Using the standard statistical definitions, the variance of $x_1, x_2, \ldots, x_n$ and the squared errors about its mean $\mu$ are given by $\sigma^2 = \sum_i(x_i - \mu)^2/n$ and $\delta_2 = \sum_i(x_i - \mu)^2 = n\sigma^2$ respectively.

Definition 1: The variance and the squared error of an integer is defined as the variance and the squared error of its positive divisors respectively.

I found that:

  • There are distinct integer pairs whose variances are equal. The smallest such pair is $(691, 817)$. Let us call them equivarient integers.
  • There are integer pairs whose squared errors are equal. The smallest such pair is $(45, 53)$.

The more interesting fact is that there are equivarient pairs which have the same number of divisors and hence their squared errors are also equal. We define:

Definition 2: Two distinct integers are said to be an intimate pair if they are have the same number of divisors and the same variance.

The first few intimate pairs are $(1403,1461)$, $(1564,1572)$,$(2068,2076)$,$(2249,2305)$,$(3397,3493)$,$(7871,8193)$,$ (23903,24101)$,$(61769, 64443)$.


  1. Are there infinitely many intimate pairs?
  2. Are there three or more integers that are intimate ( and what should we call them hahaha)

Note: Please let me know in case there is any reference in literature. I could not find any. Posted to MO since it is unanswered in MSE

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    $\begingroup$ If I am right, you can replace one of "equivarient" or "same error" by "same number of divisors", which is also "same multiplicities in the prime factorization". $\endgroup$ – Yves Daoust Apr 10 at 7:08
  • $\begingroup$ @YvesDaoust: Valid point, makes it simpler. I am updating it. $\endgroup$ – Nilotpal Kanti Sinha Apr 10 at 7:11
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    $\begingroup$ Is this a conjecture of yours? (Also, looking at your bio, you say you are a data scientist. To be that requires an extraordinary mind, so a quick well done!) $\endgroup$ – Mr Pie Apr 10 at 7:20
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    $\begingroup$ @user477343 Does it? Being one too, I'll take that as a compliment. $\endgroup$ – J.G. Apr 10 at 7:23
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    $\begingroup$ @user477343 Yes, its my conjecture. Thanks for the compliment !!! $\endgroup$ – Nilotpal Kanti Sinha Apr 10 at 7:26

For what it's worth, not an answer but a long comment:

Let $\sigma_k$ denote the sum of the $k^{th}$ powers of the divisors.

If we consider an integer with a single prime factor, say


we have $$\sigma_0=n,$$ $$\sigma_1=\frac{p^n-1}{p-1},$$ $$\sigma_2=\frac{p^{2n}-1}{p^2-1}.$$

Then with the variance $v$,


This function seems to be growing for $p>1$ and any $n$, so chances are low that two distinct $p$ yield the same variance.

Now with two prime factors, say


we have


which just seems intractable.

In the simplest case, $n=m=2$, the expression reduces to


The challenge is to find two points with prime coordinates on the iso-curves of this function. But from Wolfram Alpha, the only solution is $(1,1)$. I tried a few other exponents, to no avail.


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