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Reorganizing the section on restrictions about odd perfect numbers

If we add to the section on restrictions about odd perfect numbers the assumption p 1 < p 2 < p 3 < p k {\displaystyle p_{1}<p_{2}<p_{3}\cdots <p_{k}} then we can directly use the same framework to talk about the the results on the largest prime factor, second largest and third largest by referring to bounds on p k {\displaystyle p_{k}} , p k 1 {\displaystyle p_{k-1}} , and p k 2 {\displaystyle p_{k-2}} . Would anyone object to that tweak? JoshuaZ (talk) 15:48, 22 October 2018 (UTC)

Actually, raising an objection to my own suggestion. This notation doesn't help that much since it doesn't naturally handle the Euler prime, so we cannot just talk about bounding p k {\displaystyle p_{k}} , p k 1 {\displaystyle p_{k-1}} , and p k 2 {\displaystyle p_{k-2}} . without losing some strength. JoshuaZ (talk) 16:40, 29 October 2018 (UTC)

Pomerance and Luca's result on the radical of an odd perfect number

Luca and Pomerance have shown that if N {\displaystyle N} is an odd perfect number with N = p 1 a 1 p 2 a 2 p k a k {\displaystyle N=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}} that one must have p 1 p 2 p 3 p k < 2 N 17 26 {\displaystyle p_{1}p_{2}p_{3}\cdots p_{k}<2N^{\frac {17}{26}}} . See https://www.math.dartmouth.edu/~carlp/LucaPomeranceNYJMstyle.pdf . Should this be added to the section on restrictions on an odd perfect number? My inclination is yes, but I worry about the section becoming overly long. JoshuaZ (talk) 04:08, 9 November 2018 (UTC)

Klurman's result

I'm undoing my earlier edit since I've now gone and read Klurman's paper. Although some places say that he proved the result in the edit, in the actual paper he doesn't have an explicit constant but rather just a bound of the form C N 9 14 {\displaystyle CN^{\frac {9}{14}}} for some constant C and for N sufficiently large. I strongly suspect that this constant can be made explicit with a little work but this would be original research and shouldn't be in the article. Therefore, until someone publishes a version of Klurman's argument with an explicit constant, Pomerance and Luca's bound should be the one that stays listed in the article. My apologies for including the Klurman article without having actually read the paper carefully. JoshuaZ (talk) 19:12, 28 May 2019 (UTC)

Overcomplicated lead

I made an edit simplifying the initial paragraph of the lead and moving examples into the lead in this edit to make it accessible to younger readers or those with less formal mathematical education. I was reverted by Anita5192 with the reasoning "This not only is no clearer, it isn't even correct. 6 is divisible by -1, -2, and -3. The sum of its divisors is 0."

The assertion that what I wrote was "incorrect" is simply false. The first sentence read: In number theory, a perfect number is a positive integer that is equal to the sum of its divisors, excluding the number itself. By positive integers, we're therefore talking about natural numbers only, in which "-1" doesn't exist. I understand that in contemporary academia, number theory deals with Z {\displaystyle \mathbb {Z} } and then the set of units is { ± 1 } {\displaystyle \{\pm 1\}} rather than just 1 {\displaystyle 1} , but the point is that the lead should be understandable to the broadest possible group of people. In high school or primary school, when divisors and multiples are taught, only positive integers are considered (at least conventionally), and so the clarification that divisors must be positive doesn't add anything.

As for the reasoning behind me merging the example section into the lead, it seems fairly uncontroversial to me that it's easier for someone to learn a property about the natural numbers if they see an example or two. The current version of the lead now gives no examples of perfect numbers.

I would imagine that my eleven-year-old self would have struggled to parse the sentence In number theory, a perfect number is a positive integer that is equal to the sum of its proper positive divisors, that is, the sum of its positive divisors excluding the number itself (also known as its aliquot sum). The sentence that follows it is even worse. But eleven-year-old me did understand what perfect numbers are without much difficulty. So something needs to be done so that an eleven-year-old can look at the lead of the article and learn what a perfect number is. — Bilorv (he/him) (talk) 00:04, 30 June 2019 (UTC)

Proposed change to odd perfect number section, upper bound on second largest prime factor, new reference, COI

We currently note that the second largest prime factor of an odd perfect number must be be greater than 10. I'd like to add that the second largest prime factor is at most ( 2 N ) 1 / 5 {\displaystyle (2N)^{1/5}} . The citation for this is Zelinsky, Joshua (July 2019). "Upper bounds on the second largest prime factor of an odd perfect number". International Journal of Number Theory. 15 (6): 1183–1189. doi:10.1142/S1793042119500659. Retrieved 2 July 2019.. Since I'm the author, I have a COI for this, so I'd like to know if other people agree with adding this to the article. Note that the article also proves a few other results (in particular, an upper bound on the product of the largest two prime factors) but the other results are technical enough that they seem like they should not be cited in this article. Thoughts? JoshuaZ (talk) 12:36, 2 July 2019 (UTC)

