An Infinite Sequence of Positive Integers!

Consider a number $n$ of the form $n=p_1^{x_1}p_2^{x_2}\dots p_m^{x_m}$ where $p_k$ is a distinct prime and $x_k$ is a positive integer. There are $x_k+1$ ways to pick how many times $p_k$ divides $n$ , therefore the number of divisors of $n$ is equal to $(x_1+1)(x_2+1)\dots(x_m+1)$ . Setting $x_1=4$ gives us numbers with a number $5t$ divisors. So long as there is a prime that divides $n$ $4$ , the number of divisors of $n$ will be divisible by $5$ .

Our infinite sequence begins with $a_0 = 16$ . We can see that $16 = 2^4$ . Also, note that the next number $n$ that fits our description is $48=2^43$ . We know this has to be the next number because $2^5$ has $6$ divisors, the next greatest prime number is $3$ , and $3^4=81 > 48$ .

$48-16=32=2^5$

$a_n = a_{n-1}+2^5$

$a_n = 2^4+n2^5=2^4(2n+1)$

Each number in this sequence is divisible by $2^4$ , and $2n+1$ is clearly odd so $2$ can only divide $a_n$ $4$ times, meaning that $a_n$ must have $5t$ divisors, just like every term before it. Since $32$ was the minimum common difference with a sequence starting at $16$ , our final answer must be $\boxed{32}$ .

Due to Dirichlet's theorem, in order to have such an arithmetic progression we must have $gcd(16,d) > 1$ , where $d$ is the common difference. Also 16 is the smallest number to have 5 divisors. So, each term in the arithmetic progression must be of the form $16k$ , where $k$ is odd. This is because if $a$ and $b$ are coprime to each other,then $\tau(ab)$ = $\tau(a)$ . $\tau(b)$ . So, if $k$ is odd we would have $\tau(16k)$ = $\tau(16)$ . $\tau(k)$ = $5$ . $\tau(k)$ , which is divisible by 5. Smallest odd number greater than $1$ is $3$ . So, setting the next term $3 \times 16$ = 48, we obtain smallest value of $d$ = $\boxed{32}$ .