Lets' divide this code to 3 parts.
for (k = 3; k <= n; k++)
A[k] = 0;
Above code is assigning value $0$ to array $A$ from index $3$ till $n$. It’s interesting to note that $A[0], A[1], A[2]$ could be garbage values or some integers, but for this particular code, we dont’ need to worry about them. Since, we are operating on index $\geq 3$.
for (j = 3; j <= n; j++)
if (!A[j]) printf("%d", j);
We are just printing the values of index $j$ after processing over the array $A[j\geq3]$. Hence above two codes are not contributing to operations over $A$, rather assigning and printing.
for (k = 2; k <= TwoLog_n; k++)
for (j = k+1; j <= n; j++)
A[j] = A[j] || (j%k);
We need to analyze third line of this code. $||$ is logical OR operator which gives $True$ when atleast one value is $True$. Hence co-domain of $A[p]$ will be $\{0,1\}$. Now $A[j] == 1$ when $A[j]==1$ or $j\%k \ != 0$ (inclusive OR). $A[j]==1$ won’t reveal us with much information regarding how $A[j]$ getting updated, so we need to analyze $j\%k$.
$\text{if }j\%k ==0 \text{ then }\exists_{m \in \{1,2,\dots, \lfloor\frac{n}{k}\rfloor\}} \ j = =mk$
$\implies A[j] == 1 \text{ when j is not a multiple of k}$
when,
$k=2$, then $\forall_{n \geq j \geq 2+1} \ (2 \bcancel{|} j \implies \ A[j] == 1)$
$k=3$, then $\forall_{n \geq j \geq 3+1} \ (3 \bcancel{|} j \implies \ A[j] == 1)$
Its’ important to recall that we are updating the same array $A$ at $i^{th}$ index in every successive iteration of loop. Hence when $k=p$ then $\forall_{n \geq j \geq 3} \ (\exists_{m \in \{1,2,\dots,p\}} \ m \ \bcancel{|} \ j \implies A[j]==1)$. Its’ contrapositive can be written as $\forall_{n \geq j \geq 3} \ (A[j]==0 \implies \forall_{m \in \{1,2,\dots,p\}} \ m \ | \ j )$.
This says for a value of $j$, if $A[j]==0$ then $j$ should be divisible by every integer from $1$ to $p$. This means that $j$ should be a multiple of every integer from $1$ to $p$, hence $j = p!$ and $p$ goes till $2*\lceil\log_2{n}\rceil$.
After exiting nested loops, the $j$ value(s) for which $A[j]==0$ should be multiple of $(2*\lceil\log_2{n}\rceil)!$. But first we need to check if $(2*\lceil\log_2{n}\rceil)! \leq n$ ($\textbf{why?}$), because $j \leq n$.
$$\begin{align}(2*\lceil\log_2{n}\rceil)! &\qquad n\cr \log {2*\lceil\log_2{n}\rceil!} &\qquad \log_2{n} \tag{log both sides} \cr 2*\log_2{n}\log_2{2*\log_2{n}} &\qquad \log_2{n}\end{align}$$
Clearly $(2*\lceil\log_2{n}\rceil)! > n$, hence for no value of $j \leq n$, $A[j]==0$. So no $j$ value will be printed in the last for-loop $or$ it would be an empty set.
Hence, $\textbf{Option D}$.
