Let $n,i$ be positive integers and $C$ a strictly positive real value. Consider the equation for $f$ : $$1*\ln(f(n)) = C * \sum_{3<i* \ln(i) < \sqrt{n}} \left(\ln[f( i* \ln(i) )-1)] - \ln[(f( i* \ln(i) )-1) - 2 f( i* \ln(i) ) *i* \ln(i)] \right)$$
Where $*$ means multiply and round off with the floor function. The order for $*$ is from left to right.
Does this equation for $f(n)$ have any solution for a given $C$ such that $f(n)\ll n$ for sufficiently large $n$ ?
As stated in current form, the equation doesn't have any solution beyond $n=25$.
Notice $i*\log(i) = 0, 1, 3, 5, 8, \ldots$ for $i = 1, 2, \ldots$
For $n \le 25$, there is no $i$ that satisfy $3 < i*\log(i) < \sqrt{n}$, we have:
$$1 * \log f(n) = 0 \implies 0 \le \log f(n) < 1 \implies 1 \le f(n) < e$$
For $25 < n \le 64$, the only $i$ that satisfy $3 < i*\log(i) < \sqrt{n}$ is $4$, we have:
$$1 * \log f(n) = C * \left(\log[f(5)-1] - \log[(f(5)-1) - 2 f(5)*4*\log(4)] \right)\tag{*}$$ Since $1 \le f(5) < e$, the argument in last $\log$ of $(*)$ becomes negative!
$$(f(5)-1) - 2 f(5)*5*\log(5) < e - 1 - 2 ( 1*4*\log(4)) = e - 11 < 0$$