Got numerical support that for odd $n$, $\zeta(n)$ might be
expressed in terms of the derivatives of $\zeta(\frac12)$.


Based on *More Zeta Functions for the Riemann Zeros*, Andre Voros, p.12, Table 3:


Conjecture: For odd $n$, 
$$\zeta(n) = \left(\frac{2}{(n-1)!} (\log(|\zeta|)^{(n)} (\frac12) - 2^n \beta(n))\right)/(2^n-1)$$


$\beta(n)$ is Dirichlet beta function and it is a rational multiple of $\pi^n$
for odd $n$.
The derivative can be expressed in terms of $\zeta(\frac12),\zeta^{(k)}(\frac12)$


For $n=3$ get numerical support for:


$$\zeta(3) = (-\zeta'''(\frac12)/|\zeta(\frac12)| -3 \zeta''(\frac12) \zeta'(\frac12)/|\zeta(\frac12)|^2 -2 \zeta'(\frac12)^3/|\zeta(\frac12)|^3- \pi^3 / 4)/7 $$


The last equality holds with precision $10^4$ decimal digits.
One can eliminate the first derivative since there is closed form
for $\zeta'(\frac12)/\zeta(\frac12)$



> 
> Is this result true?
> 
> 
> 


sage/mpmath code in case of typos of the latex.



```
#run in sage

import mpmath
from mpmath import mpf
mpmath.mp.pretty=True

def zeta3test():

    n=3
    mpmath.mp.dps=10^3
    zeta3=mpmath.zeta(3)

    Pi=mpmath.pi
    gamma=mpmath.euler
    z12=mpmath.zeta(1/mpmath.mpf(2))
    z1=mpmath.zeta(1/mpmath.mpf(2),derivative=1)
    z2=mpmath.zeta(1/mpmath.mpf(2),derivative=2)
    z3=mpmath.zeta(1/mpmath.mpf(2),derivative=3)

    # eliminate the first derivative

    #rh0=1/32*(72*Pi*mpmath.log(2)*mpmath.log(Pi)*z12+144*gamma*mpmath.log(2)*mpmath.log(Pi)*z12+72*mpmath.log(2)*z12*gamma*Pi+24*mpmath.log(Pi)*z12*gamma*Pi-144*z2*mpmath.log(2)-48*z2*mpmath.log(Pi)+216*mpmath.log(2)^3*z12+8*mpmath.log(Pi)^3*z12-48*z2*gamma-24*z2*Pi+8*z12*gamma^3+z12*Pi^3+72*mpmath.log(2)*z12*gamma^2+18*mpmath.log(2)*z12*Pi^2+24*mpmath.log(Pi)*z12*gamma^2+6*mpmath.log(Pi)*z12*Pi^2+216*mpmath.log(2)^2*mpmath.log(Pi)*z12+72*mpmath.log(2)*mpmath.log(Pi)^2*z12+216*gamma*mpmath.log(2)^2*z12+24*gamma*mpmath.log(Pi)^2*z12+108*Pi*mpmath.log(2)^2*z12+12*Pi*mpmath.log(Pi)^2*z12+32*z3+12*z12*gamma^2*Pi+6*z12*gamma*Pi^2)/z12
    z12a=mpmath.fabs(z12)
    rh1= -z3/z12a -3*z2*z1/z12a**2 -2* z1**3/z12a**3

    #print 'rh',mpmath.chop(rh0-rh1)
    #rhs= (mpmath.diff( lambda y:  mpmath.log(mpmath.fabs(mpmath.zeta(y))),1/2,n) - mpmath.pi^3 / 4 )/(7)
    rhs= (rh1 - mpmath.pi**3 / 4 )/(7)
    print mpmath.chop(zeta3-rhs)

def conjecture1(n):
    """

    voros, p. 12
    """
    assert n%2==1
    a1= mpmath.zeta(n)
    a2= (2/factorial(n-1) * mpmath.diff( lambda y:  mpmath.log(mpmath.fabs(mpmath.zeta(y))),1/2,n) - 2**(n) * dirichletbeta(n))/(2**n-1)
    print mpmath.chop(a1-a2)


def dirichletbeta(s):
    """
    dirichlet beta
    """
    return 4**(-s) * (mpmath.hurwitz(s,1/4)-mpmath.hurwitz(s,3/4))

```


zeta(3) in terms of derivatives of zeta at 1/2 and pi
In fact for odd $n\ge3$ we have 


$$\Bigl.\frac{d^{n}}{ds^n}\log\zeta(s)\Bigr|\_{s=\frac12}=
\frac{(n-1)!}{2}\Bigl(2^n L(n,\chi)+(2^n-1)\zeta(n)\Bigr)$$


The proof (due to Voros) is the following:
It is well known that 
$$\frac{\zeta'(s)}{\zeta(s)}=-\frac{1}{s-1}+\sum\_\rho\Bigl(\frac{1}{s-\rho}-\frac{1}{\rho}\Bigr)
+\sum\_{n=1}^\infty \Bigl(\frac{1}{s+2n}-\frac{1}{2n}\Bigr)+B$$


Differentiating this we get
$$\frac{d^{n-1}}{ds^{n-1}}\frac{\zeta'(s)}{\zeta(s)}=\frac{(-1)^{n-2} (n-1)!}{(s-1)^{n}}-\sum\_{\rho}
\frac{(n-1)!}{(\rho-s)^{n}}+\sum\_{k=1}^\infty \frac{(-1)^{n-1} (n-1)!}{(s+2k)^{n}}.$$
Now take $s=\frac12$


We get 


$$\Bigl.\frac{d^{n}}{ds^n}\log\zeta(s)\Bigr|\_{s=\frac12}=$$


$$(n-1)! 2^{n}-
\sum\_{\rho}\frac{(n-1)!}{(\rho-\frac12)^{n}}+\sum\_{k=1}^\infty \frac{(-1)^{n-1} (n-1)! 2^{n}}
{(4k+1)^{n}}.$$


When $n$ is odd the sum on the non trivial zeros is $0$ by symmetry. 
There is only the question of recognizing the sum.


Assuming $n$ even we have


$$\Bigl.\frac{d^{n}}{ds^n}\log\zeta(s)\Bigr|\_{s=\frac12}\frac{1}{(n-1)!\,2^{n-1}}=$$


$$=2+2\sum\_{k=1}^\infty \frac{1}{(4k+1)^{n}}$$


It is easily seen that 
$$2\sum\_{k=1}^\infty \frac{1}{(4k+1)^{s}}=-2+L(s,\chi)+(1-2^{-s})\zeta(s)$$
from which the result follows.