I'm going to wait until Monday, and if no one objects, add in a citation to the paper. JoshuaZ (talk) 12:52, 5 July 2019 (UTC)

The Next Perfect Number is 28 = 1 + 2 + 4 + 7 + 14

I added... The next perfect number is 28 = 1 + 2 + 4 + 7 + 14. The next paragraph is confusing with explaining how 56 = 28 x 2. 2601:580:107:4A57:6D32:859F:D592:8ABB (talk) 11:24, 6 July 2019 (UTC)

The explanation given in the next paragraph of the lede is perfectly clear (provided of course that one takes the pain of reading the whole sentence...) Sapphorain (talk) 11:32, 6 July 2019 (UTC)
Hi there. The next paragraph is intended to illustrate a condition for perfect numbers in terms of the divisor function. The first paragraph is intended to be as accessible as possible to readers, but the divisor function is an important concept in number theory so it is helpful to mention it in the next paragraph. — Bilorv (he/him) (talk) 11:38, 6 July 2019 (UTC)

Gallardo's Result

Is Gallardo's result (in Minor results) true? In the linked paper he implicitly assumed that x + a {\displaystyle x+a} and x 2 a x + a 2 {\displaystyle x^{2}-ax+a^{2}} are coprime, but it might not be the case if both a {\displaystyle a} and x {\displaystyle x} are even. 219.78.80.30 (talk) 06:48, 2 September 2020 (UTC)

That's a good point. I don't see how he's getting that step either. Maybe raise this on Mathoverflow or contact him directly. JoshuaZ (talk) 22:25, 10 September 2020 (UTC)

Changes to odd perfect number section and COI

Same issue as before but another paper. Again, I'm the author so I have a clear COI, so this needs to be okayed before I make any edits. This paper has multiple possibly relevant inequalities. Paper is here.

First, the page currently cites Grun's bound that the smallest prime factor must be less than 2 k + 8 3 . {\displaystyle {\frac {2k+8}{3}}.} The paper has much better than linear bounds in general, but those bounds are long and technical, and so probably shouldn't be on this page by themselves. However, Corollary 4 on page 43, is equivalent to in the notation on this page that the smallest prime is at most 1 2 k 1 2 {\displaystyle {\frac {1}{2}}k-{\frac {1}{2}}} which is tighter than Grun's bound. Should that be included? My inclination is to include that bound but *not* the more technical non-linear bounds.

Second, the page currently has the bound that α + 2 e 1 + 2 e 2 + 2 e 3 + + 2 e k ( 21 k 18 ) / 8 {\displaystyle \alpha +2e_{1}+2e_{2}+2e_{3}+\cdots +2e_{k}\geq (21k-18)/8} . This paper improves that bound to α + 2 e 1 + 2 e 2 + 2 e 3 + + 2 e k 66 x 191 25 . {\displaystyle \alpha +2e_{1}+2e_{2}+2e_{3}+\cdots +2e_{k}\geq {\frac {66x-191}{25}}.} This is stronger when k 9 {\displaystyle k\geq 9} and thus for all odd perfect numbers. It would make sense to include this tighter bound.

Are there objections to making these two changes? JoshuaZ (talk) 01:45, 3 August 2021 (UTC)

Joshua, is the latter supposed to have k rather than x? In that case yes, let's include both. Otherwise, what is x in this context?
P.S. Feel free to ping me if this sort of question comes up in the future, the OPN community is pretty small (not that I'm a member, but I dabble, and I don't know of any other Wikipedians who do more than the two of us in the area). - CRGreathouse (t | c) 18:21, 5 August 2021 (UTC)
Sorry, yes, that should have read 66 k 191 25 . {\displaystyle {\frac {66k-191}{25}}.} I'll wait another day then and if no one has any objections, I'll make these changes. JoshuaZ (talk) 20:01, 5 August 2021 (UTC)

Addition for the odd perfect number and COI- (III?)

