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Let $P$ be the prime zeta function

\begin{aligned} {\displaystyle P (s) = \sum _ {p\, \in\mathrm {\, primes}} {\frac {1} {p^{s}}} = {\frac {1} {2^{s}}} + {\frac {1} {3^{s}}} + {\frac {1} {5^{s}}} + {\frac {1} {7^{s}}} + {\frac {1} {11^{s}}} + \cdots } \end{aligned}

and define the semiprime zeta function as

\begin{aligned} & \zeta_{\Omega_ 2} (s) = \exp \sum _k^\infty \frac{P (k s)^2+P(2 k s)}{2 k} \end{aligned}

Then let $\gamma$ be the imaginary part the zeros of this function. When these then are plugged into

\begin{aligned} &\sum _{\gamma} -\frac{\cos (\gamma \log t)}{\ \sqrt{\gamma}} \end{aligned}

the sum presumably then should peak at the semiprimes, (just as peaks are formed at the primes when summed over the imaginary parts of the Riemann Zeta function).

But the zeros are hard to find. I assumed that the zeros are along the line $ (\frac{1}{2}+it)$ (a huge assumption, I know, given that the zeros of the prime zeta function are not on this line). I then calculated the near-zeros along this line with

func[t_] := 1/Abs[Exp[(PrimeZetaP[(1/2 + I t)]^2 + PrimeZetaP[2 (1/2 + I t)])/2]];
data = Table[{t, func[t]}, {t, .01, #, .01}] &[(**)10^4];
peaks = Pick[data, PeakDetect[data[[;; , 2]], .01, .001], 1];

in Mathematica, which takes a long time to evaluate, so I uploaded the data here:

peaks = ToExpression[Import["https://raw.githubusercontent.com/martinq321/peaks/main/omega2"]];

Below is a plot of the absolute value of the first term of $\zeta_{\Omega_ 2} (s)$

$\left| \exp \left(P \left (1/2+ i t \right)^2/2+P \left(1+ 2i t \right)/2\right)\right|$ with the red lines marking near-zero points:

Given the above, the plot below shows

\begin{aligned} &\sum _{k \in \text{peaks}} -\frac{\cos (k \log t)}{\ \sqrt{k}}, \quad 200 \geq t \geq 250 \end{aligned}

where, despite the noise, the peaks show at the semiprimes and the primes:

and with the noise cleared up a bit:

li2 = Select[peaks, #[[2]] > 5 &][[All, 1]]; x1 = 200; x2 = 250;
data1 = Table[(Sum[-Cos[li2[[p]] Log[t]]/Sqrt@p, 
{p, Length@li2}]), {t, x1, x2, .1}];
ListLinePlot[BilateralFilter[data1, 1., 1, MaxIterations -> 25], 
PlotStyle -> {AbsoluteThickness[3]}, DataRange -> {x1, x2}, 
ImageSize -> 700, Frame -> True, AspectRatio -> 1/8]

What is going on here? I presume the peaks also showing at the primes because many of the 'zeros' are very close to the zeros of the zeta function. Is there a way to filter them out? Is there a better way to find the zeros of $\zeta_{\Omega_ 2} (s)$?

Let $P$ be the prime zeta function

$$ P (s) = \sum_{p\, \in\text{ primes}} \frac 1 {p^s} = \frac {1} {2^s} + \frac {1} {3^s} + \frac 1 {5^s} + \frac 1 {7^s} + \frac 1 {11^s} + \cdots $$

and define the semiprime zeta function as

$$ \zeta_{\Omega_2} (s) = \exp \sum _k^\infty \frac{P (k s)^2+P(2 k s)}{2 k} $$

Then let $\gamma$ be the imaginary part the zeros of this function. When these then are plugged into

$$ \sum_\gamma -\frac{\cos(\gamma \log t)}{\sqrt \gamma} $$

the sum presumably then should peak at the semiprimes, (just as peaks are formed at the primes when summed over the imaginary parts of the Riemann Zeta function).

