Diagonal  of the double sequence $(n+1)v_{h,n+1}-(2h+1)v_{h,n}-nv_{h,n-1}=0$ Update: it is not possible to reply to this question without additional information. 
My comment below: "I have to agree with you that one "cannot derive (2) from (1) alone". Now it seems to me that one must consider how the sequences $v_{h,n}^{\prime }$ and $v_{h,n}$ are constructed. That is described in the first part of the paper, but unfortunately it is not easy for me to summarize it. As I understand Apéry transformed repeatidely a continued fraction whose approximants are $\dfrac{u_{h,n}^{\prime }}{u_{h,n}}$, iterating on $h$. The above sequences are $v_{h,n}^{\prime }=\dfrac{u_{h,n}^{\prime }}{h!n!}, v_{h,n}=\dfrac{u_{h,n}}{h!n!}$."  

In Irrationalité de Certaines Constantes, Bull. section des sciences du C.T.H.S., n.º3, p.37-53, Roger Apéry derives rational approximations $\dfrac{v_{h,n}^{\prime }}{v_{h,n}}$ for $\ln (1+t)$, 
$\zeta (2)$ and $\zeta (3)$, the simplest being the one for the $\ln $. The
sequences $v_{h,n}^{\prime }$ and $v_{h,n}$, whose ratio converges
to $\ln (2)$, satisfy the recursive relation
$$(n+1)v_{h,n+1}-(2h+1)v_{h,n}-nv_{h,n-1}=0.\qquad (1)$$
The diagonal sequences $w_{n}^{\prime }=v_{n,n}^{\prime },w_{n}=v_{n,n}$
satisfy
$$(n+1)w_{n+1}-3\left( 2n+1\right) w_{n}-nw_{n-1}=0.\qquad (2)$$
Remarks: 


*

*The initial conditions for $v_{h,n}^{\prime }$, $v_{h,n}$, $w_{n}^{\prime
}$, $w_{n}$ are not indicated in the paper. 

*Recurrences $(1)$ and $(2)$ are the particular case for $t=1$ of, respectively,
$$(n+1)v_{h,n+1}-\left( \left( n+1\right) -nt+h\left( 1+t\right) \right)
v_{h,n}-ntv_{h,n-1}=0\qquad (\ast)$$
(to simplify the notation the index $h$ was deleted in the original) and
$$(n+1)w_{n+1}-\left( 2n+1\right) \left( 2+t\right) w_{n}-nt^{2}w_{n-1}=0.\qquad (\ast\ast)$$
Question: How do you derive $(2)$ from $(1)$?

Copy of the mentioned paper

 A: I hope I'm not misunderstanding your question; please correct me if I am.
I think the answer must be that you cannot derive (2) from (1) alone. For each $h$, (1) is a separate recurrence relation linking only values with the same $h$. If you assume only (1), you can freely choose two initial values $v_{h,0}$ and $v_{h,1}$ and use them to generate a two-dimensional vector space of sequences (for fixed $h$) that can be made to take any value for a given $n$. But (2) contains only members for different values of $h$, so each of the members occurring in (2) can separately be chosen arbitrarily by choosing suitable initial values for its value of $h$. But there can't be a linear relationship between three quantities that can all be independently made to take an arbitrary value.
(It's possible that for some values of $h$ and $n$ (1) forces $v_{h,n}$ to be zero, but that could only imply (2) if this were the case on the entire diagonal, i.e. if the diagonal sequence were in fact zero -- if it is non-zero somewhere, we can always take a multiple of the sequence for that $h$ and thereby disturb (2) but not (1).)
