Zeta zero spectrum comb asymptotics

(*start*)Clear[n, k, t, A, nn];
nn = 50;
A = Table[
   Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
     nn}], {n, 1, nn}];
MatrixForm[A];
g1 = ListLinePlot[
   Table[Total[
     1/Table[n, {n, 1, nn}]*
      Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000,
     60, N[1/6]}], DataRange -> {0, 60}, PlotRange -> {-0, 10}];
f[t_] = D[RiemannSiegelTheta[t], t];
g2 = Plot[(f[t] + HarmonicNumber[nn]), {t, 0, 60},
   PlotRange -> {0, 15}, PlotStyle -> {Red}];
Show[g1, g2]
(*end*)

(*start*)
Clear[n, k, t, A, nn];
nn = 50;
A = Table[
   Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
     nn}], {n, 1, nn}];
MatrixForm[A];
g1 = ListLinePlot[
   Table[Total[
     1/Table[n*t, {n, 1, nn}]*
      Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000,
     60, N[1/6]}], DataRange -> {0, 60}, PlotRange -> {-0, 1}];
f[t_] = D[RiemannSiegelTheta[t], t];
g2 = Plot[(f[t] + HarmonicNumber[nn])/t, {t, 0, 60},
   PlotRange -> {0, 1}, PlotStyle -> {Red}];
Show[g1, g2]
(*end*)

(*start*)f[t_] = D[RiemannSiegelTheta[t], t];
nnn = 60
cc = 10;
g1 = Plot[(f[t] + cc + EulerGamma), {t, 0, nnn},
   PlotStyle -> {Thickness[0.004], Red}, PlotRange -> {-2, cc + 5}];
c = 1 + 1/cc;
g2 = Plot[
   Re[Zeta[1/2 + I*t]*Zeta[c]/Zeta[1/2 + I*t + c - 1]], {t, 0, nnn},
   PlotRange -> {-2, cc + 5}, PlotStyle -> Thickness[0.02]];
Show[g2, g1, ImageSize -> Large]
(*end*)


(*Better agreement between asymptotics*)

(*start*)
Clear[n, k, t, A, nn];
nn = 60;
A = Table[
   Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
     nn}], {n, 1, nn}];
MatrixForm[A];
g1 = ListLinePlot[
   Table[Total[
     1/Table[n, {n, 1, nn}]*
      Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000,
     60, N[1/6]}], DataRange -> {0, 60}, PlotRange -> {-0, 10}];
f[t_] = D[RiemannSiegelTheta[t], t];
g2 = Plot[(f[t] + HarmonicNumber[nn]), {t, 0, 60},
   PlotRange -> {0, 15}, PlotStyle -> {Red}];
gg1 = Show[g1, g2]
(*end*)

(*start*)
f[t_] = D[RiemannSiegelTheta[t], t];
nnn = nn
cc = Log[nn];
g1 = Plot[(f[t] + cc + EulerGamma), {t, 0, nnn},
   PlotStyle -> {Thickness[0.002], Red}, PlotRange -> {-2, cc + 5}];
c = 1 + 1/cc;
g2 = Plot[
   Re[Zeta[1/2 + I*t]*Zeta[c]/Zeta[1/2 + I*t + c - 1]], {t, 0, nnn},
   PlotRange -> {-2, cc + 5}, PlotStyle -> Thickness[0.002]];
gg2 = Show[g2, g1]
Show[gg1, gg2]
(*end*)


(* Latex *)

In [this answer](https://math.stackexchange.com/a/4199789/8530) the following asymptotic relation was found:

$$\Re \frac{{\zeta\! \left( {1 + \frac{1}{c}} \right)\zeta\! \left( {\frac{1}{2} + it} \right)}}{{\zeta\! \left( {\frac{1}{2} + it + \frac{1}{c}} \right)}} = c + \frac{{d\vartheta (t)}}{{dt}} + \gamma  + \mathcal{O}\!\left( {\frac{1}{c}} \right)$$

$$\Re\left(\sum _{n=1}^N \frac{\zeta \left(\frac{1}{2}+i t\right) \sum \frac{\mu (d(n))}{d(n)^{\frac{1}{2}+i t-1}}}{n}\right)=\Re\left(\frac{\zeta \left(\frac{1}{2}+i t\right) \zeta \left(1+\frac{1}{\log (N)}\right)}{\zeta \left(\frac{1}{2}+i t+\frac{1}{\log (N)}+1-1\right)}\right)$$

(*start*)
Clear[n, k, t, A, nn];
nn = 60;
A = Table[
   Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
     nn}], {n, 1, nn}];
MatrixForm[A];
g1 = ListLinePlot[
   Table[Total[
     1/Table[n, {n, 1, nn}]*
      Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000,
     60, N[1/6]}], DataRange -> {0, 60}, PlotRange -> {-0, 10}];
f[t_] = D[RiemannSiegelTheta[t], t];
g2 = Plot[(f[t] + HarmonicNumber[nn]), {t, 0, 60},
   PlotRange -> {0, 15}, PlotStyle -> Black];
gg1 = Show[g1, g2]
(*end*)

(*start*)
f[t_] = D[RiemannSiegelTheta[t], t];
nnn = nn
cc = Log[nn];
g3 = Plot[(f[t] + cc + EulerGamma), {t, 0, nnn},
   PlotStyle -> {Thickness[0.002], Black}, PlotRange -> {-2, cc + 5}];
c = 1 + 1/cc;
g4 = Plot[
   Re[Zeta[1/2 + I*t]*Zeta[c]/Zeta[1/2 + I*t + c - 1]], {t, 0, nnn},
   PlotRange -> {-2, cc + 5}, PlotStyle -> Red];
gg2 = Show[g4, g3]
Show[gg1, gg2]
Show[g1, g4]
(*end*)