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Let $ n $ be a $ k $-multiperfect number. Denote by $ d_m $ its $ m $ smallest divisor, and $ n_{m} $ the number of divisors of $ n $ divisible by $ d_m $. Is there for all $ 2\leq m\leq\tau(n) $ an integer $ s(m) $ such that $ n_m=d_{s(m)} $? Does the function $ s $ admit a fixed point ?

Edit : actually $ n_{m}=\tau(n/d_{m}) $. A sufficient condition for a number $ n $ to fulfill $ d=\tau(n/d) $ is to take $ n=\prod_{r}p_{r}^{p_{\sigma(r)}-1} $ with $ \sigma $ a permutation of the set of primes $ p $ such that $v_{p}(n)\geq 1 $ and $d=\prod_{p, v_{p}(n)\geq 1}p $ .

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