Solve and asymptotic expansion of $\sum_{a=1}^{H} \sum_{b=a+1}^{H} \left\lfloor{\frac{H}{a\, b}}\right\rfloor$

Question

I am solving constrained polynomial systems resulting in constrained sums. I am looking to see if $$\sum_{a=1}^{H} \sum_{b=a+1}^{H} \left\lfloor{\frac{H}{a\, b}}\right\rfloor$$ is expressible in terms of known functions and very importantly the asymptotic form as $H \rightarrow \infty$. If needed we can make the sum over $b$ symmetric to the sum over $a$.

Answer 1

I get $\frac14 n\ln^2(n)+O(n\ln(n)) $.

If

$\begin{array}\\ s(n) &=\sum_{a=1}^{n} \sum_{b=a+1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\ \text{then}\\ s(n) &=\sum_{b=1}^{n} \sum_{a=1}^{b-1} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\ &=\sum_{a=1}^{n} \sum_{b=1}^{a-1} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\ \text{so}\\ 2s(n) &=\sum_{a=1}^{n} (\sum_{b=1}^{a-1} \left\lfloor{\frac{n}{a\, b}}\right\rfloor+ \sum_{b=a+1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor)\\ &=\sum_{a=1}^{n} (\sum_{b=1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor-\left\lfloor{\frac{n}{a^2}}\right\rfloor)\\ &=\sum_{a=1}^{n} \sum_{b=1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor -\sum_{a=1}^{n}\left\lfloor{\frac{n}{a^2}}\right\rfloor\\ &= u(n)-v(n)\\ v(n) &= \sum_{a=1}^{n}\left\lfloor{\frac{n}{a^2}}\right\rfloor\\ &= \sum_{a=1}^{\lfloor \sqrt{n} \rfloor}\left\lfloor{\frac{n}{a^2}}\right\rfloor\\ &\le \dfrac{n\pi^2}{6}\\ \text{and}\\ u(n) &=\sum_{a=1}^{n} \sum_{b=1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\ &=\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\ &=\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor} \left\lfloor{\frac{\lfloor n/a \rfloor}{b}}\right\rfloor\\ &\le \sum_{a=1}^{n} \lfloor n/a \rfloor\sum_{b=1}^{\lfloor n/a \rfloor} \left\lfloor{\frac1{b}}\right\rfloor\\ &\le \sum_{a=1}^{n} \lfloor n/a \rfloor\sum_{b=1}^{\lfloor n/a \rfloor} \frac1{b}\\ &\lt \sum_{a=1}^{n} \lfloor n/a \rfloor (\ln(n/a)+1)\\ &\le \sum_{a=1}^{n} (n/a)\ln(n/a)+ \sum_{a=1}^{n} (n/a)\\ &= \sum_{a=1}^{n} (n/a)\ln(n)-\sum_{a=1}^{n} (n/a)\ln(a)+ n\sum_{a=1}^{n} \frac1{a}\\ &= n\ln(n)\sum_{a=1}^{n} \frac1{a}-n\sum_{a=1}^{n} \frac{\ln(a)}{a}+ n\sum_{a=1}^{n} \frac1{a}\\ &= n\ln^2(n)+O(n\ln(n))-n(\frac12 \ln^2(n)+O(\ln(n))+ n(\ln(n)+O(1))\\ &= \frac12 n\ln^2(n)+O(n\ln(n))\\ \text{so}\\ s(n) &\lt \frac14 n\ln^2(n)+O(n\ln(n))\\ \text{and}\\ u(n) &=\sum_{a=1}^{n} \sum_{b=1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\ &=\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\ &\gt\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor} ({\frac{n}{a\, b}}-1)\\ &=\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor} {\frac{n}{a\, b}}-\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor}1\\ &=\sum_{a=1}^{n} \frac{n}{a}\sum_{b=1}^{\lfloor n/a \rfloor} {\frac1{b}}-\sum_{a=1}^{n} \lfloor n/a \rfloor\\ &>\sum_{a=1}^{n} \frac{n}{a}\ln(\lfloor n/a \rfloor)-n\ln(n)\\ &>\sum_{a=1}^{n} \frac{n}{a}(\ln( n/a)-1)-n\ln(n)\\ &= \frac12 n\ln^2(n)+O(n\ln(n)) \quad\text{by the same argument as above}\\ \end{array} $

Answer 2

Using the symmetric form of the summations we have $$\sum_{a=1}^{H} \sum_{b=1}^{H} \left\lfloor{\frac{H}{a\, b}}\right\rfloor = \sum_{a=1}^{H}\left\lfloor{\frac{H}{a}}\right\rfloor+\sum_{a=1}^{\left\lfloor{H/2}\right\rfloor}\left\lfloor{\frac{\left\lfloor{H/2}\right\rfloor}{a}}\right\rfloor+\sum_{a=1}^{\left\lfloor{H/3}\right\rfloor}\left\lfloor{\frac{\left\lfloor{H/3}\right\rfloor}{a}}\right\rfloor+\cdots+\sum_{a=1}^{\left\lfloor{H/H}\right\rfloor}\left\lfloor{\frac{\left\lfloor{H/H}\right\rfloor}{a}}\right\rfloor=\sum_{a=1}^{H} D \left({\left\lfloor{\frac{H}{a}}\right\rfloor}\right)$$

where $D \left({\left\lfloor{\frac{H}{a}}\right\rfloor}\right)$ is the sum over divisors. The asymptotic follows from the divisor sum expansions.