In other words, to solve a more general problem we just need to change initial $\nu_i = 1$ to $\nu_i = c_{i-1}$. Nice!

Ok, you've done most of the work, so I accept your answer.

Maybe you mean $(qk+1)$? In this case, we have $(0,\beta,\gamma,\alpha,-\alpha,0)=(0,q,1,p,-p,0)$. Below we have for F1b $c_{2k-1} = \gamma + (k-1)\beta, c_{2k} = k\alpha'x$ compared with $c_{2k-1} = w + (k-1)u, c_{2k} = y + (k-1)v$ and since $(w,y,u,v) = (1,p,q,p)$ we have the same result (if $x$ is not taken into account).

Thank you very much again! I completely agree that proposition 3.3 gives $a_1=a_4$. Could you point out where it is shown that $a_4=a_5$?

Thank you for answer! Could you add the details of the calculations performed in Maple? Also thank you for reference. Please see proposition 3.3. Could you add this result to your answer?