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Let $n$ be a natural number. $\Omega = \{ d : d | n \}$, For $A\subset \Omega$ define the probability for $A$ as : $$P(A) = \frac{1}{\sigma(n)}\sum_{d\in A}{d}$$ Consider the set $B = \{d \in \Omega |d \equiv 0 (2) \}$. Let $a = v_2(n)$. Then (first conjecture) $$P(B) = \frac{2^{a+1}-2}{2^{a+1}-1}$$ We have $E(d) = \sum_{d|n} {d \cdot P(d)} = \sum_{d|n} { d \cdot \frac{d}{\sigma(n)}} = \frac{\sigma_2(n)}{\sigma(n)}$ Second conjecture: $|B| = a \cdot b$, where $n = 2^a \cdot p_1^{a_1}\cdots p_r^{a_r}$ is the factorization of $n$, with $a$ possibly $=0$ and $b = (a_1+1)\dots(a_r+1)$. Consider $Z = \sum_{d \in B} {d}$ and consider the random variable: $Y_d = 1$ if $d \equiv 0 ( 2) $, otherwise $= 0$. Then $Z = \sum_{d|n} Y_d d$ is a random variable. Then on the one hand we have: $E(Z) = \sum_{d \in B}{E(d)} = \sum_{d \in B}{\frac{\sigma_2(n)}{\sigma(n)}} = a b \frac{\sigma_2(n)}{\sigma(n)}$ On the other hand we have $P(B) = \frac{Z}{\sigma(n)}$ hence solving for $Z$ we get: $$Z = \sum_{d|n}{d} \cdot \frac{2^{a+1}-2}{2^{a+1}-1}$$ From this it follows that: $$E(Z) = \frac{2^{a+1}-2}{2^{a+1}-1} \sum_{d|n}{E(d)} = \frac{2^{a+1}-2}{2^{a+1}-1} \sum_{d|n}{\frac{\sigma_2(n)}{\sigma(n)}} = \frac{2^{a+1}-2}{2^{a+1}-1} \tau(n) \cdot \frac{\sigma_2(n)}{\sigma(n)}$$ Hence we get: $$a b \frac{\sigma_2(n)}{\sigma(n)} = \frac{2^{a+1}-2}{2^{a+1}-1} \tau(n) \cdot \frac{\sigma_2(n)}{\sigma(n)}$$ from which it follows: $$ab = \tau(n) \cdot \frac{2^{a+1}-2}{2^{a+1}-1}$$ But for example for $n=6$ this is wrong, so where is the mistake in the argument?

If you happen to have a proof for one of the conjectures, that would also be fine.

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    $\begingroup$ Thank you for your comment. I will edit the question to clarify what I mean. $\endgroup$ – orgesleka May 1 '18 at 12:01
  • $\begingroup$ @m_t_: does this make sense what I wrote in the question, or do you see any mistake in the argument? $\endgroup$ – orgesleka May 1 '18 at 12:03
  • $\begingroup$ It doesn't make sense to me, except in the trivial sense that $Z$ is the random variable always taking the value $\sum_{d \in B} d$, and I think this is your mistake. $\endgroup$ – Matthew Towers May 1 '18 at 12:05
  • $\begingroup$ Ok, Thanks! That was what I suspected, since I tried to program it on a computer, but could not get anywhere because $Z$ is not a random variable but a constant. Thanks again for the clarification. $\endgroup$ – orgesleka May 1 '18 at 12:07
  • $\begingroup$ If this is from a source, rather than something you created yourself, sharing the source would probably help clarify what is intended. $\endgroup$ – Matthew Towers May 1 '18 at 12:10
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So both conjectures are true, let's prove conjecture two first.

The factors of $n$ are of the form $2^{u}p_1^{k_1}p_2^{k_2}...p_r^{k_r}$ where $u\leq a$ and $k_i\leq a_i$. So there are $(a+1)(a_1+1)...(a_k+1)$ divisors, you don't want the ones where $u=0$ of which there are $(a_1+1)...(a_k+1)$, ergo, the conjecture.

Similarly conjecture one, your desired expression is seen to be $\frac{\sigma(n) - \sigma(n/2^a)}{\sigma(n)} = 1 - \frac{1}{2^{a+1}-1}$ and the conjecture follows.

Sorry I can't spot your mistake.

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    $\begingroup$ Thank you for your answer. That is helpfull! $\endgroup$ – orgesleka May 1 '18 at 11:00

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