Can you prove that the $13$ first zeros of $(r\zeta(r+\Im i))^2$ have real part $r=\frac{1}{2}$, assuming LeClaire's approximation?

Question

Can you prove using double series reversion that the $13$ first zeros of $(r\zeta(r+\Im i))^2$ have real part $r=\frac{1}{2}$ (as their convergent), with initial guess for the real part $r$ to be somewhere in the interval $0<r<1$, and with LeClaire's approximation as initial guess for the imaginary part:

$$\Im(\rho(n)) = \frac{2 \pi \exp (1) \left(n-\frac{11}{8}\right)}{\exp (1) W\left(\frac{n-\frac{11}{8}}{\exp (1)}\right)}$$

$$n=1,2,3,4,5,6,7,8,9,10,11,12,13$$

where $W$ is the Lambert W function, and $\rho(n)$ is $n$-th Riemann zeta zero?

Program before this edit 18.1.2014: Mathematica program:

(*Jan 5 2014*)
Clear[n, rr, rrr]
rrr = {1, 3/4, 1/2, 1/4, 1/100};
Do[
 rr = rrr[[k]];
 Print["r = ", rr]
  Do[
   Clear[a, b, c, a, x, r, i];
   b = 14;
   r = rr;
   c = 40;
   Clear[a, x];
   (*André LeClaire's approximation*)
   i = N[2*Pi*Exp[1]*(n - 11/8)/Exp[1]/LambertW[(n - 11/8)/Exp[1]], c];
   a = Normal[InverseSeries[Series[r*(Zeta[x + I*i]), {x, r, b}], x]];
   x = 0;
   r = N[Re[a], 14];
   i;
   (*Print[{r,i,N[ZetaZero[1],12]}]*)

   Clear[a, x];
   (*André LeClaire's approximation*)
   i = N[2*Pi*Exp[1]*(n - 11/8)/Exp[1]/LambertW[(n - 11/8)/Exp[1]], c];
   a = Normal[InverseSeries[Series[r*(Zeta[r + I*x]), {x, i, b}], x]];
   x = 0;
   r;
   i = N[Re[a], 14];
   (*Print[{r,i,N[ZetaZero[1],12]}]*)

   Monitor[Do[
      Clear[a, x];
      a = 
       Normal[InverseSeries[Series[r*(Zeta[x + I*i]), {x, r, b}], x]];
      x = 0;
      r = N[Re[a], 14];
      i;
      (*Print[{r,i,N[ZetaZero[1],12]}]*)

      Clear[a, x];
      a = 
       Normal[InverseSeries[Series[r*(Zeta[r + I*x]), {x, i, b}], x]];
      x = 0;
      r;
      i = N[Re[a], 14];
      (*Print[{r,i,N[ZetaZero[1],12]}]*)
      , {j, 1, 8}], j]
    Print[{r, i, N[ZetaZero[n], 12]}], {n, 1, 5}], {k, 1, 
  Length[rrr]}]

New program 18.1.2014:

(*Jan 18 2014 klockan 13:23*)
Clear[n, rr, rrr]
rrr = {99/100, 3/4, 1/2, 1/4, 1/100};
Do[rr = rrr[[k]];
 Print["r = ", rr] Do[Clear[a, b, c, a, x, r, i];
   b = 14;
   r = rr;
   c = 40;
   Clear[a, x];
   (*André LeClaire's approximation*)
   i = N[2*Pi*Exp[1]*(n - 11/8)/Exp[1]/LambertW[(n - 11/8)/Exp[1]], c];
   a = Normal[
     InverseSeries[Series[(r*Zeta[x + I*i])^2, {x, r, b}], x]];
   x = 0;
   r = N[Re[a], 14];
   i;
   (*Print[{r,i,N[ZetaZero[1],12]}]*)
   Clear[a, x];
   (*André LeClaire's approximation*)
   i = N[2*Pi*Exp[1]*(n - 11/8)/Exp[1]/LambertW[(n - 11/8)/Exp[1]], c];
   a = Normal[
     InverseSeries[Series[(r*Zeta[r + I*x])^2, {x, i, b}], x]];
   x = 0;
   r;
   i = N[Re[a], 14];
   (*Print[{r,i,N[ZetaZero[1],12]}]*)Monitor[Do[Clear[a, x];
      a = Normal[InverseSeries[Series[Zeta[x + I*i], {x, r, b}], x]];
      x = 0;
      r = N[Re[a], 14];
      i;
      (*Print[{r,i,N[ZetaZero[1],12]}]*)Clear[a, x];
      a = Normal[InverseSeries[Series[Zeta[r + I*x], {x, i, b}], x]];
      x = 0;
      r;
      i = N[Re[a], 14];, {j, 1, 1}], j] 
    Print[{r, i}], {n, 1, 13}], {k, 1, Length[rrr]}]

