How do number shapes relate to k-perfect numbers? 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
 A: 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.
