Let $\psi(n,x)$ denote the polygamma function.

In this answer Lucia gave linear relations for $\psi(m,1/3),\psi(m,1/6),\zeta(m+1)$.

The computer managed to find closed form for $\psi(2,1/6)$ and $\psi(4,1/6)$ in terms of $\zeta, \pi$ and $\sqrt{3}$. As a byproduct this give another closed form for $\zeta(3),\zeta(5)$.

Working with precision 1000 decimal digits we have:

$$\psi(2,1/6)= -2/3 \sqrt{3}(6\pi^3 + 91\sqrt{3}\zeta(3))$$ $$\psi(4,1/6)=-176 \sqrt{3}\pi^5 - 90024\zeta(5)$$ $$ \zeta(3)=-1/546\sqrt{3}(12 \pi^3 + \sqrt{3}\psi(2, 1/6)) $$

$$\zeta(5)= -2/1023 \sqrt{3} \pi^5 - 1/90024 \psi(4, 1/6) $$

Are these experimental results correct? Do they generalize for $\psi(m,1/6)$ for other $m$?

**Added** Looks like for all $m$, $\pi,\psi(m,1/6),\psi(m,1/3)$
are algebraically dependent and if we add $\sqrt{3}$ they are
linearly dependent. Proving this and Lucia's answer will answer the question.

Adding linear dependencies.

In machine readable form:

```
p_2_16= -2/3*sqrt(3)*(6*pi^3 + 91*sqrt(3)*zeta(3)) # psi(2,1/6)
p_4_16=-176*sqrt(3)*pi^5 - 90024*zeta(5) # psi(4,1/6)
zeta3=-1/546*sqrt(3)*(12*pi^3 + sqrt(3)*psi(2, 1/6))
zeta5= -2/1023*sqrt(3)*pi^5 - 1/90024*psi(4, 1/6)
print 'psi(2,1/6)',(p_2_16 - psi(2,1/6))
print 'psi(4,1/6)',p_4_16 - psi(4,1/6)
print 'zeta(3)',zeta3 - zeta(3)
print 'zeta(5)',zeta5 - zeta(5)
#linear depend.
m= 2 ; x0=psi(2, 1/6); x1=psi(2, 1/3); x3=sqrt(3)*pi^3 ;f= 9*x0 - 63*x1 + 8*x3
m= 3 ; x0=psi(3, 1/6); x1=psi(3, 1/3); x2=pi^4 ;f= 3*x0 - 51*x1 + 16*x2
m= 4 ; x0=psi(4, 1/6); x1=psi(4, 1/3); x3=sqrt(3)*pi^5 ;f= 3*x0 - 93*x1 + 32*x3
m= 5 ; x0=psi(5, 1/6); x1=psi(5, 1/3); x2=pi^6 ;f= 9*x0 - 585*x1 + 832*x2
m= 6 ; x0=psi(6, 1/6); x1=psi(6, 1/3); x3=sqrt(3)*pi^7 ;f= 3*x0 - 381*x1 + 896*x3
m= 7 ; x0=psi(7, 1/6); x1=psi(7, 1/3); x2=pi^8 ;f= -3*x0 + 771*x1 - 10496*x2
m= 8 ; x0=psi(8, 1/6); x1=psi(8, 1/3); x3=sqrt(3)*pi^9 ;f= 27*x0 - 13797*x1 + 414208*x3
m= 9 ; x0=psi(9, 1/6); x1=psi(9, 1/3); x2=pi^10 ;f= 3*x0 - 3075*x1 + 687104*x2
m= 10 ; x0=psi(10, 1/6); x1=psi(10, 1/3); x3=sqrt(3)*pi^11 ;f= -3*x0 + 6141*x1 - 3782656*x3
m= 11 ; x0=psi(11, 1/6); x1=psi(11, 1/3); x2=pi^12 ;f= 9*x0 - 36873*x1 + 206614528*x2
```