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 6 of 8: The middle case cannot occur in the sequence

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v_2(x)=1,\,5\mid x\ \Longrightarrow\ \sigma(x)=\tfrac{31}{30}x\ \text{is odd, contradicting that every term is even}

Detailed analysis

Suppose some term has $4\nmid x$ but $5\mid x$ (so $6\mid x$ , $5\mid x$ , $v_2(x)=1$ ); write $x=30k$ with $k$ odd (since $v_2(x)=1$ and $30=2\cdot15$ with $15$ odd forces $v_2(k)=0$ ). Then $\sigma(x)=\tfrac{31}{30}x=31k$ , a product of two odd numbers, hence odd. But step 3 shows every term of the sequence must be even, a contradiction. So this middle case never actually occurs among the $a_i$ : whenever $6\mid a_n$ and $4\nmid a_n$ , automatically $5\nmid a_n$ as well.