20
votes
Accepted
Kindda-Perfect number: Is there a sequence of numbers which are equal to the sum of its proper divisors excluding itself as well as 1?
Such an abundant number with abundance 1 is called a quasiperfect number (which is a more professional way to say "kindda-perfect"). None have been found, according to Wikipedia. This 1982 article ...
- 188k
20
votes
Perfect numbers and perfect powers
Using Joe Silvermann’s notations, and under the assumption that any perfect number is even, we are trying to find the integer solutions of $$2^{p-1}(2^p-1)=x^m+1,$$ where $x \geq 2, m \geq 2, p\geq 2$ ...
- 201
17
votes
Accepted
Density of perfect numbers
Yes - Wirsing has shown that the number of odd perfect numbers in $[1,n]$ is at most $n^{O(1/\log\log n)}$. By Euler's correspondence between even perfect numbers and Mersenne primes it is certainly ...
- 7,003
9
votes
Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$?
All these identities can indeed be proved essentially trivially using modular forms and quasi-modular forms (those involving $E_2$), and the fact that the dimension
of such spaces is $1$ for weight 4,...
- 13.1k
9
votes
Accepted
Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$?
Numerical experiments suggest that
$$A_2(n) := \sum_{k=1}^{n-1} k^2\sigma(k)\sigma(n-k) = \frac{n^2}{8}\sigma_3(n) - \frac{4n^3-n^2}{24}\sigma(n).$$
PS. In fact, it directly follows from the quoted ...
- 34.3k
9
votes
Accepted
On a GCD approach to odd perfect numbers
As I mentioned in another answer, using GCDs and fractions is usually a bad idea. Here is how I would interpret all of this material.
Let $N$ be an odd perfect number. Write its prime factorization ...
- 18.7k
8
votes
Perfect numbers, Galois groups and a polynomial
The second factor is $P(2t)$ where $P=X^{p-1}+\cdots+X+1$, the $p$-th cyclotomic polynomial.
Hence the Galois group of $P(2t)$ is the same as the Galois group of $P(t)$, which is simply $(\mathbb{Z}/p\...
- 1,766
8
votes
Perfect numbers and perfect powers
If you accept the conjecture that there are no odd perfect numbers, then the finiteness follows quite easily from the $ABC$ conjecture. Thus an example has the form
$$ 2^{p-1}(2^p-1) = x^m + 1. $$
...
- 47.4k
8
votes
Accepted
Can perfect numbers be seen $p$-adically?
(Not a complete answer but a bit too long for a comment.)
There's a fundamental difficulty here in proving the sort of result you envision. If there were some prime $p$ which had to divide every odd ...
- 6,969
7
votes
Accepted
A possible axiomatic characterization of the set of divisors of a perfect number
Try this.
$$
S = \{1,2,3,12,18,36\}
$$
then
$$
\sum_{k \in S}\frac{1}{k} = 2 \in S
$$
and
$$
\frac{1}{6}\sum_{k \in S} k \pi(k) = 36 \in S
$$
when $\pi$ is the permutation of $S$ that reverses the ...
- 41.1k
6
votes
Perfect Runs of Consecutive Integers
523776, 523777 is another example of a 2-run. Again, 523776 is a triperfect number.
An example of a 3-run is 5829840, 5829841, 5829842. No apparent pattern in these three numbers:
...
- 588
6
votes
A conjecture regarding odd perfect numbers
It is rare that complicating an expression leads to an insight. There are, of course, important counter-examples to this principle. But generally speaking, one seeks to simplify an expression, ...
- 18.7k
6
votes
Generalized quasi-perfect numbers
See the abstract of my PhD thesis, "Generalised quasiperfect numbers", Bulletin Australian Math. Soc., 27 (1983), 153-156, where I consider numbers $n$ with $\sigma(n) = 2n + k^2$, $k$ odd, $(n,k)=1$.
...
- 61
6
votes
Accepted
Can an even perfect number be a sum of two cubes?
Here is a proof that 28 is the only even perfect number that is the sum of two positive cubes. The proof in Gallardo's article must be adapted in the case $x,a$ are even.
Write $N=2^{p-1}(2^p-1) = x^3+...
- 20.8k
5
votes
Accepted
The action of the unitary divisors group on the set of divisors and odd perfect numbers
Here are some general comments:
You don't need to bring these actions of abelian groups on various sets of divisors. The identity
$$\sigma(n)=\sum_{d^2|n}d\sigma^{*}(\frac{n}{d^2})$$
is easy to check ...
- 85.5k
5
votes
Accepted
If $p^k m^2$ is an odd perfect number with special prime $p$, then must $m^2 - p^k = s^2 - t^2$ hold for some $s$ and $t$?