But the zeros are hard to find. I assumed that the zeros are along the line $ (\frac{1}{2}+it)$ (a huge assumption, I know, given that the zeros of the prime zeta function are not on this line). I then calculated the near-zeros along this line with

func[t_] := 1/Abs[Exp[(PrimeZetaP[(1/2 + I t)]^2 + PrimeZetaP[2 (1/2 + I t)])/2]];
data = Table[{t, func[t]}, {t, .01, #, .01}] &[(**)10^4];
peaks = Pick[data, PeakDetect[data[[;; , 2]], .01, .001], 1];

in Mathematica, which takes a long time to evaluate, so I uploaded the data here:

peaks = ToExpression[Import["https://raw.githubusercontent.com/martinq321/peaks/main/omega2"]];

Below is a plot of the absolute value of the first term of $\zeta_{\Omega_2} (s)$

$\left| \exp \left(P \left (1/2+ i t \right)^2/2+P \left(1+ 2i t \right)/2\right)\right|$ with the red lines marking near-zero points:

Given the above, the plot below shows

\begin{aligned} \sum_{k \in \text{peaks}} -\frac{\cos (k \log t)}{\sqrt k}, \quad 200 \geq t \geq 250 \end{aligned}

where, despite the noise, the peaks show at the semiprimes and the primes:

and with the noise cleared up a bit:

li2 = Select[peaks, #[[2]] > 5 &][[All, 1]]; x1 = 200; x2 = 250;
data1 = Table[(Sum[-Cos[li2[[p]] Log[t]]/Sqrt@p, 
{p, Length@li2}]), {t, x1, x2, .1}];
ListLinePlot[BilateralFilter[data1, 1., 1, MaxIterations -> 25], 
PlotStyle -> {AbsoluteThickness[3]}, DataRange -> {x1, x2}, 
ImageSize -> 700, Frame -> True, AspectRatio -> 1/8]

What is going on here? I presume the peaks also showing at the primes because many of the 'zeros' are very close to the zeros of the zeta function. Is there a way to filter them out? Is there a better way to find the zeros of $\zeta_{\Omega_2} (s)$?

Let $P$ be the prime zeta function

\begin{aligned} {\displaystyle P (s) = \sum _ {p\, \in\mathrm {\, primes}} {\frac {1} {p^{s}}} = {\frac {1} {2^{s}}} + {\frac {1} {3^{s}}} + {\frac {1} {5^{s}}} + {\frac {1} {7^{s}}} + {\frac {1} {11^{s}}} + \cdots } \end{aligned}

and define the semiprime zeta function as

\begin{aligned} & \zeta_{\Omega_ 2} (s) = \exp \sum _k^\infty \frac{P (k s)^2+P(2 k s)}{2 k} \end{aligned}

When these zeros are plugged into

\begin{aligned} &\sum _{\gamma} -\frac{\cos (\gamma \log t)}{\ \sqrt{\gamma}} \end{aligned}

where $\gamma$ is the imaginary part of the zeros of $\zeta_{\Omega_ 2} (\frac{1}{2}+it)$, the sum presumably then should peak at the semiprimes, (just as peaks are formed at the primes when summed over the imaginary parts of the Riemann Zeta function).

But the zeros are hard to find. I calculated the near-zeros along the line $1/2 +it$ with

func[t_] := 1/Abs[Exp[(PrimeZetaP[(1/2 + I t)]^2 + PrimeZetaP[2 (1/2 + I t)])/2]];
data = Table[{t, func[t]}, {t, .01, #, .01}] &[(**)10^4];
peaks = Pick[data, PeakDetect[data[[;; , 2]], .01, .001], 1];

in Mathematica, which takes a long time to evaluate, so I uploaded the data here:

peaks = ToExpression[Import["https://raw.githubusercontent.com/martinq321/peaks/main/omega2"]];

Below is a plot of the absolute value of the first term of $\zeta_{\Omega_ 2} (s)$

$\left| \exp \left(P \left (1/2+ i t \right)^2/2+P \left(1+ 2i t \right)/2\right)\right|$ with the red lines marking near-zero points:

Given the above, the plot below shows

\begin{aligned} &\sum _{k \in \text{peaks}} -\frac{\cos (k \log t)}{\ \sqrt{k}}, \quad 200 \geq t \geq 250 \end{aligned}

where, despite the noise, the peaks show at the semiprimes and the primes:

and with the noise cleared up a bit:

li2 = Select[peaks, #[[2]] > 5 &][[All, 1]]; x1 = 200; x2 = 250;
data1 = Table[(Sum[-Cos[li2[[p]] Log[t]]/Sqrt@p, 
{p, Length@li2}]), {t, x1, x2, .1}];
ListLinePlot[BilateralFilter[data1, 1., 1, MaxIterations -> 25], 
PlotStyle -> {AbsoluteThickness[3]}, DataRange -> {x1, x2}, 
ImageSize -> 700, Frame -> True, AspectRatio -> 1/8]

What is going on here? I presume the peaks also showing at the primes because many of the 'zeros' are very close to the zeros of the zeta function. Is there a way to filter them out? Is there a better way to find the zeros of $\zeta_{\Omega_ 2} (s)$?

