Finding the Integer

Let $q_k$ denote the $k$ -th prime. Assign to the positive integer $m=\prod_{k=1}^\infty q_k^{a_k}$ the infinite string $S(m)=\overline{a_1,a_2,\dots}$ (of course, for some $K$ , all $k>K$ have $a_k=0$ ). Look at how we get $S(x_{n+1})$ from $S(x_n)$ . Recall that $x_{n+1}=\frac{x_n p(x_n)}{X(x_n)}$ . Let the index of the prime $p(x_n)$ in $(q_k)$ be $\ell$ , then $a_{\ell}$ is the index of the first number in $S(x_n)$ for which $a_\ell=0$ . Hence $S(x_{n+1})$ is the same as $S(x_n)$ with its first $\ell-1$ numbers decreased by $1$ , and its $\ell$ -th number increased by $1$ . Let's prove by induction that $S(x_n)$ is none other than the binary representation of $n$ written backwards. It's easy to check that this holds for small $n$ , now assume it holds for some $n>0$ . Then the binary representation of $n+1$ can be derived from $S(x_n)$ by changing the first few $1$ digits to $0$ , until we hit a $0$ , and we switch that $0$ to $1$ . But since this is the same way to reach $S(x_{n+1})$ , the two strings are the same. This establishes our claim. Since $S(2090)=S(2\cdot 5\cdot 11\cdot 19)=S(p_1\cdot p_3\cdot p_5\cdot p_8)=\overline{10101001'000\dots}$ , this means that the index $n$ we are looking for is $2^{7}+2^4+2^2+2^0=128+16+4+1=\boxed{149}$