[math-fun] derangements divisible by 3

given the many alternatives of writing (!n), cfr. OEIS A000166, and considering there is a nice way to find the powers of p in n! by writing n in base p (if I remember correctly), it is funny to observe the powers of 3 in !n. I get, using a[ n ] for (!n) and p as representing 3 (to avoid immediate evaluation in Mathematica): {Mod[a[1 + k*p], k*p]==0, Mod[a[1 + k*p^2], k*p^3]==0, Mod[a[1 + k*p^2], k*p^4]==0, Mod[a[1 - p^2 + k*p^3], -p^5 + k*p^6]==0, Mod[a[1 - p^2 - p^3 + k*p^4], -p^6 - p^7 + k*p^8]==0, Mod[a[1 - p^2 - p^3 + k*p^5], -p^7 - p^8 + k*p^10]==0, Mod[a[1 - p^2 - p^3 + k*p^6], -p^8 - p^9 + k*p^12]==0, ......... } and I wonder where these lead to. W.