The denominator of a term in the expansion of $F(x)^{2}$ is equal to $n^{x}$ if and only if it is a product of two terms of $F$ of the form $\frac{d(n / k)}{(n / k)^{x}}$ and $\frac{d(k)}{k^{x}}$ for some divisor $k$ of $n$. Thus $a\left(105^{m}\right)=\sum_{k \mid 105^{m}} d(k) d\left(\frac{105^{m}}{k}\right)$. We can write $k=3^{a} 5^{b} 7^{c}$ with $a, b, c \leq m$ for any divisor $k$ of $105^{m}$, and in this case $d(k)=(a+1)(b+1)(c+1)$. Thus the sum becomes $\sum_{0 \leq a, b, c \leq m}(a+1)(b+1)(c+1)(m-a+1)(m-b+1)(m-c+1)$. For a fixed $b$ and $c$, we can factor out $(b+1)(c+1)(m-b+1)(m-c+1)$ from the terms having this $b$ and $c$ and find that the sum is equal to $a\left(105^{m}\right) =\sum_{0 \leq b, c \leq m}(b+1)(c+1)(m-b+1)(m-c+1)\left(\sum_{a=1}^{m+1} a(m-a+2)\right) =\sum_{0 \leq b, c \leq m}(b+1)(c+1)(m-b+1)(m-c+1)\left((m+2) \frac{(m+1)(m+2)}{2}-\frac{(m+1)(m+2)(2 m+3)}{6}\right) =\sum_{0 \leq b, c \leq m}(b+1)(c+1)(m-b+1)(m-c+1)\left(\frac{(3 m+6-2 m-3)(m+1)(m+2)}{6}\right) =\sum_{0 \leq b, c \leq m}(b+1)(c+1)(m-b+1)(m-c+1)\binom{m+3}{3}$. Fixing $c$ and factoring out terms again, we find by a similar argument that $a\left(105^{m}\right)=\binom{m+3}{3}^{3}$.