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 7 of 8: Only two behaviors remain, giving the closed form

Previous step Next step

a_1=(\tfrac{12}{13})^{T-1}a_T=12^{T-1}\cdot6\ell,\qquad \gcd(\ell,10)=1

Detailed analysis

By steps 3–5, only two behaviors ever occur: $a_{n+1}=\tfrac{13}{12}a_n$ whenever $4\mid a_n$ , or $a_{n+1}=a_n$ whenever $4\nmid a_n$ (in which case necessarily $5\nmid a_n$ too, by step 5). Since the terms cannot keep being multiplied by $\tfrac{13}{12}$ forever (they would grow without the divisibility structure needed to stay in $S$ in the required form), there is a smallest index $T$ with $a_T=a_{T+1}=\cdots$ , constant from then on, and $a_n=a_1\cdot(\tfrac{13}{12})^{\min(n,T)-1}$ . Writing $a_T=6\ell$ with $\ell$ coprime to $10$ (since $4\nmid a_T$ , $5\nmid a_T$ , and $6\mid a_T$ ), we get $a_1=(\tfrac{12}{13})^{T-1}a_T=12^{T-1}\cdot6\ell$ .