Problem 4.  A proper divisor of a positive integer N is a positive divisor of N other than N itself.The infinite sequence a1,a2,... consists of positive integers, each of which has at least three proper divisors. For each n≥1, the integer an+1 is the sum of the three largest proper divisors of an.

Determine all possible values of a1

Solution 1Solution 2

Solution 1

The answer is all integers of the form $\boxed{6\cdot12^a m}$, where $a\geq 0$ and $m$ is relatively prime with $10$.

Let $f(n)$ be the sum of the three largest proper divisors of $n$. Note that $f(n)>n$ if the smallest four divisors of $n$ are $1,\,2,\,3,\,4$ or $1,\,2,\,3,\,5$, $f(n)=n$ if the smallest four divisors of $n$ are $1,\,2,\,3,\,6$, and $f(n)<n$ otherwise.

Claim 1. $a_1$ is even.
Proof. Suppose FTSOC $a_1$ is odd and its smallest divisors are $1,\,a,\,b,\,c$ where $a,\,b,\,c$ are odd. Then $a_2=f(a_1)<a_1$ and


\[ a_2=a_1\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{a_1(ab+bc+ca)}{abc}. \]

Since $ab+bc+ca$ is odd, $a_2$ is odd. Eventually the sequence terminates, a contradiction. $\square$

Claim 2. $a_1$ is a multiple of $3$.
Proof. Suppose FTSOC $3\nmid a_1$ and its smallest divisors are $1<2<a<b$ where $3\nmid a,\,b$. Then $a_2=f(a_1)<a_1$ and


\[ a_2=a_1\left(\frac{1}{2}+\frac{1}{a}+\frac{1}{b}\right)=\frac{a_1(2a+2b+ab)}{2ab}. \]

Doing casework on $(a,\,b)\mod 3$ yields that only $a\equiv b\equiv 2\pmod{3}$ gives $3\mid 2a+2b+ab$. If $3\nmid 2a+2b+ab$, then $3\nmid a_2$ so eventually the sequence terminates, a contradiction.

Hence assume $a\equiv b\equiv 2\pmod{3}$. Note that neither $a$ nor $b$ is $4$, so $4\nmid a_1$. If neither $a$ nor $b$ is even, then $2a+2b+ab$ is odd and $\nu_2(a_1)=\nu_2(2ab)=1$, so $a_2$ is odd, a contradiction. Thus $b$ is even and $a=b/2$. This is a contradiction with $a\equiv b\equiv 2\pmod{3}$. $\square$

Thus let $a_1=:6n$.

Case 1: $n$ is an odd multiple of $5$. Write $n=:5m$. Then $f(30m)=31m$ is odd so the eventually the sequence terminates.

Case 2: $n$ is even. Write $n=:2m$. Then $f(12m)=13m$ so $6\mid m$. Thus $n$ works iff $m$ works. This yields the solution set.

Case 3: $n$ is relatively prime with $10$. Then $f(6n)=6n$ so this works.

Solution 2

For any $n$ integer let $f_1(n),f_2(n),f_3(n)$ the first, second and third minor divisors of $n$, diferent to $1$.
So

\[a_{n+1} = a_n(\frac{1}{f_1(a_n)}+\frac{1}{f_2(a_n)}+\frac{1}{f_3(n)})\]


If exist $n$ such that $a_n$ odd:
$f_1(n)\geq 3,f_2(n)\geq 5, f_3(n)\geq 7$, so $a_{n+1}<a_n$ and is easy to see that $a_{n+1}$ it's also odd.
So the sequence is strictly decreasing from now on, which is a contradiction, because the sequence is infinite.

So $a_n$ is even forall $n$ integer: so $f_1(a_n) = 2$, forall $n$ integer.


\[a_{n+1} = a_n(\frac{1}{2}+\frac{1}{f_2(a_n)}+\frac{1}{f_3(a_n)})\]

Now if $f_2(a_n)\geq 4$, for some $n$ integer, the sequence $a_n$ is strictly decreasing from now on, which is a contradiction.
So $f_2(a_n)=3$, forall $n$ integer.



\[a_{n+1} = a_n(\frac{1}{2}+\frac{1}{3}+\frac{1}{f_3(a_n)})\]


Since $2|a_n$ and $3|a_n$, we have $6|a_n$, so $f_3(a_n) \in {4,5,6}$

if $f_3(a_n) = 5$: so $a_{n+1} = a_n\cdot\frac{31}{30}$, so $30\cdot a_{n+1}|a_n$, but $a_{n+1}$ is even, so $4|a_n$ but $f_2(a_n)=3<4<5 = f_3(a_n)$.

If $f_3(a_n) = 4$, we have $a_{n+1} = a_n\cdot\frac{13}{12}$
If $f_3(a_n) = 6$, we have $a_{n+1} = a_n$

Then we can say that $a_n = a_1\cdot (\frac{13}{12})^{x_n}$, forall $n$ integer, where $x_n$ is the number of $n$ such that $4|a_n$.
Now $a_n$ is an integer, so $v_2(12^{x_n})\leq v_2(a_1)$, so $x_n$ is a bounded sequence, so it's eventually constant, i.e. forall $n>N$
we have $x_n=c$, so $a_n = a_1\cdot (\frac{13}{12})^c$, so $a_1 = 12^c\cdot X$, since $a_n$ is constant for $n>N$, we have that $4\nmid a_n$, but $a_n = X\cdot 13^c$ and $6|a_n$, so $X = 6\cdot A$ with $(A,2) =1$, so $a_1 = 6\cdot 12^c\cdot A$.
Finally also $(A,5) =1$, so we have $a_1 = 6\cdot 12^c \cdot A$, with $(A,10)=1$.

Solution 3 Solution3
Solution 4 Solution4