The answer is yes. Since $p\equiv 1\pmod{4}$, and $m$ is odd, we have $m^2-p^k\equiv 0\pmod{4}$. The only integers that are not differences of squares are exactly those that are $2\pmod{4}$.
- 18.7k
4
votes
Accepted
Bounds for the number of prime numbers less than the Euler's factor, the radical and the greatest prime factor, respectively, of an odd perfect number
Since we have good asymptotics for $\pi(n)$ by the prime number theorem (and can get good explicit bounds on that from Rosser and Schoenfeld's work as well as later work such as that by Dusart) this ...
- 6,969
4
votes
Accepted
What work can be done to study the solutions of $\varphi\left(x^{\sigma(x)}\sigma(x)^x\right)=2^{x-1} x^{3x-1}\varphi(x)$?
Here is a proof that if an odd integer $x>1$ satisfies (1), then $x$ is a perfect number.
First, by using the property that $\varphi(nm)=n\varphi(m)$ whenever $\mathrm{rad}(n)\mid\mathrm{rad}(m)$, ...
- 34.3k
4
votes
Conjecture on odd perfect numbers
The earliest result appears to be
$$\frac{1}{2}<\sum_{p\mid N}\frac{1}{p}<2 \log \left (\frac{\pi}{2} \right ),$$
as proved by Perisastri in 1958. Other authors have sharpened this result (eg D. ...
- 141
4
votes
Accepted
Steuerwald's theorem
Here is a proof of this fact.
We start with a standard
Lemma 1. Any prime divisor $q$ of $1+x+x^2$ for an integer $x$ is either equal to 3 or is congruent to 1 modulo $3$.
Proof. If, on the contrary, ...
- 109k
3
votes
Accepted
Proving $k = 1 \implies q = 5$, if $q^k n^2$ is an odd perfect number with Euler prime $q$
The equivalence :
$(k=1 \lor q> 5) \iff (q=5 \implies k=1)$
is not correct on its own, except if you have some other information (I don't know the problem, I'm simply evaluating the "logical" ...
- 180
3
votes
Has it been proved that odd perfect numbers cannot be triangular?
Not an answer, but I just want to point out some thoughts that recently occurred to me, which are related to this problem.
By this answer, we know that every odd perfect number $N = q^k n^2$ can be ...
3
votes
Perfect numbers and perfect powers
Actually a stronger result (for cube powers) is proved here:
Luis H. Gallardo, "On a remark of Makowski about perfect numbers"
Elem. Math. 65 (2010) 121 – 126
The only even perfect number ...
- 3,866
3
votes
Accepted
Inductively computing Mersenne primes / perfect numbers?
The conjecture fails for $n=8128$, which can be verified in matter of seconds as explained below. I used PARI/GP for my verification.
First, since the conjecture concerns only values of at $x$'s being ...
- 34.3k
3
votes
Bounds for two arithmetic functions, when one assumes that $n$ are odd perfect numbers
As far as I'm aware, we don't have any substantially non-trivial bounds on the behavior of $\psi(n)$ when $n$ is an odd perfect number.
We can at least prove the following but none of these are ...
- 6,969
3
votes
Accepted
On the nearest-square function and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number
Middle of page 6 of
https://arxiv.org/pdf/1312.6001v10.pdf
" we always have $0 < n−\lceil\sqrt{n^2−q^k}\rceil$ "
No, this requires that $q^k\ge 2n-1$,
an helpful assumption when the goal ...
- 572
3
votes
Accepted
Number of distinct near-squares primes dividing an odd perfect number
In general, very few prime factors in an odd perfect number can be of the form $n^2+1$.
In particular, if N is an odd perfect number then $\frac{\sigma(N)}{N}=2$, and for any $m$ (perfect or not), $\...
- 6,969
3
votes
Accepted
Divisibility relation with a specific sum of divisors
From the paper of Touchard that is linked in the question on we get the relation $$3nS_0(n)-\frac{n(n-1)\sigma(n)}{6}=\frac{10}{n}S_2(n)
....(1)$$
here $S_i(n)=\sum_{k=1}^{n-1}k^i\sigma(k)\sigma(n-k)$....
- 1,755
3
votes
Accepted
On the largest prime factor and the largest component of an odd perfect number
Not a complete or as organized as an answer should be, too long for a comment. (Also I suspect that most of what I write here is going to be stuff you already know.)
One naïve approach here is to try ...
- 6,969
2
votes
On the nearest-square function and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number
You are asking if $m\lt p^k$ can be proved in the following way :
We have
$$\Bigg(m + \left\lceil{\sqrt{m^2 - p^k}}\right\rceil\Bigg)\Bigg(m - \left\lceil{\sqrt{m^2 - p^k} }\right\rceil\Bigg) = p^k +...
- 4,757
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