Let $ \forall n \in N, n\geq 1$ $$ R_n(x)=(-1)^n n! \displaystyle \frac{(x-1)...(x-n)}{(x(x+1)..(x+n))^2}$$ thus by decomposition in simple element it's easy to see that

$$ (1): \quad R_n(x)= \displaystyle \sum_{k=0}^{n} \displaystyle \frac{\alpha_{k,n}}{(x+k)^2}+\displaystyle \sum_{k=0}^{n} \displaystyle \frac{\beta_{k,n}}{(x+k)}$$

with $ \forall 0\leq k \leq n,$ $$ \alpha_{n,k}= (C_n^k)^2 C_{n+k}^n \quad \quad\beta_{n,k}= (C_n^k)^2 C_{n+k}^n( \displaystyle \sum_{j=1}^n \displaystyle \frac{1}{-k-j}-\displaystyle \sum_{j=0, j \neq k}^n \displaystyle \frac{2}{-k+j}) $$

Now let $$ \forall n \geq 1,P_n(x)=\displaystyle \sum_{j=0}^{n} \displaystyle \alpha_{j,n}x^j \quad \quad Q_n(x)=\displaystyle \sum_{j=0}^{n} \displaystyle \beta_{j,n}x^j$$ and $$P_0=1, \quad Q_0=0$$

it's easy to see that the relation $(1)$ gives us the following relation $(2)$ $$ (2): \quad \forall t \in \{1,..n \} \quad \int_{0}^1 (-P_n(x).\log(x)+Q_n(x))x^{t-1} dx=R_n(t)=0 $$

Maple using Zeilberger algorithm gives us for all $$ n \in N, A_{n,0}(x)P_n(x)+ A_{n,1}(x)P_{n+1}(x)+A_{n,2}(x)P_{n+2}(x)+A_{n,3}(x)P_{n+3}(x)=0$$ and I checked using Maple that this relation is also satisfied by $Q_n:$ (with the same $A_{n,j}$)

$$ \forall 0 \leq n \leq 20, A_{n,0}(x)Q_n(x)+ A_{n,1}(x)Q_{n+1}(x)+A_{n,2}(x)Q_{n+2}(x)+A_{n,3}(x)Q_{n+3}(x)=0$$ with $$A_{n,0}(x)=66+159n-142x+27n^3+120n^2-260xn^2-343xn+184xn^2-59xn^3+76n^2+140x^2n^2+32x^2n^3$$


I can give $A_{n,1}$ and $A_{n,2}$ but I think that this is not important. So my question is the following: is it possible WITHOUT USING Zeilberger algorithm, using for example $(1)$ or $(2)$ (or other argument such that relation between $R_n(x),R_{n+1}(x),R_{n+1}(x+1),R_{n+2}(x),...$) to prove the existence of such recurrence satisfied by both the family $(P_n)$ and $(Q_n)$ and is there a way to compute the coefficient $A_{n,0}(x),..A_{n,3}(x)$ of the recurrence (using Euclidian division, scalar product for example etc..) but WITHOUT using Zeilberger algorithm. Thanks for your help.


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