A conjecture for an inequality for every odd number.

Question

Let $E(n) = \sum_{d|n} d \log(d) $. Then for coprime $n,m$ we have $$E(mn) = \sigma(m) E(n) + \sigma(n) E(m).$$ Suppose there exists an odd perfect number $n$. Then$$E(2n) = E(2)\sigma(n)+E(n)\sigma(2) = 4 n \log(2) + 3 E(n).$$ But looking at a few numerical examples, one can see that the right hand side seems to be alway bigger than the left hand side. Consider the upper bound for logarithm:$$\log(x) \le x^s/s. \quad \forall s>0$$ Making use of this upper bound we can derive: $$E(2n) = 2 \sum_{d|n} d \log(2d) + E(n) \le 2 \sum_{d|n} \frac{(2d)^s}{s} + E(n). \quad \forall s > 0$$ Now the question is, how do we choose $s>0$ such that we have $$2 \sum_{d|n} \frac{(2d)^s}{s} + E(n) < 4 n \log(2) + 3 E(n).$$ My conjecture is, that it is always possible to choose one such $s>0$ such that the last inequality is valid.

Answer 1

The equality $E(2n) = 2\sigma(n)\log 2+3E(n)$ holds for any odd $n$.

You're trying to show that it is always less than $4n\log 2+3E(n)$, i.e. that all odd numbers are "deficient" (i.e. $\sigma(n) < 2n$).

This is not true, there are "abundant" ($\sigma(n) > 2n$) odd numbers as well.

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You'd have to exploit further the assumption that $n$ is perfect for this method to work.