Compute $x_1$ through $x_8$ and factor each; compare the set of primes appearing in $x_n$ with the binary digits of $n$

The recursion is binary counting: dividing by $X(x_n)$ clears the trailing prime “ones” and multiplying by $p(x_n)$ carries a $1$ into the next bit

List the primes in order as $\rho_0 = 2,$ $\rho_1 = 3,$ $\rho_2 = 5, \ldots.$ Every $x_n$ is squarefree, so it is described by the set of primes dividing it, and we claim this set encodes $n$ in binary: if $n = \sum_i d_i 2^i$ with $d_i \in \{0, 1\},$ then $x_n = \prod_i \rho_i^{d_i}.$

Indeed, suppose $x_n = \prod_i \rho_i^{d_i}$ and let $j$ be the smallest index with $d_j = 0.$ Then $p(x_n) = \rho_j,$ and $X(x_n) = \rho_0 \rho_1 \cdots \rho_{j-1}$ is exactly the product of the primes for the trailing $1$-bits (with $X(x_n) = 1$ when $j = 0$). So $$x_{n+1} = \frac{x_n \, \rho_j}{\rho_0 \rho_1 \cdots \rho_{j-1}}$$ removes the trailing ones and inserts $\rho_j$ — precisely adding $1$ in binary. Since $x_0 = 1$ corresponds to $0,$ induction proves the claim.

Now $2090 = 2 \cdot 5 \cdot 11 \cdot 19 = \rho_0 \rho_2 \rho_4 \rho_7,$ which corresponds to binary digits at positions $0,$ $2,$ $4,$ $7.$ Hence $t = 2^0 + 2^2 + 2^4 + 2^7 = 149.$