I have this next equation:
$c=\left( p^n\cdot c_0\right)\cdot \prod_{j=0}^{n-1}f(j)$
where:
- $c, p, c_0$ are known constants.
- $n$ is to be determined from this equation.
The function $f(j)$ is given by: $f(j)=\dfrac{1}{n-1}j+1$, where j is a natural number inside the range $\left[0, n-1\right]$
Plugging $f(j)$ into the main equation gives:
$c=\left( p^n\cdot c_0\right)\cdot \prod_{j=0}^{n-1}\left( \dfrac{1}{n-1}j+1\right)$
Is there any easy way to solve for $n$? Or do I need a computer?
Assuming that $x=j$ in the product, the solutions of the equation
$$c_n=\left( p^n\cdot c_0\right)\cdot \prod_{j=0}^{n-1}(a\sqrt j+b)$$
must be found by exhaustive comparisons
$$c_0=c_0b=:t_0\\ c_1=pc_0b(a+b)=p(a+b)t_0=:t_1\\ c_2=p^2c_0b(a+b)(a\sqrt2+b)=p(a\sqrt2+b)t_1=:t_2\\ c_3=p^3c_0b(a+b)(a\sqrt2+b)(a\sqrt3+b)=p(a\sqrt3+b)t_2=:t_3\\ \cdots$$
There is apparently no limit of $n$ (hence the search is infinite), and equality may very well never arise or arise several times.
Update:
Due to a late changes in the problem statement, $a=\frac1{\sqrt{n-1}},b=1$, and the recurrence relations do not hold.
With these definitions, the products can be computed once for all as a function of $n$ alone, but that doesn't change the conclusion regarding the resolution.