We have a function $$ f(n,i)=floor \left(\frac{n}{i} \right)\qquad \text{ for } n,i>0$$
Given any n such that $ 10^{11}< n<10^{12}$, provide a methodology to calculate; $$ \sum_{i=1}^{\infty} f(n,i) $$ Solution should involve less than $ 10^7 $operations
You can use any code or excel for the same
let $$F(n) = \sum_{i=1}^{\infty} f(n,i)$$ Notice that $$f(n,i) = \lfloor\frac{n}{i}\rfloor = Card\{j | j\cdot i\leq n\} $$ So $$F(n) = \sum_{i=1}^{\infty} Card\{j | j\cdot i\leq n\} = Card\{(i,j) | j\cdot i\leq n\}$$ The trick is to split this set in two not disjoint parts : $$\{(i,j) | j\cdot i\leq n\} = \{(i,j) | j\leq \lfloor\sqrt n\rfloor,j\cdot i\leq n\} \cup \{(i,j) | i\leq \lfloor\sqrt n\rfloor,j\cdot i\leq n\} $$ Name these sets : $$ S = A \cup B$$ So $$F(n) = Card(S) = Card(A\cup B) = Card(A)+Card(B)-Card(A\cap B)$$ A and B are symetrical, so their cardinality is the same, and $Card(A\cap B) = \lfloor\sqrt n\rfloor^2$, So $$F(n) = 2\sum_{i=1}^{\lfloor\sqrt n\rfloor} \lfloor\frac{n}{i}\rfloor - \lfloor\sqrt n\rfloor^2 $$
Hint:
Assuming $i$ a continuous variable for a second, note that the slope of $\dfrac ni$ is $-\dfrac n{i^2}$ and its absolute value drops below $1$ when $i\approx\sqrt n$. This means that from $1$ to $\sqrt n$ the ratio $\dfrac ni$ changes by several units every time, and then stays constant longer and longer.
So for $n$ like $10^{12}$, you will compute a sum of $10^6$ varying terms, followed by a sum of roughly $10^6$ constant "runs" of increasing length, for a total of $2\cdot10^6$ operations.
