Why Only Semi-Perfect

First we note that $n = 1$ satisfies the condition, if in fact the condition is that the "divisor product" equals $n^{3}.$

To achieve a "divisor product" of $n^{3}$ for $n \gt 1$ , we will need to have three distinct pairs of divisors that have a product of $n.$ This implies that $n$ must have precisely $6$ divisors. So we will need to look for the least $n$ which have prime decompositions of either the form $p^{5}$ or $p^{2}q$ for primes $p,q.$ This gives us

$2^{2}*3 = 12, 2*3^{2} = 18, 2^{2}*5 = 20, 2^{2}*7 = 28$

$2^{5} = 32, 2^{2}*11 = 44, 3^{2}*5 = 45.$

The desired sum is then $1 + 12 + 18 + 20 + 28 + 32 + 44 + 45 = \boxed{200}.$

Using the Prime Factorization Theorem we can find a charatherization of the semi multiplicative perfect numbers. Suppose $n$ to have three distinct primes in its factorization: $p,q,r$ . Then $\dfrac{n}{p}, \dfrac{n}{q}, \dfrac{n}{r}$ and $\dfrac{n}{pq}, \dfrac{n}{qr}, \dfrac{n}{pr}$ are distinct divisors of $n$ (possibly and likely not all the divisors) Their products $\rho$ equals to $\dfrac{n^6}{p^2q^2r^2} \geq n^4 > n^3$ so this implies $n$ can't be a semi-multiplicative perfect number (the product of ALL its divisors being obviously not lower than $rho$ and hence strictly greater than $n^3$ ).

So we can limit our search of semi-multiplicative perfect numbers to numbers like $p^n$ and numbers like $p^nq^m$ .

About the type $p^n$ we have easily the list of its divisors $1, p, p^2, \ldots, p^{n-1},p^n$ their products being $\displaystyle{p^{\frac{n(n+1)}{2}}}$ If $p^n$ is a semi-multiplicative perfect number we have ${\frac{n(n+1)}{2}} = 3n$ and this lead necessarily to $n = 0$ or $n = 5$ .

About the type $a= p^nq^m$ (with $m,n \geq 1$ ) we can see that if BOTH $m,n \geq 1$ than we have among the $a$ 's divisors $a, \dfrac{a}{p}, \dfrac{a}{q}$ and $\dfrac{a}{pq}, \dfrac{a}{p^2}, \dfrac{a}{q^2}$ when just their products $\rho$ equals to $\dfrac{a^6}{p^4q^4} \geq \dfrac{a^6}{a^2} = a^4 > a^3$ and this means $mn \in \{m,n\}$ . If both $m=n=1$ we have easily the list of the divisors $1,p,q,pq$ and their product is $a^2$ (in this case we have a multiplicative perfect number, but not semi!). \ So we have one among $m$ and $n$ is $1$ (say $m=1$ ) and the other is greater (say $n \geq2$ ). How greater? \ If $n \geq 3$ we have the following distinct divisors $a, \dfrac{a}{p}, \dfrac{a}{p^2}, \dfrac{a}{p^3}, \dfrac{a}{q}, \dfrac{a}{pq}, \dfrac{a}{p^2q}$ whose product $\rho$ equals to $\dfrac{a^7}{p^9q^3} > \dfrac{a^7}{a^3} = a^4 > a^3$ and hence necessarily $n = 2$ . \ Now for naturals of the type $a = p^2q$ we have easily the list of the divisors and their products is in fact $a^3$ , as showed below $1 \cdot p \cdot p^2 \cdot q \cdot pq \cdot p^2q = p^6q^3 = a^3.$

Finally the semi-multiplicative perfect numbers are exactly the positive integers whose prime factorisation is of the type $p^0=1$ , $p^5$ or $p^2q$ .

Writing down the smallest numbers of this type we get the list

$1, 12, 18, 20, 28, 32, 44, 45$

and the sum is $200$ .

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

public static void main(String args[]){
    int sum = 0;
    for(int i = 1, n = 0; n != 8; i++)
        if(factorProduct(i) == i * i * i){
            sum += i; n++;
        }
    System.out.println(sum);
}

private static int factorProduct(int n){
    int prod = n;
    for(int i = 2; i <= n / 2; i++)
        if(n % i == 0) prod *= i;
    return prod;
}

It was tagged with Computer Science...