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The variance $v_n$ of a natural number $n$ is defined as the variance of its divisors. There are distinct integer pairs whose variances are equal; the smallest such pair is $(691, 817)$. However I observed that for $n \le 10^8$ the ratio $\dfrac{n}{v_n}$ is different for each $n$. Is this true in general?

Unique variance conjecture: The ratio of a number to the variance of its divisors is injective. There does not exist two distinct positive integers $m$ and $n$ such that $$ \frac{m}{v_m} = \frac{n}{v_n} $$

A variation of this question with with standard deviation was posted here was posted here and verified for $n \le 10^8$.

Nilotpal Sinha
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  • I would imagine that this is a strictly harder problem than similar questions about perfect numbers, for which some basic results remain unsolved. Also, the sequence of perfect numbers explodes incredibly quickly, so I think that checking up to "only" $10^8$ provides not very good evidence. – Him Aug 02 '19 at 07:31
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    Well Skewes number has taught us that checking even up to $10^{100}$ would not have been good enough. So lets conjecture anyways :) – Nilotpal Sinha Aug 02 '19 at 08:44
  • $(1)$ It might be helpful to know which formula for the variance you use $(2)$ The limit $10^8$ is impressing ! How did you manage to show that all the values are distinct in this range ? – Peter Sep 17 '19 at 08:18
  • @Peter You can find the definition of variance here. https://math.stackexchange.com/questions/3176228/does-the-average-primeness-of-natural-numbers-tend-to-zero . To check up to $10^8$ I calculated the ratio for $n \le 10^8$ and then did a check duplicate to verify that none of the values repeated for $n \le 10^8$. So if there is a non-unique value, it must be greater than $10^8$. This was we can verify in $O(n)$ time. But if we are to compare each ratio with all the previous values then the program will run inefficiently in $O(n^3)$ time. – Nilotpal Sinha Sep 17 '19 at 10:29
  • By sorting the values , I guess. Because comparing every value with every other should take a very long time. – Peter Sep 17 '19 at 10:32
  • @Peter We can do without sorting. The outputs were all stored in a dataframe file and I just did a count unique on the file. – Nilotpal Sinha Sep 17 '19 at 10:42
  • I am pretty sure that the software used in this tool uses also sorting, how else could we efficiently check for duplicates ? The complexity without sorting would be $O(n^2)$ – Peter Sep 17 '19 at 10:45
  • @Peter yes all software do highly optimized internal sorting for giving unique counts. So at a user level we don't have to do it manually because it is taken care of by the programming language internally in the fastest way possible. E.g the function $unique(x)$ in $R$ removes all duplicate rows in the file $x$. – Nilotpal Sinha Sep 17 '19 at 10:49

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