached to your user page in the OEIS wiki) or elsewhere (eg as a question on mathoverflow).

Also, the definition seems rather arbitrary (eg why not sigma(k)-k or 3*sigma(k)+1 or whatever?) -- What is the motivation?  Why is this property interesting?

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[Charles Greathouse]

The question of “why is this interesting?” is a general one which is applied to each submission. There are uncountable infinitely many integer sequences, but we only have room for finitely many.

Spending the time on a sequence such that you can convey the interest to a reader is an important part of the process. This is why I usually suggest spending (at least) an hour on each new submission:

https://oeis.org/wiki/User:Charles_R_Greathouse_IV/Rule_of_thumb

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[Michel Marcus]

You could add a comment into A155085.

[David Radcliffe]

Leo asked if there are numbers k such that sigma(k)+k is a perfect number, other than k=10.

I have found one other example, as I will explain below.

Let P be an even perfect number. We wish to find k such that sigma(k)+k = P.

We will suppose (optimistically) that k = q*r where r is small and q is a prime not dividing r.

Then P = q*r + (q+1)*sigma(r) = q*(r+s) + s, where s = sigma(r).

So the strategy is to loop through the candidate values of r and s until 

q := (P-s)/(r+s) is prime, and return k := q*r.

For P = 2^30 * (2^31 - 1), I found r=187268, s=327726, and q=4477417228433.

So k = q*r = 838476969534191044 satisfies sigma(k)+k = 2^30 * (2^31 - 1).

- David

[Leo Hennig]

Thanks, this is great.

Leo A. Hennig

Website: https://www.leohennig.minkus.org/

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[Leo Hennig]

Well, I try my best, for the general audience, to explain why I find this number (and now after we have another, this sequence, much to my delight) interesting. 

There is a thing with bases, that we do not know why we should prefer one over the other.  We have base10 but is there any reason we should prefer base10 over another base, a priori , that is. 

There are obvious ways in which some numbers are interesting to people, ranging from numerologists to number theorists, and most start with something like the perfect numbers. They are, in our number system, with how we operate with numbers on a daily basis a good starting point, since they are not too rare, their relative freq is just right.  

SOme people delve deeper into the matter, and they then come across sublime numbers, which have an absolutely curious pattern, a small (12) and a very large number (>10^50),  and lots of mystery around them,

Now what we found here in ten, this puny little number,  that is in itself mathematically not interesting, its not prime or abundant or anything like that, that has become our base now for a few hundred years, and will probably continue to be our main base, is a number that has properties of both the perfect (in the construction of the sequence, in that the sum of the divisors as well as ) and the sublime numbers in that, it is a small, everyday number, and a very large one that comes after it. 

To phrase it in EUclidean terms: A number that the sum of its parts and the number itself added to it gives a perfect number may be called a superbly perfect number.  

Som Som Som. 

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[Leo Hennig]

Ok, as a note: David created a python script for finding the terms we interested in:  https://gist.github.com/Radcliffe/84b69e09606e193a1446b6cb0971c54c 

But we might miss smaller terms. Maybe we can improve on this?

[M F Hasler]

On Sat, Mar 8, 2025 at 8:46 AM Leo Hennig <leoh...@gmail.com> wrote:

Ok, as a note: David created a python script for finding the terms we interested in:  https://gist.github.com/Radcliffe/84b69e09606e193a1446b6cb0971c54c 

Nice, thanks to David for making this public!

I notice that it tests  q%r != 0, but it should be r%q != 0.

I understand that most P = (2^p-1)<< p-1 ~ 10^(0.6p) are much larger than r,

and even  q=(P-s)/(r+s)  will be much larger than r in most cases; 

which may suggest the first form to avoid more primality tests, 

but I don't think that's useful - at least, technically we would still have to check whether r%q > 0.

There's a much simpler "sieving condition", namely  q & 1, that would avoid more than half of the primality checks.

Indeed,  q = (P-s)/(r+s)  must be odd (because we assume r even), which occurs not very often.

But it turns out that the bottleneck is (r+s) | (P-s) :

for example, using p's up to 5000 and r up to 10^6, there are only 49 cases where this is satisfied,

and q can even be computed -- and in 27 of these, q is even, in only 22 it's odd.

So, for larger p, it might be useful to compute P first only mod r+s to see whether it equals s,

and only in that case, compute the (possibly huge) quotient q = (P-s)/(r+s).

I did so using p's up to 10^6 and r up to 10^7: there are only 105 cases where r+s | P-s,

and in only 38 cases, q  is odd.

We can directly test whether  P-s  is an odd multiple of  r+s  

by computing  P (mod 2(r+s)), which must equal  s +- (r+s),  or P+r = 0 (mod 2(r+s)).

Finally, there are only very few candidates (r,p) where the primality check takes considerable time.

I don't know for sympy's isprime, but PARI/GP's is(speudo)prime()  took more than a second for

{ [37088, 86243], [83416, 21701], [211392, 756839] } (in the above mentioned range).

Actually, I don't know why it takes so long for these, because I found they have small factors:

[37088, 86243]: numdigits(q)=51919, prime: NO (small factors : 1091 · 7487)
[83416, 21701]: numdigits(q)=13060, prime: NO (small factor : 3373)
[211392, 756839]: numdigits(q)=455657, prime: NO (small factor : 143287).

It seems that trial division is done only for really small primes (up to 100 or so)

Maybe it can speed up the search by doing  factor(q, 0)  which in PARI factors

up to the precomputed primes (by default 10^6 or so).

Also, to avoid the memory voracious pre-computation of sigma(r),

I suggest to use " for r  ... : s=sigma(r); ...  for p ... : ..." 

with sigma = int(divisor_sigma imported from sympy).

That allows me to find the second solution  (p = 31, r = 187 268),

in a few seconds on my laptop, with no need for much memory.

But we might miss smaller terms. Maybe we can improve on this?

David's method uses only that the solution is of the form  k = q*r  with prime q not dividing r.

That means, the only condition is that k has at least one prime factor that does not appear to a higher power.

So we can miss a solution below the one found by David (p=31, r=187268) only if it has all prime factors to a power >= 2.

That appears to be rather improbable, but could be studied by replacing "q+1" in David's formula with 1+q+q² (+...),

(viz., (q^m-1)/(q-1), where m-1 is the (least) multiplicity of a prime factor q of k).

That said, I think that even if we know the second smallest solution of k+sigma(k)=(perfect number) for sure, I'm not sure this should lead to a new entry, but rather be added as comment to existing "popular" sequences such as A396 (perfect numbers, already somewhat overcharged) or A155085 (= n +sigma(n)), with a link to this discussion.

- Maximilian

Leo Hennig schrieb am Mittwoch, 5. März 2025 um 22:13:49 UTC+1:

Hi Guys,