Let $n>1$ be an integer and $F_n:=2^{2^n}+1$.
Result: $n$ even: $F_n\equiv17\pmod{210}$
Result: $n$ odd: $F_n\equiv47\pmod{210}$
Proof: Suppose $n$ even, $F_n - 1\equiv16\pmod{210}$. Then $2^{2^{n+2}}=({2^{2^n}})^4\equiv16^4\pmod{210}\equiv16\pmod{210}$ and $F_{n+2}\equiv17\pmod{210}$. The proof is the same if $n$ is odd.
I found that result using primoradic (see stub OEIS: https://oeis.org/wiki/Primorial_numeral_system).
That's the way I found that result see here, which I did't know before and I wonder if there are other results with $2310$, $30030,\ldots$ (primorials)