John Baez question mathoverflow my answer

Clear[t, m, s, M, r, b, M, v, d, x, k, q, z, nn, H, T, TT, g, t, L];
(*b=N[Sum[Sum[(BernoulliB[2*r]/((2*r)!))*(-d k^(1-2 r) E^(I x) \
Abs[StirlingS1[2*r-1,m]] Gamma[1+m,s Log[d*k]] \
Log[d*k]^(-1-m)),{m,1,2*r-1}],{r,1,q-1}],30];*)
b = 4;
M = 7;
v = 2;
k = 20;
q = 8;
c = 1;
dd = 2;
nn = 1;

z[t_] = Sum[1/n^(1/2 + I*t), {n, 1, k}] +
   k^(1 - (1/2 + I*t))/((1/2 + I*t) - 1) - (k^(-(1/2 + I*t)))/2;
z1 = ListLinePlot[Table[N[Re[z[t]]], {t, 0, 60, 1/10}],
   DataRange -> {0, 60}, ImageSize -> Large,
   PlotStyle -> {Thickness[0.004]}];
z2 = Table[
   Graphics[{Arrowheads[0.025],
     Arrow[{{Im[ZetaZero[n]], -1/2}, {Im[ZetaZero[n]], 0}}]}], {n, 1,
    13}];
z3 = Table[
   Graphics[
    Style[Rotate[Text[N[Im[ZetaZero[n]]], {Im[ZetaZero[n]], -1}],
      90 Degree], Medium]], {n, 1, 13}];
Show[z1, z2, z3]

h[t_] = Sum[(I d^(1/2 - I t) E^(I t x) n^(-(1/2) - I t))/(2 (-x +
          Log[d] +
          Log[n])) - (I d^(1/2 + I t) E^(I t x) n^(-(1/2) +
           I t))/(2 (x + Log[d] + Log[n])), {n, 1,
     k}] + -(1/2) I E^(x/2) ExpIntegralEi[
     1/2 I (I + 2 t) (x - Log[d k])] +
   1/2 I E^(-x/2) ExpIntegralEi[
     1/2 (1 + 2 I t) (x +
        Log[d k])] - (I d^(1/2 - I t) E^(I t x) k^(-(1/2) -
         I t))/(4 (-x + Log[d] +
        Log[k])) + (I d^(1/2 + I t) E^(I t N[x]) k^(-(1/2) +
         I t))/(4 (x + Log[d] + Log[k]));
t1 = -100;
t2 = 100;
L = Table[N[(h[t2] - h[t1])], {x, 1/200, dd, 1/200}];
l1 = ListLinePlot[
   Re[Sum[Sum[L*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]],
   DataRange -> {1/200, dd}, PlotRange -> {-80, 80},
   ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
l2 = Table[
   Graphics[{Arrowheads[0.025],
     Arrow[{{Log[n], -35}, {Log[n], 0}}]}], {n, 1, 7}];
l3 = Table[
   Graphics[
    Style[Rotate[Text[Log[n], {Log[n], -52}], 90 Degree], Large]], {n,
     1, 7}];
Show[l1, l2, l3]

nn = k;
m[t_] = Sum[
  Sum[(Sum[1/n^(1/2 + I*t), {n, 1, k}] +
       k^(1 - (1/2 + I*t))/((1/2 + I*t) - 1) - (k^(-(1/2 + I*t)))/2)*
     MoebiusMu[d]/N[d]^(1/2 + I*t - 1), {d, Divisors[z]}]/z^c, {z, 1,
   nn}]; m1 =
 ListLinePlot[Table[N[Re[m[t]]], {t, 1/10, 60, 1/10}],
  DataRange -> {0, 60}, PlotRange -> {-3, 7 + 1/2},
  ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
m2 = Table[
   Graphics[{Arrowheads[0.025],
     Arrow[{{Im[ZetaZero[n]], -1}, {Im[ZetaZero[n]], 0}}]}], {n, 1,
    13}];
m3 = Table[
   Graphics[
    Style[Rotate[Text[N[Im[ZetaZero[n]]], {Im[ZetaZero[n]], -2}],
      90 Degree], Medium]], {n, 1, 13}];
Show[m1, m2, m3]

nn = k;
t1 = -100;
t2 = 100;
T = Table[N[(h[t2] - h[t1])], {x, 1/200, dd, 1/200}];
f1 = ListLinePlot[
   Re[Sum[Sum[T*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]],
   DataRange -> {1/200, dd}, PlotRange -> {-80, 80},
   ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
f2 = Table[
   Graphics[{Arrowheads[0.025],
     Arrow[{{Log[n], -35}, {Log[n], 0}}]}], {n, 1, 7}];
f3 = Table[
   Graphics[
    Style[Rotate[Text[Log[n], {Log[n], -52}], 90 Degree], Large]], {n,
     1, 7}];
Show[f1, f2, f3]