Intereting Posts

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Finding all possible $n\times n$ matrices with non-negative entries and given row and column sums.

What is the average value for $\mathrm{lcm}(a,b)$, with $ 1\le a \le b \le n$, for a

given $n$, and what is the asymptotic behavior? The $\mathrm{lcm}$ is the least common multiple.

I have calculated, as $(n, avg)$:

- Determining the Value of a Gauss Sum.
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- Norm of Gaussian integers: solutions to $N(a) = k$ for $k \in \mathbb{N}$ and $a$ a Guassian integer

$$(10, 19.836)$$

$$(100, 1826.859)$$

$$(1000, 182828.976)$$

The values appear to converge quadratically to some constant.

This generalizes to $$\sum_{1 \le j \le k \le n} \mathrm{lcm}(j,k)$$

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- Perfect squares formed by two perfect squares like $49$ and $169$.

Here is a back-of-the-envelope calculation.

Start with the average value of $ab$, which is $(n+1)^2/4$.

Consider powers of $2$ only. Three-quarters of the $(a,b)$ pairs lack $2$ as a common factor. 3/16 have $2^1$ as a common factor, $3/64$ have $4$ as a common factor and so on. So, on average, dealing with powers of $2$ reduces the average by a factor $$\frac341+\frac3{16}\frac12+\frac3{64}\frac14+…\\

=\frac34\left[1+\frac18+\frac1{64}+…\right]\\=\frac{3/4}{7/8}=6/7$$

Consider powers of $3$ only, which will be independent from powers of $2$. The reduction is

$$\frac891+\frac8{81}\frac13+\frac8{729}\frac19+…=\frac{8/9}{26/27}=\frac{12}{13}$$

For any prime $p$, the factor is $\frac{p^3-p}{p^3-1}$, so my final answer is

$$\frac{(n+1)^2}4\prod_{p\,\text{prime}}\frac{p^3-p}{p^3-1}=\frac{(n+1)^2}4\frac{\zeta(3)}{\zeta(2)}$$

That is the Riemann $\zeta$ function, and the constant ratio is

$$\frac{\zeta(3)}{4\zeta(2)}=0.18269074235…$$

Theorem 6.3 of Olivier Bordelles, Mean values of generalized gcd-sum and lcm-sum functions, Journal of Integer Sequences, Vol. 10 (2007), Article 07.9.2, says, for any real number $x$ sufficiently large, $$\sum_{n\le x}\sum_{j=1}^n[n,j]={\zeta(3)\over8\zeta(2)}x^4+O\bigl(x^3(\log x)^{2/3}(\log\log x)^{4/3}\bigr)$$

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