Is it possible in principle (but not in practice) to recursively factor away the Riemann zeta zeros as they are computed?

Question

Let: $$f_0(x)=\frac{\zeta (x)}{\sin \left(\frac{\pi x}{2}\right)}$$ and let the seed point be: $$s=\sqrt{-1}$$ which is the input into the limit: $$\rho_1=s+\lim\limits_{n \rightarrow \infty}\left(1-\frac{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f_0\left(\frac{k}{n}-\frac{1}{n}+s\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f_0\left(\frac{k}{n}+s\right)}}\right)^{-1}$$ Then for the truncated limit at $n=170$ we seem to numerically get the first Riemann zeta zero:

$\rho_1 \approx $ 0.500000000809315424 + 14.134725141999078436 I

Now factor away the first zeta zero pair $\rho _{+j}$ and $\rho _{-j}$ from $f_0(x)$ by dividing with the product: $$\prod _{j=1}^m (x-\rho _j) (x-\rho _{-j})$$ where $m=1$,
that is:

$$f_m(\text{x}) = \frac{\zeta (x)}{\sin \left(\frac{\pi x}{2}\right) \prod\limits_{j=1}^{m} (x-\rho _j) (x-\rho _{-j})}$$

and with $s=\sqrt{-1}$ then feed $f_m (x)$ into the binomial sum limit:

$$\rho_{m+1}=s+\lim\limits_{n \rightarrow \infty}\left(1-\frac{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f_m\left(\frac{k}{n}-\frac{1}{n}+s\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f_m\left(\frac{k}{n}+s\right)}}\right)^{-1}$$

which again for the truncated limit at $n=170$ seems to numerically give the first, second, third, fourth, fifth... Riemann zeta zero:

$m = 0$:
$\rho_1 \approx $ 0.500000000809315424 + 14.134725141999078436 I
$m = 1$:
$\rho_2 \approx $ 0.500003240725315219 + 21.022038108654950356 I
$m = 2$:
$\rho_3 \approx $ 0.500044971947739555 + 25.010893102992784503 I
$m = 3$:
$\rho_4 \approx $ 0.500605374548171145 + 30.424287649143723384 I
$m = 4$:
$\rho_5 \approx $ 0.500830274970246344 + 32.936917518238269262 I
and so on with $m \rightarrow \infty$ and $n \rightarrow \infty$.

Is it then possible in principle (but not in practice) to recursively factor away the Riemann zeta zeros as they are computed? Or will this pattern of consecutive zeros eventually fail? Is the Lehmer pair a problem?

The Mathematica program by which the approximations of zeros were computed:

(*Mathematica start, takes a minute to run*)
Clear[f, s, X, n, k, x, m];
m = 0; (* try setting: m = 0, m = 1, m = 2, m = 3, m = 4,... *)
f[x_] := Zeta[x]/Sin[Pi*x/2]/
   Product[(x - ZetaZero[j])*(x - ZetaZero[-j]), {j, 1, m}];
(* f[x_]:= x^2-2 *)
s = I;
n = 170;
Block[{$MaxExtraPrecision = 500}, 
 X = N[(1/(1 - 
        Sum[((-1)^(k - 1)*Binomial[n - 1, k - 1])/
           f[k/n + s - 1/n], {k, 1, n}]/
         Sum[((-1)^(k - 1)*Binomial[n - 1, k - 1])/f[k/n + s], {k, 1, 
           n}]) + s), 20]]
f[X]
(*end*)

$f(x)$ can probably be any polynomial, and the binomial sum ratio comes from a tweaked version of the ratio of higher order derivatives of a function, with the formula "left unevaluated" until the desired highest order of the derivatives is chosen. I have the partial derivation somewhere on Mathematics stack exchange.