2

We know that all perfect numbers are a Mersenne prime, multiplied with the corresponding power of 2 for that prime, and then halved.$$2^{n}-1(2^{n-1})$$ It is also true that all perfect numbers are triangular numbers.

I'm trying to find other patterns. We know that tri-perfect numbers exist. The sum of tri-perfect number's factors equals 3 times the tri-perfect number.

I was trying to think what patterns there are? 120, a tri-perfect number is hexagonal, but 672 is not. Is there some easy formula to find the shape of a k-perfect number? (where k is 2 for perfect, 3 for tri-perfect etc). Does the number of dimensions needed to display k-perfect numbers increase as k does?

Also, even perfect numbers are closely related to Mersenne primes. Is there another type of prime number for tri-perfect numbers? Do you have to do something else to a Mersenne prime to get a tri-perfect number?

I find perfect numbers perfectly interesting but man they are confusing. Thanks, Andy

Spammer
  • 91
  • 3
  • 3
    Not true. We only know that all even perfect numbers are of that form, and that all even perfect numbers are triangular. Nobody has proven the non-existence of an odd perfect number yet, funnily enough. – merelymyself Sep 17 '21 at 09:29
  • 1
    @merelymyself is correct. It is currently unknown whether odd perfect numbers are triangular or not. So without a proof for the nonexistence of an odd perfect number, or a proof that odd perfect numbers are also triangular, you cannot say that all perfect numbers are triangular. – Jose Arnaldo Bebita Dris Sep 17 '21 at 10:12
  • @Opti_byte: I will have more to say about the shape of a $k$-perfect number (also known as multiperfect numbers), which is derived from work of Holdener and one of her past undergraduate research mentees. Let me pull up that information for you real quick. – Jose Arnaldo Bebita Dris Sep 17 '21 at 10:15

1 Answers1

1

From the undergraduate research project titled The Form of Perfect and Multiperfect Numbers by Judy Holdener and Kaitlin Rafferty (Kenyon College, 2009), we have the following:

Euler's Characterization of Odd Perfect Numbers

  • If an odd perfect number exists, then it is of the form $$n = p^{\alpha} {q_1}^{2\beta_1} \cdots {q_r}^{2\beta_r}$$ where $p$ and the $q_i$'s are distinct primes, and $p = 1 + 4m_1$ and $\alpha = 1 + 4m_2$.

Generalization of Euler's Characterization

  • Theorem: Let $n$ be a positive integer with unique factorization $$n = 2^r \prod_{i=1}^{k}{p_i}^{\alpha_i}\prod_{j=1}^{l}{q_j}^{\beta_j},$$ with $p_i \equiv 1 \pmod 4$ and $q_j \equiv 3 \pmod 4$. If at least one $\beta_j$ is odd, then $4 \mid \sigma(n)$. If all the $\beta_j$'s are even then $$\sigma(n) \equiv \begin{cases}{ \prod_{i=1}^{k}{\Bigg(\alpha_i + 1\Bigg)} \pmod 4 \text{ if } n \text{ is even } \\ 3\prod_{i=1}^{k}{\Bigg(\alpha_i + 1\Bigg)} \pmod 4 \text{ if } n \text{ is odd.}} \end{cases}$$
  • Corollary: If $n \equiv 1 \pmod 4$, then $$\sigma(n) \equiv \prod_{i=1}^{k}{\Bigg(\alpha_i + 1\Bigg)} \pmod 4,$$ and if $n$ is multiperfect with multiplicity $K$, then $$K \equiv \prod_{i=1}^{k}{\Bigg(\alpha_i + 1\Bigg)} \pmod 4.$$
  • Theorem: If $n$ is an odd multiperfect number with multiplicity $K$ and $2 \parallel K$, then $n = p^{\alpha} m^2$ where $p$ is prime and $p \equiv \alpha \equiv 1 \pmod 4$.

The paper is available via JSTOR.

Jose Arnaldo Bebita Dris
  • 8,471