Factors in a sequence

Observe that the first few terms of the sequence is $1, 2, 3, 6, 9, 18, 27, 54, \ldots$ . We seem to be getting the powers of 3 and twice the powers of 3.

Claim: The closed form of the sequence is

$a_n = \begin{cases} 3^k & n = 2k+1 \\ 2 \times 3 ^ k & n = 2k + 2 \end{cases}$

Proof: We will prove this by induction. The base case $n = 1 = 2 \times 0 + 1$ is true since $1 = a_1 = 3^0$ .
If $n = 2k$ , then $a_{2k+1 } = a_{2k} + 3^ { \lfloor \frac{ 2k-1 } { 2} \rfloor } = 2 \times 3^{k-1} + 3^{k-1 } = 3^ k$ .
If $n = 2k+1$ , then $a_{2k+2} = 1_{2k+1} + 3 ^ { \lfloor \frac{2k}{2} \rfloor } = 3 ^k + 3^k = 2 \times 3 ^ k$ .
Hence, we are done. $_\square$

Now, let's consider the factors:
$a_{2k+1} = 3^k$ has factors $1, 3, 3^2 , \ldots 3^{k-1}, 3^k$ .
$a_{2k+2} = 2 \times 3^k$ has factors $1, 3, 3^2, \ldots, 3^{k-1} , 3^k, 2\times 1, 2\times 3, 2\times3^2 , \ldots , 2\times3^{k-1}, 2\times3^{k}$ .
Hence, $a_n$ satisfies the condition of "The factors of $a_n$ are $a_i$ for $i \leq n$ " if and only if $n$ is even.

Out of $1 \leq n \leq 10$ , there are 5 such cases.

n=1, sequence is 1, and since 1 has only factor 1, this satisfies

From now onwards only even numbers are possible as if number is odd, 2nd term 2 cant satisfy.

Surprisingly, each even term has this property. I am listing the sequences.

1,2

1,2,3,6

1,2,3,6,9,18

1,2,3,6,9,18,27,54

So in total there are 5 values of n, hence the answer.

Note

The pattern is indeed true for all even n's. Proof:

Series is $2^03^0,2^13^0,2^03^1,2^13^1,2^03^2,2^13^2,...$

Let for some $n$ , all its factors are listed and $a_{n}$ is even.

$a_{n+1}$ is odd. So $2$ cant be its factor. So it cant be true for $n+1$ .

$a_{n+2}$ is even, and from the above series we know that $a_{n+2}=2a_{n+1}=3a_{n+2}$ .

So $a_{n+2},a_{n+2}/2,a_{n+2}/3,a_{n}'s$ factors all are listed and these are the only possible factors for $a_{n+2}$ .

So if for any even $n$ , if our supposition is true, it is true for all even numbers by Principle of Mathematical Induction. And we can see $n=2$ clearly satisfies. Hence proved.