Let $P$ be the prime zeta function

\begin{aligned} {\displaystyle P (s) = \sum _ {p\, \in\mathrm {\, primes}} {\frac {1} {p^{s}}} = {\frac {1} {2^{s}}} + {\frac {1} {3^{s}}} + {\frac {1} {5^{s}}} + {\frac {1} {7^{s}}} + {\frac {1} {11^{s}}} + \cdots } \end{aligned}

and define the semiprime zeta function as

\begin{aligned} & \zeta_{\Omega_ 2} (s) = \exp \sum _k^\infty \frac{P (k s)^2+P(2 k s)}{2 k} \end{aligned}

Then let $\gamma$ be the imaginary part the zeros of this function. When these then are plugged into

\begin{aligned} &\sum _{\gamma} -\frac{\cos (\gamma \log t)}{\ \sqrt{\gamma}} \end{aligned}

the sum presumably then should peak at the semiprimes, (just as peaks are formed at the primes when summed over the imaginary parts of the Riemann Zeta function).

But the zeros are hard to find. I assumed that the zeros are along the line $ (\frac{1}{2}+it)$ (a huge assumption, I know, given that the zeros of the prime zeta function are not on this line). I then calculated the near-zeros along this line with

func[t_] := 1/Abs[Exp[(PrimeZetaP[(1/2 + I t)]^2 + PrimeZetaP[2 (1/2 + I t)])/2]];
data = Table[{t, func[t]}, {t, .01, #, .01}] &[(**)10^4];
peaks = Pick[data, PeakDetect[data[[;; , 2]], .01, .001], 1];

in Mathematica, which takes a long time to evaluate, so I uploaded the data here:

peaks = ToExpression[Import["https://raw.githubusercontent.com/martinq321/peaks/main/omega2"]];

Below is a plot of the absolute value of the first term of $\zeta_{\Omega_ 2} (s)$

$\left| \exp \left(P \left (1/2+ i t \right)^2/2+P \left(1+ 2i t \right)/2\right)\right|$ with the red lines marking near-zero points:

Given the above, the plot below shows

\begin{aligned} &\sum _{k \in \text{peaks}} -\frac{\cos (k \log t)}{\ \sqrt{k}}, \quad 200 \geq t \geq 250 \end{aligned}

where, despite the noise, the peaks show at the semiprimes and the primes:

and with the noise cleared up a bit:

li2 = Select[peaks, #[[2]] > 5 &][[All, 1]]; x1 = 200; x2 = 250;
data1 = Table[(Sum[-Cos[li2[[p]] Log[t]]/Sqrt@p, 
{p, Length@li2}]), {t, x1, x2, .1}];
ListLinePlot[BilateralFilter[data1, 1., 1, MaxIterations -> 25], 
PlotStyle -> {AbsoluteThickness[3]}, DataRange -> {x1, x2}, 
ImageSize -> 700, Frame -> True, AspectRatio -> 1/8]

What is going on here? I presume the peaks also showing at the primes because many of the 'zeros' are very close to the zeros of the zeta function. Is there a way to filter them out? Is there a better way to find the zeros of $\zeta_{\Omega_ 2} (s)$?

Let $P$ be the prime zeta function

\begin{aligned} {\displaystyle P (s) = \sum _ {p\, \in\mathrm {\, primes}} {\frac {1} {p^{s}}} = {\frac {1} {2^{s}}} + {\frac {1} {3^{s}}} + {\frac {1} {5^{s}}} + {\frac {1} {7^{s}}} + {\frac {1} {11^{s}}} + \cdots } \end{aligned}

and define the semiprime analogue of the Riemann zeta function as

\begin{aligned} & \zeta_{\Omega_ 2} (s) = \exp \sum _k^\infty \frac{P (k s)^2+P(2 k s)}{2 k} \end{aligned}

When these zeros are plugged into

\begin{aligned} &\sum _{\gamma} -\frac{\cos (\gamma \log t)}{\ \sqrt{\gamma}} \end{aligned}

where $\gamma$ is the imaginary part of the zeros of $\zeta_{\Omega_ 2} (\frac{1}{2}+it)$, the sum presumably then should peak at the semiprimes, (just as peaks are formed at the primes when summed over the imaginary parts of the Riemann Zeta function).

