Divisors of a $ k $-multiperfect number












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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 $ 2leq mleqtau(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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  • $begingroup$
    I am not sure, but maybe you could use a result in Dagal's A Lower Bound for $tau(n)$ of $k$-Multiperfect Number?
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Feb 4 at 4:53
















1












$begingroup$


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 $ 2leq mleqtau(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 $ .










share|cite|improve this question











$endgroup$












  • $begingroup$
    I am not sure, but maybe you could use a result in Dagal's A Lower Bound for $tau(n)$ of $k$-Multiperfect Number?
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Feb 4 at 4:53














1












1








1


1



$begingroup$


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 $ 2leq mleqtau(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 $ .










share|cite|improve this question











$endgroup$




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 $ 2leq mleqtau(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 $ .







number-theory perfect-numbers






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edited Dec 23 '18 at 22:19







Sylvain Julien

















asked Dec 23 '18 at 21:02









Sylvain JulienSylvain Julien

1,135918




1,135918












  • $begingroup$
    I am not sure, but maybe you could use a result in Dagal's A Lower Bound for $tau(n)$ of $k$-Multiperfect Number?
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Feb 4 at 4:53


















  • $begingroup$
    I am not sure, but maybe you could use a result in Dagal's A Lower Bound for $tau(n)$ of $k$-Multiperfect Number?
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Feb 4 at 4:53
















$begingroup$
I am not sure, but maybe you could use a result in Dagal's A Lower Bound for $tau(n)$ of $k$-Multiperfect Number?
$endgroup$
– Jose Arnaldo Bebita Dris
Feb 4 at 4:53




$begingroup$
I am not sure, but maybe you could use a result in Dagal's A Lower Bound for $tau(n)$ of $k$-Multiperfect Number?
$endgroup$
– Jose Arnaldo Bebita Dris
Feb 4 at 4:53










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