Half Perfect

Solution using Python(Jupyter Notebook):

INPUT:-

---

y=1000
for i in range(1,y+1):
    s=0
         for m in range(1,i):
          x=i%m
          if x==0:
              s=s+m
     if s==2*i:
     print(i)

---

OUTPUT:-

---

120
672

---

Out of which 120 is the smallest and hence it is the solution :) i hope python as a solution is accepted.

The equation we are trying to solve is $\sigma(n) = 3n,$ where $\sigma(n)$ is the sum of the divisors of $n.$ It's not hard to see that $\fbox{120}$ is the answer by inspection. Here's an indication of how I got to it.

The fraction $\frac{\sigma(n)}{n}$ is multiplicative , and it's not hard to see that, for any prime $p,$ we have $\frac{p+1}{p} \le \frac{\sigma(p^k)}{p^k} < \frac{p}{p-1}.$ (The left inequality is because $\sigma(p^k) \ge p^k + p^{k-1},$ and the right inequality is because $\frac{\sigma(p^k)}{p^k} = \frac{p^{k+1}-1}{(p-1)p^k} < \frac{p^{k+1}}{(p-1)p^k} = \frac{p}{p-1}.$ ) So $\begin{aligned} \frac32 \le \frac{\sigma(2^k)}{2^k} &< 2 \\ \frac43 \le \frac{\sigma(3^k)}{3^k} &< \frac32 \\ \frac65 \le \frac{\sigma(5^k)}{5^k} &< \frac54 \end{aligned}$ so if $n$ has only $2$ s and $3$ s in its prime factorization, we always have $\frac{\sigma(n)}{n} < 2 \cdot \frac32 = 3.$ This argument also shows that a solution $n$ must have at least three distinct prime factors.

So let's try $n$ of the form $2^a 3^b 5^c.$ In this case, we will need $\sigma(2^a)$ or $\sigma(3^b)$ to be divisible by $5.$ The first such $b$ is $b=3,$ which leads to a number that's at least $270.$ The first such $a$ is $a=3.$ In this case, $\sigma(2^3)/2^3 = 15/8.$ In this case, $\frac{15}8 \frac43 \frac65 \le \frac{\sigma(2^3 3^b 5^c)}{2^3 3^b 5^c},$ but the left side equals 3, and equality holds precisely when $b=c=1,$ which gives us $n=2^3 \cdot 3 \cdot 5 = 120.$

There aren't very many numbers less than $120$ with three distinct prime factors, so it's easy to check them all directly to see that $\sigma(n) < 3n$ for all of them.

Using Mathematica: i = 1; While[True, If[i == (Total@Divisors[i] - i)/2, Print[i]; Break[]]; i++ ]

We are looking for $n$ where $σ(n) = 3n$

(where $σ(n)$ = sum of divisors of $n$ including $n$ )

A perfect number has $σ(n) = 2n$ and therefore a half perfect number is $σ(n) = 3n$ .

Note: we are looking for proper divisors of $n$ so we discount $n$ and therefore $σ (n) = 3n$ in order for an integer to be exactly half the sum of its proper divisors.

Example: the proper divisors of $12$ are $1, 2, 3, 4, 6$ where the sum $= 16$ . $σ(12) = 28$ .

The formula for $σ(p^k) = \frac {p^{k+1} - 1} {p - 1}$

For example, consider $n = 28$ , which is a perfect number:

$n = 2^2 * 7^1$

=>

$σ(28) = σ(2^2 * 7^1) = σ(2^2) * σ(7^1) = (\frac {2^{3} - 1} {2 - 1}) * (\frac {7^{2} - 1} {7 -1}) = 56$

=>

$σ(28) = 56 = 2n$ = perfect number.

A Solution

With the knowledge that $σ(n) = 3n$ , I first started by considering the values of $\frac {2^{k} - 1} {2-1} = 2^{k} - 1$ where $3$ is a factor. So $2^{4} - 1$ , $2^{6} - 1$ , $2^{8} - 1$ etc etc

Take $σ(2^{3}) = \frac {2^{4} - 1} {2 - 1} = 15 = 3 * 5$ which is a factor of $σ(n)$ if $n = 2^{3} * a * b* c......$

So in our value of $σ(n)$ we will have $\frac {2^{4} - 1} {2 - 1} = 15 = 3 * 5$ as a factor.

Therefore, as $σ(n)$ is $3$ times $n$ , I began by letting $n = 2^{3} * 5 * a * b .....$ not including the factor $3$ from $σ(2^{4})$ because of the fact $σ(n) = 3n$ .

Note that we have not yet accounted for $2^{3}$ in our value of $n$ which is our aim.

With $5$ added as a factor of $n$ we consider $σ(5^{k}) = \frac {5^{k} - 1} {5-1}$ which will be a factor of $σ(n)$ .

Let's consider $σ(5^{1}) = \frac {5^{2} - 1} {5-1} = 6 = 2 * 3$

Now let $n = 2^{3} * 3 * 5 * .....$ and we have now accounted for $2^{1}$ from our value of $σ(5^{1})$

With $3$ added as a factor of $n$ we consider $σ(3^{k}) = \frac {3^{k} - 1} {3-1}$ which will be a factor of $σ(n)$ .

Let's consider $σ(3^{1}) = \frac {3^{2} - 1} {3-1} = 4 = 2^{2}$ .

From this result we add no more distinct prime factors to the value of $n$ and have accounted for the remaining factorisation of $2^{3}$ .

Therefore $n = 2^{3} * 3 * 5$ .

I hope this explanation does make sense. It is one way to solve the question.

I had initially got the question half-wrong, in that I discovered a half perfect number but not the smallest. This is because I actually first considered $2^{6} -1 = 63$ and in my haste rushed to produce an answer.

The solution for that is as follows:

Take $σ(2^{5}) = \frac {2^{6} - 1} {2 - 1} = 63 = 3 * 3 * 7$ which is a factor of $σ(n)$ if $n = 2^{5} * a * b* c......$

Let $n = 2^{5} * 7 * 3 * a * b .....$ removing a factor $3$ .

Consider $σ(7^{1}) = \frac {7^{2} - 1} {7-1} = 8 = 2^{3}$

Now let $n = 2^{5} * 7 * 3 * a * b .....$ with no distinct primes being added and $2^{3}$ being accounted for.

Consider $σ(3^{1}) = \frac {3^{2} - 1} {3-1} = 4 = 2^{2}$

Now let $n = 2^{5} * 7 * 3$ be our final answer with no distinct primes being added and the remaining $2^{2}$ being accounted for.

Therefore $n = 2^{5} * 7 * 3 = 672$ and is a half perfect number.

$σ(672) = σ(2^{5}) * σ(7^{1}) * σ(3^{1})$

$= \frac {2^{6} - 1} {2 - 1} * \frac {7^{2} - 1} {7 - 1} * \frac {3^{2} - 1} {3 - 1} = 63 * 8 * 4 = 2016 = 3 * 672$

=>

$672$ is a half perfect number.