Time complexity of a recursive enumeration in the problem of finding n-tuples of naturals greater than 1 with bounded product
I thought about this problem a lot and I think that I devised a suitable 
solution. I focus here on asymptotic analysis with respect to $K$, and my aim 
is to provide tighter upper bound than one that I suggested in the
question. Therefore, I now observe slightly different problem 
where $k_i \geq 1$. In that case the number of steps $T(K,n)$ 
required to enumerate all n-tuples of naturals greater or equal one 1 with 
product less than or equal to $K$ is:
\begin{equation}
T(K, n) = \sum_{k=1}^{K} T(\lfloor \frac{K}{k} \rfloor, n - 1)
\end{equation}
\begin{equation}
T(K, 0) = 1
\end{equation}
where $T(K, 0)$ is the number of elementary operations in the basic case.

For any bound $x$, we know that: 
\begin{equation}
T(x, 1) = \sum_{k=1}^{x} T(x, 0) = x  
\end{equation}

Moreover, for $T(x, 2)$ we have:
\begin{equation}
T(x, 2) = T(\frac{x}{1}, 1) + T(\frac{x}{2}, 1) + T(\frac{x}{3}, 1)... + T(\frac{x}{x}, 1) = \frac{x}{1} + \frac{x}{2} + \frac{x}{3} ... + \frac{x}{x} 
\end{equation}
We can see that this is in fact finite partial sum of harmonic series:
\begin{equation}
T(x, 2) = x \sum_{k=1}^{x} \frac{1}{k} = x H_x
\end{equation}
where $H_x = \sum_{k=1}^{x} \frac{1}{k}$ is the x-th harmonic number.

Now with $T(x, 2) = x H_x$, $T(x, 3)$ can be expressed as:
\begin{equation}
T(x, 3) = T(\frac{x}{1}, 2) + T(\frac{x}{2}, 2) + ... + T(\frac{x}{x}, 2) = x H_x + \frac{x}{2} H_{x/2} + ... + \frac{x}{x} H_1
\end{equation}
Now we can bound $T(x, 3)$:
\begin{equation}
T(x, 3) = x H_x + \frac{x}{2} H_{x/2} + ... + \frac{x}{x} H_1 \leq x H_x (1 + \frac{1}{2} + ... + \frac{1}{x}) = x H_x^2
\end{equation}

By induction, we get that $T(K, n) = K H_K^{n-1}$. Harmonic number can be 
approximated with integral:
\begin{equation}
H_K = \int_{1}^{K} \frac{1}{t} dt = \ln K
\end{equation}

Therefore, an upper bound on complexity is given with $O(K \log^{n-1} (K))$.