Different maddening numbers

The world is simply full of numbers of all kinds. All the clutter on numbers was made by none but man. Fibonacci and Lucas numbers; perfect numbers; Fermat numbers; Multiply-perfect numbers; abundant and super-abundant numbers, practical and impractical (!) numbers, etc. However, there is an interesting correlation between all these junk of numbers. In this entry, I try to let us get an insight into Perfect, Multiply-perfect, Abundant and Super-abundant and Practical numbers.
As usual, all of us have realized from the definition of a perfect number that what a multiply-perfect number is.
Definition and properties of the \sigma function
Let us, for the sake of clarity, define our notations first. We denote \sigma (n) as the sum of the divisors of n, and \tau (n) as the number of the divisors of n. We try to express \tau(n) in terms of the prime divisors of n. Note that, by the fundamental theorem we have that any n can be expressed in the form of (where p_i‘s are the primes arranged in order or not in order)
n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k};
A simple combinatorial argument leads to the obvious and well-known fact that
\tau (n)=(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1) (including 1 as a factor).
Now, for a formula of \sigma(n) we see that all the divisors d_i are of the form
d_i=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}; and 0\leq \beta_i\leq \alpha_i. Thence the sum, can be factored into (or rather, found out to be) terms in the expansion of \prod_{i=1}^n(1+p_i+\cdots+p_i^{\alpha_i})=\prod_{i=1}^n\frac{p_i^{\alpha_i+1}-1}{p_i-1}.
Now it is evident that when \text{gcd}(m,n)=1 we have the multiplicity property of this \sigma function, ie \sigma(mn)=\sigma(m)\sigma(n). The (obvious) proof is left to the reader.
Multiply-Perfect Numbers
Before discussing the general multiply-perfect numbers, we dive into the perfect numbers for a little time. Let us try to list some of the perfect numbers. They are 6,28,496,8128,\cdots written in order. The next values can be generated with a computer, but what do we see common in these numbers?
(1) (Obviously) All are even and \sigma(n)=2n; which makes them perfect.
(2) 6=2\cdot (2^2-1); and 3=2^2-1 is a prime.
28 = 2^2\cdot (2^3-1); and 2^3-1=7 is a prime.
496 = 2^4\cdot(2^5-1); and 2^5-1=31 is a prime.
8128 = 2^6\cdot(2^7-1); and 2^7-1=127 is a prime.
Proceeding in this manner, Euler proved (again, how many theorems did he prove?) that if n is a given number with the property that 2^n-1 is a prime, that 2^{n-1}\cdot(2^{n}-1) is going to be a perfect number.
With this (extremely) brief idea of perfect numbers we are able to give a generalization. We consider a number n such that its \sigma (n)=kn where k is some natural number.
Before getting some information and problems on multiply-perfect numbers, let us try a problem.
\boxed{E1.}If n is a perfect number and n+1, n-1 are primes, find all possible values of n
Solution.
Obviously considering modulo 6 we have that 6|n\implies n=6n_1 for some n_1\in\mathbb N. If so, we consider n_1>3. In this case,
\sigma(6n_1)>1+2+3+6+n_1+2n_1+3n_1+6n_1>12n_1+12, which is way too bizarre for a perfect number 6n_1. So n_1\leq 3. So we get n=6,12,18. Indeed, 5,7 and 11,13 and 17, 18 are all primes.\Box
Let us denote a multiply perfect number of the order k by x_k whereas \sigma(x_k)=kx_k. We play around with a few multiply-perfect numbers of some orders. Just as in the previous case, we try to list some properties of a x_k number. At the first glance into a k- tuply perfect number like 120 or 672 (Verify with a calculator or the prime factorization formulae for \sigma(n)), we might try to conjecture a few things. Like, all x_k are even. We might try to construct a sequence of these numbers for different k, and conjecture that there does or does not exist such a sequence. These conjectures are needed in mathematics, but we still need a proof of those, that is why they are called conjectures. :lol:
Very few has been known about large k‘s, so let us start by noting some examples.
\boxed{E2.}Let n be a number such that 3 does not divide n yet it satisfies \sigma(n)=3n, show that O\sigma(3n)=12n, ie 3n is a perfect number of order 4.
Solution.
It is obvious from the multiplicative property of \sigma(n) that
\sigma(3n)=3\sigma(n)=12n; so that 3n is an x_4-type number.
\boxed{E3.}Every number such that \sigma(n)=5n has at least 6 different prime factors.
Solution.
This, also is obvious on using the prime decomposition of n, which is
n=p_1^{\alpha_1}\cdots p_k^{\alpha_k} gives us
\prod_{i=1}^n\frac{p_i^{\alpha_i+1}-1}{p_i^{\alpha_i}(p_i-1)}=5.
Now using \frac{p_1^{\alpha_1+1}-1}{p_1^{\alpha_1}(p_1-1)}>\frac{p_1}{p_1-1}; which is obvious on expanding, we have
\frac{p_1}{p_1-1}\cdots\frac{p_n}{p_n-1}>5.
Now with the same old technique, we consider the product upto 5 terms, which is surely going to be <5 (because the statement is modified to contradict this event by origin); therefore we are surely done.
Superabundant Numbers
A number n is known as superabundant if and only if \dfrac{\sigma(n)}{n} exceeds \dfrac{\sigma(k)}{k} for all k<n.
\boxed{E4}. The sequence of numbers t_n=\dfrac{\sigma(n)}{n} does not have any upper bound..
Solution.
If d_i are the divisors of n then we have, d_i=\dfrac n{d_{n-i}} (why?) and summing up on this sequence we get \sigma(n) as,
\sigma(n)=\sum_{d_i|n}\dfrac 1{d_i}\implies \dfrac{\sigma(n)}{n}=\sum_{d_i|n}\dfrac 1{d_i}.
Since we have to prove that this has no upper bound, let n=m! for our convenience. Then we have,
\dfrac{\sigma(n)}{n}=\sum_{d_i|n}\dfrac 1{d_i}=\sum_{d_i|m!}\dfrac1{d_i}\geq 1+\frac 12+\frac 13+\cdots+\frac 1m.
Then we have t_n\geq H_m where H is the harmonic series. As we know the harmonic series on natural numbers does not have any upper bound as n increases, and therefore t_n is unbounded from above.
Abundant Numbers
The numbers n with the property that \sigma(n)\geq 2n are abundant ones.
12, 18,20,24\cdots are abundant at our first instance, and we can also find out that all abundant numbers \leq 100 are even.
We have a property regarding the abundant numbers.
\boxed{E5} Every number n is abundant implies that mn is abundant, m\ge 2
Solution to be found by the reader.

We will discuss on some Practical, Highly Composite Numbers and Quasi-perfect, Semi-perfect numbers soon.
Hope you do not get insane with a lot of numbers, we will get some exercises on these too.

(To Be Updated Soon)

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