Problem 4

A proper divisor of a positive integer

N

is a positive divisor of

N

other than

N

itself. The infinite sequence

a_1,a_2,\ldots

consists of positive integers, each of which has at least three proper divisors. For each

n\ge1

, the integer

a_{n+1}

is the sum of the three largest proper divisors of

a_n

. Determine all possible values of

a_1

Step 2 of 8: Sufficiency: these values of a1 generate a valid sequence

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x=12^e\cdot6\ell,\ \gcd(\ell,10)=1:\quad \sigma(x)=\tfrac{13}{12}x\ (e>0),\qquad \sigma(x)=x\ (e=0)

Detailed analysis

Take $x=12^e\cdot6\ell$ with $\gcd(\ell,10)=1$ . If $e>0$ then $12\mid x$ , so the three smallest divisors of $x$ exceeding $1$ are $2,3,4$ , giving $\sigma(x)=\tfrac{x}{2}+\tfrac{x}{3}+\tfrac{x}{4}=\tfrac{13}{12}x=12^{e-1}\cdot6(13\ell)$ , again of the same form with $e$ decreased by $1$ and $\ell$ replaced by $13\ell$ (still coprime to $10$ ). If $e=0$ then $x=6\ell$ with $\ell$ coprime to $10$ , so $4\nmid x$ and $5\nmid x$ ; the three smallest divisors of $x$ exceeding $1$ are $2,3,6$ , giving $\sigma(x)=\tfrac{x}{2}+\tfrac{x}{3}+\tfrac{x}{6}=x$ , a fixed point. By downward induction on $e$ , the sequence starting at any such $x$ stays in $S$ forever, eventually becoming constant, so every $a_1=12^e\cdot6\ell$ with $\gcd(\ell,10)=1$ is attainable.