Someone called James Davis (see here) found that $$ 13532385396179 = 13\times 53^2\times 3853\times 96179, $$ showing that a composite number can be equal the concatenation of the primes and exponents in its canonical prime factorization. In general, if a composite number is equal the concatenation in base $b$ of the primes and exponents in its prime factorization, then let's call it a Davis number to base $b$. For example, $255987$ is a Davis number to base $2$: $$ 111110011111110011=11^{11}\times10011\times111110011\, (\text{in decimal}:255987=3^{3}\times19\times499); $$ $4617$ is a Davis number to base $11$: $$ 3518=3^{5}\times18\,(\text{in decimal}:4617=3^5\times 19); $$ $29767$ is a Davis number to base $12$: $$ 15287=15^{2}\times87\,(\text{in decimal}:29767=17^2\times 103). $$
Using symbols, let $n$ be a composite number, and $d_1,\cdots,d_r$ be the list of numbers $>1$ that appear in order in the canonical prime factorization of $n$. For example, for $n=13532385396179$, we have $$ (d_1,d_2,d_3,d_4,d_5) = (13,53,2,3853,96179). $$ By definition $n$ is a Davis number to base $b$ iff $$ n = d_1\times b^{r-1+\lfloor\log_b d_2 \rfloor+\cdots+\lfloor\log_b d_r \rfloor} + d_2\times b^{r-2+\lfloor\log_b d_3 \rfloor+\cdots+\lfloor\log_b d_r \rfloor} + \cdots + d_{r-1}\times b^{1+\lfloor\log_b d_r \rfloor} + d_r. $$ In particular, $b$ must divide $n-d_r$.
My question is: Can a composite number $n$ be a Davis number to more than one base? Of course, the question is meaningless if $b$ is very large (for example if $b>\max\{d_r\}$), but the behavior of $b\mapsto b^{1+\lfloor\log_b d_i\rfloor}$ is rather chaotic for $b$ very small. What's more, it seems difficult to represent $n$ using $d_1,\cdots,d_r$. Thank you for any help in advance.