I have a recent paper with Sean Bibby and Pieter Vyncke where we prove that an odd perfect number N {\displaystyle N} with third largest prime factor a {\displaystyle a} must satisfy a < 2 N 1 6 . {\displaystyle a<2N^{\frac {1}{6}}.} paper here(pdf). Since I'm an author, there's an obvious COI issue. I'm also just not sure that this should be included or not. The current version of the article has a lower bound on the third largest prime factor, but not upper bound. (I'm not aware of a non-trivial upper bound in the literature prior to our work, but our upper bound is pretty weak.) Should this result be included in the section? JoshuaZ (talk) 12:39, 6 December 2021 (UTC)

Only one of the three largest primes can be the special prime, and so the best case is that two have exponent 2 and the third has exponent 1. That gives a < N 1 5 . {\displaystyle a<N^{\frac {1}{5}}.} This is a respectable improvement over that naive exponent, and effective to boot. The only other information we have on a, to my knowledge, is Iannucci's 20+ year old lower bound a 101. {\displaystyle a\geq 101.} So I think this is worth including. Joshua, are you aware of similar upper bounds for other prime factors? I believe my argument generalizes, with the n-th largest prime factor p n {\displaystyle p_{n}} having the trivial bound p n < N 1 2 n 1 . {\displaystyle p_{n}<N^{\frac {1}{2n-1}}.} I'm not aware of any aside from yours, but I haven't been following OPNs closely for a while.
I'll give the paper a look and give it a go later today if I have a chance.
Disclaimer: I'm an admin who has asked to be notified in cases like this where authors have work relevant to this page but are wary of COI concerns.
- CRGreathouse (t | c) 17:50, 6 December 2021 (UTC)
Oh yes, you cite bounds for the largest and second-largest primes -- I think those should be added to the article as well. The combined bounds like your bound on the product abc are also interesting to me but should probably be left out of a general-interest article like this. A more focused OPN article written for specialists would certainly cover these in some consistent way but that's not the way we're organized at the moment. - CRGreathouse (t | c) 17:55, 6 December 2021 (UTC)
The best bounds for the largest and second largest (both upper and lower) are actually already in the article. I agree that the product abc bound should not be included in this article. (For the same reason there was a product bc bound which we also haven't included in this article.). JoshuaZ (talk) 18:59, 6 December 2021 (UTC)
 Done Perfect, makes my life easier! - CRGreathouse (t | c) 19:08, 6 December 2021 (UTC)

conjecture regarding the divisors of Perfect Numbers (PN)

I conjecture that the product of the divisors a PN derived from 2(2 - 1) will equal PN. For instances: 1×2×4×7×14 = 28 and 1×2×4×8×16×31×62×124×248 = 496. Unfortunately, I cannot prove this. Also, in the first several such PN, there is only 1 odd divisor (> 1) and which is a prime number (the first several are 3, 7, 31, 127, 8191 and 131071)--does this persist? Wmsears (talk) 01:54, 26 December 2021 (UTC)

This will be true in general. It follows from a more general theorem that the geometric mean of the divisors of a positive integer is exactly the square root of the number. However, your observation, and the observation that this would follow from this are both Original research and therefore not suitable for Misplaced Pages. In the future, if you have similar math questions, I suggest checking out Math Stack Exchange. JoshuaZ (talk) 02:39, 26 December 2021 (UTC)
@Wmsears: Note that the term divisor includes the number itself so their product becomes PN. The divisors of a number can be listed in pairs like 28 = 1×28 = 2×14 = 4×7 (a square n also has one unpaired divisor n). Your formula can be worked out from this by considering the number of divisors. Hint: They are all of form 2 or 2×(2 - 1). The divisors without the number itself are called the proper divisors.
Regarding your other observation, it's known that 2(2 - 1) is a perfect number if and only if 2 - 1 is prime (called a Mersenne prime). The only odd divisors of 2(2 - 1) are 1 and 2 - 1. Your observation follows from this so it persists. PrimeHunter (talk) 04:47, 26 December 2021 (UTC)

I think the Odd Cubes section is incorrect

For odd cubes to work they only seem to work for every second even perfect number, I don't think the article makes that clear, unless I'm miss reading the explanation ( which I may be but since I didn't quite understand it maybe it could be made clearer):

In : def ffs(x):
     ...:   x = gmpy2.mpz(x)
     ...:   return gmpy2.bit_length(x&-x)-1


     ...: def extractoddfactor(N):
     ...:   return N//(2**ffs(N))


In : def checkifperfectnum(N):
     ...:    a = ffs(N)
     ...:    e = extractoddfactor(N)
     ...:    ex = 2**(a+1)-1
     ...:    if e == ex: return True
     ...:    else: return False
In : a = pow(1, 3)
     ...: for x in range (3,8192,2):
     ...:    a += pow(x,3)
     ...:    b = extractoddfactor(a)
     ...:    if checkifperfectnum(a):
     ...:         print (a,b, gmpy2.is_prime(b), checkifperfectnum(a))
     ...: 
     ...: 

Answer:

28 7 True True
496 31 True True
8128 127 True True
130816 511 False True
2096128 2047 False True
33550336 8191 True True
536854528 32767 False True
8589869056 131071 True True
137438691328 524287 True True
2199022206976 2097151 False True
35184367894528 8388607 False True
562949936644096 33554431 False True

you'll see the 15, 63, 1023, etc do not work with the odd cube method.

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