But the zeros are hard to find. I calculated the near-zeros along the line $1/2 +it$ with

func[t_] := 1/Abs[Exp[(PrimeZetaP[(1/2 + I t)]^2 + PrimeZetaP[2 (1/2 + I t)])/2]];
data = Table[{t, func[t]}, {t, .01, #, .01}] &[(**)10^4];
peaks = Pick[data, PeakDetect[data[[;; , 2]], .01, .001], 1];

in Mathematica, which takes a long time to evaluate, so I uploaded the data here:

peaks = ToExpression[Import["https://raw.githubusercontent.com/martinq321/peaks/main/omega2"]];

Below is a plot of the absolute value of the first term of $\zeta_{\Omega_ 2} (s)$

$\left| \exp \left(P \left (1/2+ i t \right)^2/2+P \left(1+ 2i t \right)/2\right)\right|$ with the red lines marking near-zero points:

Given the above, the plot below shows

\begin{aligned} &\sum _{k \in \text{peaks}} -\frac{\cos (k \log t)}{\ \sqrt{k}}, \quad 200 \geq t \geq 250 \end{aligned}

where, despite the noise, the peaks show at the semiprimes and the primes:

and with the noise cleared up a bit:

li2 = Select[peaks, #[[2]] > 5 &][[All, 1]]; x1 = 200; x2 = 250;
data1 = Table[(Sum[-Cos[li2[[p]] Log[t]]/Sqrt@p, 
{p, Length@li2}]), {t, x1, x2, .1}];
ListLinePlot[BilateralFilter[data1, 1., 1, MaxIterations -> 25], 
PlotStyle -> {AbsoluteThickness[3]}, DataRange -> {x1, x2}, 
ImageSize -> 700, Frame -> True, AspectRatio -> 1/8]

What is going on here? I presume the peaks also showing at the primes because many of the 'zeros' are very close to the zeros of the zeta function. Is there a way to filter them out? Is there a better way to find the zeros of $\zeta_{\Omega_ 2} (s)$?

Let $P$ be the prime zeta function

\begin{aligned} {\displaystyle P (s) = \sum _ {p\, \in\mathrm {\, primes}} {\frac {1} {p^{s}}} = {\frac {1} {2^{s}}} + {\frac {1} {3^{s}}} + {\frac {1} {5^{s}}} + {\frac {1} {7^{s}}} + {\frac {1} {11^{s}}} + \cdots } \end{aligned}

and define the semiprime zeta function as

\begin{aligned} & \zeta_{\Omega_ 2} (s) = \exp \sum _k^\infty \frac{P (k s)^2+P(2 k s)}{2 k} \end{aligned}

When these zeros are plugged into

\begin{aligned} &\sum _{\gamma} -\frac{\cos (\gamma \log t)}{\ \sqrt{\gamma}} \end{aligned}

where $\gamma$ is the imaginary part of the zeros of $\zeta_{\Omega_ 2} (\frac{1}{2}+it)$, the sum presumably then should peak at the semiprimes, (just as peaks are formed at the primes when summed over the imaginary parts of the Riemann Zeta function).

But the zeros are hard to find. I calculated the near-zeros along the line $1/2 +it$ with

func[t_] := 1/Abs[Exp[(PrimeZetaP[(1/2 + I t)]^2 + PrimeZetaP[2 (1/2 + I t)])/2]];
data = Table[{t, func[t]}, {t, .01, #, .01}] &[(**)10^4];
peaks = Pick[data, PeakDetect[data[[;; , 2]], .01, .001], 1];

in Mathematica, which takes a long time to evaluate, so I uploaded the data here:

peaks = ToExpression[Import["https://raw.githubusercontent.com/martinq321/peaks/main/omega2"]];

Below is a plot of the absolute value of the first term of $\zeta_{\Omega_ 2} (s)$

$\left| \exp \left(P \left (1/2+ i t \right)^2/2+P \left(1+ 2i t \right)/2\right)\right|$ with the red lines marking near-zero points:

Given the above, the plot below shows

\begin{aligned} &\sum _{k \in \text{peaks}} -\frac{\cos (k \log t)}{\ \sqrt{k}}, \quad 200 \geq t \geq 250 \end{aligned}

where, despite the noise, the peaks show at the semiprimes and the primes:

and with the noise cleared up a bit:

li2 = Select[peaks, #[[2]] > 5 &][[All, 1]]; x1 = 200; x2 = 250;
data1 = Table[(Sum[-Cos[li2[[p]] Log[t]]/Sqrt@p, 
{p, Length@li2}]), {t, x1, x2, .1}];
ListLinePlot[BilateralFilter[data1, 1., 1, MaxIterations -> 25], 
PlotStyle -> {AbsoluteThickness[3]}, DataRange -> {x1, x2}, 
ImageSize -> 700, Frame -> True, AspectRatio -> 1/8]

What is going on here? I presume the peaks also showing at the primes because many of the 'zeros' are very close to the zeros of the zeta function. Is there a way to filter them out? Is there a better way to find the zeros of $\zeta_{\Omega_ 2} (s)$?