Conjecture: The only intersection with
A385115 is at k = 3 where 2^4 * 3^3 = 432 =
A027856(8).
Idea: For odd k > 3, covering systems ensure mutual exclusion:
If k = 1, 9, 13, 19, 25, 31, 37, 39, 43, 49, 55 (mod 60), then 7 or 31 divides (16*3^k+1).
If k = 5, 7, 11, 17, 23, 27, 29, 35, 41, 47, 53, 57, 59 (mod 60), then 11 or 13 divides (16*3^k-1).
If k = 15, 21, 33, 45, 51 (mod 60), various primes including {11,31,43,109,277,433,...} ensure at least one of 16*3^k +- 1 is composite.
If k = 3 (mod 60) and k > 3, the probability of intersection becomes vanishingly small.
Only k = 3 escapes all divisibility conditions. Verified to k = 10^5.
