Usbrth

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 $delta_2 = sum_i(x_i - mu)^2 = nsigma^2$ 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)$.

Questions:

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

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

$$p^n-1,$$

we have
$$sigma_0=n,$$
$$sigma_1=fracp^n-1p-1,$$
$$sigma_2=fracp^2n-1p^2-1.$$

Then with the variance $v$,

$$v,sigma_0^2=sigma_0sigma_2-sigma_1^2=nfracp^2n-1p^2-1-left(fracp^n-1p-1right)^2=(p^n-1)fracn(p-1)(p^n+1)-(p+1)(p^n-1)(p+1)(p-1)^2.$$

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

$$p^n-1q^m-1,$$

we have

$$v,sigma_0^2=sigma_0sigma_2-sigma_1^2=nmfracp^2n-1p^2-1fracq^2m-1q^2-1-left(fracp^n-1p-1fracq^m-1q-1right)^2,$$

which just seems intractable.

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

$$4(p^2+1)(q^2+1)-(p+1)^2(q+1)^2.$$

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.