Helmut Hasse has proved that for $s \in \mathbb{C}-\{1\}$ the Riemann zeta function can be written as:


$$\zeta(s)=\frac{1}{1-2^{1-s}}\sum\_{n=0}^\infty\frac{1}{2^{n+1}}\sum\_{k=0}^n(-1)^k\ {n \choose k}\ \frac{1}{(k+1)^{s}}$$


Let $K(a,b) = \frac{\gcd(a,b)}{a+b}$. Then we might define:


$$\hat{\zeta}(s,a)=\frac{1}{1-2^{1-s}}\sum\_{n=0}^\infty\frac{1}{2^{n+1}}\sum\_{k=0}^n(-1)^k\ {n \choose k} K(k,a)^{s}$$


Then $\hat{\zeta}(s,1) = \zeta(s)$.


I looked at some non-trivial zeros of $\zeta$ and computed numerically in SAGEMATH:


**If $\zeta(s)=0$ with $\Re(s)=\frac{1}{2}$, is then $|\hat{\zeta}(s,3)|^2=\frac{1}{2}$ or is this just some bug / numerical curiosity in my code?**


Here is the SAGE Code:



```
def K(a,b):
    return gcd(a,b)/(a+b)

def zzeta(s,a=1,N=400):
        return 1/(1-2**(1-s))*sum([ 1/2**(n+1)*sum([ (-1)**k*binomial(n,k)*K(k,a)**s for k in range(n+1)]) for n in range(0,N+1)])


zz = zeta_zeros()

for z in zz[0:100]:
    print( abs(zzeta(1/2+z*I,3))**2)

```

And here is some of the ouptut:



```
0.5000000001016516
0.49999999980871757
0.49999999990048116
0.4999999999150287
0.5000000001425052
0.5000000003259728
0.4999999999688171
0.5000000000737418
0.5000000004006991
0.5000000003452242
0.49999999982210547
0.5000000000784613
0.49999999939079287
0.5000000012082603
0.5000000000591772
0.5000000002545129
0.4999999996870637

```

Some data for $a=2$:



```
0.5027599264329976
0.4120370789607451
0.3869873346989051
0.2811880366841188
0.2594539123560252
0.7742279955909737
0.18310290214474273
0.4006464279725155
0.2723910071065707
0.16780149194046892

```


If $\zeta(s)=0$ with $\Re(s)=\frac{1}{2}$, is then $|\hat{\zeta}(s,3)|^2=\frac{1}{2}$?
This is just a simple calculation. Note that 
$$
\sum\_{n=k}^{\infty} \frac{1}{2^{n+1}} \binom{n}{k} =1 
$$ 
for any non-negative integer $k$, which immediately gives 
$$ 
\frac{1}{1-2^{1-s}} \sum\_{n=0}^{\infty} \frac{1}{2^{n+1}} \sum\_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^s} \binom{n}{k} = \frac{1}{1-2^{1-s}}\sum\_{k=0}^{\infty} 
\frac{(-1)^k}{(k+1)^s} = \zeta(s).
$$ 


Similarly, 
$$ 
{\hat \zeta}(s,3) = \frac{1}{1-2^{1-s}} \sum\_{k=0}^{\infty} (-1)^k K(k,3)^s = \frac{1}{1-2^{1-s}} \sum\_{k=0}^{\infty} \frac{(-1)^k (k,3)^s}{(k+3)^s}. 
$$
Separating out terms with $3|k$, we may rewrite the sum over $k$ above as 
\begin{align\*}
&\sum\_{k=0}^{\infty} \frac{(-1)^k}{(k+3)^s} + (3^s-1)\sum\_{k=0}^{\infty} \frac{(-1)^{3k}}{(3k+3)^s}\\
 = &-1 + \frac{1}{2^s} +\sum\_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^s} + \frac{3^s-1}{3^s} \sum\_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^s}.
\end{align\*} 
Therefore 
$$
{\hat \zeta}(s,3) = \frac{-1+2^{-s}}{1-2^{1-s}} + \zeta(s) + \frac{3^s-1}{3^s} \zeta(s).
$$
Now evaluate this at a zero of $\zeta(s)$ on the critical line (or anywhere else).