Outputs for different initial values of the real part $r$:

$r = 99/100$ $$0.5000000,14.13473$$ $$0.5000000,21.022040$$ $$0.5000000,25.01086$$ $$0.50000,30.425$$ $$0.50000,32.935$$ $$0.5000000,37.586178$$ $$0.50000000,40.918719$$ $$0.50000,43.327$$ $$0.50000,48.005$$ $$0.500000,49.7738$$ $$0.50000000,52.970321$$ $$0.5000000,56.446248$$ $$0.500000,59.3470$$ $r = 3/4$ $$0.5000000,14.13473$$ $$0.5000000,21.022040$$ $$0.5000000,25.01086$$ $$0.50000,30.425$$ $$0.50000,32.935$$ $$0.5000000,37.586178$$ $$0.50000000,40.9187190$$ $$0.50000,43.327$$ $$0.5000,48.01$$ $$0.500000,49.7738$$ $$0.50000000,52.9703215$$ $$0.5000000,56.446248$$ $$0.500000,59.3470$$ $r = 1/2$ $$0.50000000,14.134725$$ $$0.50000000,21.022040$$ $$0.5000000,25.01086$$ $$0.50000,30.425$$ $$0.50000,32.935$$ $$0.50000000,37.586178$$ $$0.50000000,40.9187190$$ $$0.50000,43.327$$ $$0.5000,48.01$$ $$0.500000,49.7738$$ $$0.50000000,52.9703215$$ $$0.50000000,56.446248$$ $$0.5000000,59.34704$$ $r = 1/4$ $$0.5000000,14.13473$$ $$0.5000000,21.022040$$ $$0.5000000,25.01086$$ $$0.500000,30.425$$ $$0.500000,32.935$$ $$0.5000000,37.586178$$ $$0.50000000,40.918719$$ $$0.500000,43.3271$$ $$0.5000,48.01$$ $$0.500000,49.7738$$ $$0.50000000,52.970321$$ $$0.5000000,56.446248$$ $$0.500000,59.34704$$ $r = 1/100$ $$0.5000000,14.13473$$ $$0.5000000,21.02204$$ $$0.500000,25.01086$$ $$0.500000,30.425$$ $$0.500000,32.935$$ $$0.5000000,37.58618$$ $$0.5000000,40.91872$$ $$0.500000,43.327$$ $$0.50000,48.005$$ $$0.500000,49.7738$$ $$0.5000000,52.970321$$ $$0.5000000,56.44625$$ $$0.500000,59.3470$$

As we see the above values are very close to the $13$ first zeta zeros:

$$0.5000000000 + 14.1347251417 I$$ $$0.5000000000 + 21.0220396388 I$$ $$0.5000000000 + 25.0108575801 I$$ $$0.5000000000 + 30.4248761259 I$$ $$0.5000000000 + 32.9350615877 I$$ $$0.5000000000 + 37.5861781588 I$$ $$0.5000000000 + 40.9187190121 I$$ $$0.5000000000 + 43.3270732809 I$$ $$0.5000000000 + 48.0051508812 I$$ $$0.5000000000 + 49.7738324777 I$$ $$0.5000000000 + 52.9703214777 I$$ $$0.5000000000 + 56.4462476971 I$$ $$0.5000000000 + 59.3470440026 I$$