Sigma(n)-2n=20 Has Many Solutions

[Alex Violette]

Hello SeqFans,

I wanted to announce that the equation sigma(n)-2n=20 or A=20 has "many" solutions. This is mainly as the factorization needed to find these solutions is a special case for sigma(n)-2n=20 as sometimes it can lead to a buttload of solutions. 2^176*q*r and 2^226*q*r have 191 and 236 solutions respectively where A=20. To put into perspective on how crazy this is, there are more solutions of the form n=2^p*q*r where p=176 than for all smaller values of p combined. I put them into two google docs as otherwise it would take up too much room here. I've only seen something like this once prior.

https://docs.google.com/document/d/1OV_3aIhEFk18XEgpEWk4SfP_6ORlGOZEb69WXyfxgyY/edit?usp=sharing (2^176*p*q solutions)

https://docs.google.com/document/d/1zt4s7OsGWHoyQiQ5DLPZzJY2qAx1jqvQ0BoXK657bHI/edit?usp=sharing (2^226*p*q solutions)

Other solutions not of this form I managed to obtain too for both sigma(n)-2n=-20/20, the last of which is the largest squarefree weird number known to date:

-20: 22676188020687854193666634934837575690
-20: 2831124349658085626574918972050788187735050
-20: 2*5*11*83*157*62639*128939*1161317221477111*372649202822024389*26369203106435203377997931091644814326530242391*5022601292214037732998095974513774764183804502564640362438339006897691869

20: 475483761216097553026174788305625590
20: 38019150288298600519398940689802388039456978868856287470013230216516108807619437287575373054544130047990

Additionally, here are two solutions for sigma(n)-2n=-4 I have not shown yet as well as some either big or massive solutions for other A of which the 28 solution required me to "go the mile" just to obtain it:
-4: 2^146*178405961588244985132356118144328223588639017*452292158828981328840409680830644834867931266027294773712710677881
-4: 2^138*696898287454081973172991196020261297190297*3782160447445154331670940555298494873842299836124336830512605941316811807458313 (Note: larger than the other one despite a smaller power of 2, was not expecting it for the sequence they're in given how rare these are)

28: 2^180*3064991081731777716716694054300618367237571899483230929*100306002807464356447341699106964967409128000481349630696381582657562194582895848107106715877409
100: 2^129*1361129467683785459262206613717306477761*58618872639902481774861222662687432745938411654657677
100: 2^141*5575186299632655785383929654921098375957881*358264841799529714419425103905219474936522341029091238361570045680357

-126: 11439963301310178066495 (First term of A222263 where sigma(n)-2n=-126, been trying to hunt this down for years)
150: 31303272930292763135925

150: 1735981361725444339092624499461489648074096565

Best,
Alex Violette

[Allan Wechsler]

This is pretty darned cool, in my opinion. The relevant sequence is oeis.org/A223611. Alex already knows this, since he has contributed to that entry. Alex, I hope you are planning to include these "engineered" solutions in an auxiliary page to that entry.

Are there other magic A-numbers, like 20? I note that A=1 has no known solutions.

-- Allan

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[Jack Brennen]

Also see A188486, to which I contributed some elements years ago.

Numbers with an abundance equal to an odd square.  They are quite rare.

A=1 would be a special case of this.  Years ago, I extended the search for A=1 out to about 10^80 but never published my results; I did send the results to R. K. Guy, but never followed up with the details of my search.

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[Allan Wechsler]

The closely-related sequence oeis.org/A054024 blandly states, "a(p) = 1 for p prime.".

This is certainly true, but you have to look twice before you realize that it is not claiming that a(k) = 1 implies that p is prime. Any number with A=1 would be a counterexample to that claim, and as has been mentioned a couple of times in this thread, whether there is any number with A=1 is still open.

If we allow sigma(n) = kn + 1 for any k (not just 2), you would think that the problem would be easier. Do we know any composite n for which A054024(n) = 1?

-- Allan

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[M F Hasler]

You may be interested in our paper:

Farideh Firoozbakht and MFH: Variations on Euclid’s Formula for

Perfect Numbers 

Journal of Integer Sequences, Vol. 13 (2010), 3.6.1

https://cs.uwaterloo.ca/journals/JIS/VOL13/Hasler/hasler2.pdf

Theorem 1 gives solutions to sigma(x) = 2(x+m) which is your case for m=10.

-Maximilian

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[Max Alekseyev]

On Thu, May 14, 2026 at 12:40 AM Allan Wechsler <acw...@gmail.com> wrote:

whether there is any number with A=1 is still open.

Btw, there is a Wikipedia article on that - https://en.wikipedia.org/wiki/Quasiperfect_number

 

If we allow sigma(n) = kn + 1 for any k (not just 2), you would think that the problem would be easier. Do we know any composite n for which A054024(n) = 1?

Per a quick check, there are none below 10^28.

Regards,

Max