Laisant’s recurrence relation is a formula which is as follows:
$$\left(n-1\right)A_{n+1}=\left(n^{2}-1\right)A_{n}+\left(n+1\right)A_{n-1}+4\left(-1\right)^{n}$$
Where $A_n$ are menage numbers.
A proof of this formula has been given here.
In page 3,I have a problem with $C$-arrangements,I don' understand what exactly is going on,can someone explain it better?
A C-arrangement takes the form
$$M_1F_1X_1F_2X_2\ldots F_nX_nF_{n+1}\;.\tag{1}$$
Suppose first that $X_1=M_{n+1}$, so that $(1)$ becomes
$$M_1F_1M_{n+1}F_2X_2\ldots F_nX_nF_{n+1}\;.\tag{2}$$
If we now remove the first couple, $M_1F_1$, from the arrangement, what remains is the arrangement
$$M_{n+1}F_2X_2\ldots F_nX_nF_{n+1}\;.\tag{3}$$
We can rotate it around the table one place to the right to get
$$F_{n+1}M_{n+1}F_2X_2\ldots F_nX_n$$
and then renumber couple $n+1$ as couple $1$ to get
$$F_1M_1F_2X_2\ldots F_nX_n\;,$$
which is a C-arrangement, but this time of only $n$ couples: $M_1$ sits to the left of $F_1$, and exactly one of the other men sits next to his wife. By definition there are $C_n$ such arrangements, so the case $X_1=M_{n+1}$ contributes $C_n$ C-arrangements of $n+1$ couples. Moreover, each C-arrangement of $n+1$ couples in which $X_1=M_{n+1}$ is uniquely derivable from one of these $C_n$ C-arrangements of $n$ couples.
Now suppose that $X_1\ne M_{n+1}$. One of the $n$ men $M_2,M_3,\ldots,M_{n+1}$ is sitting next to his wife. Looking at $(1)$, we see that $M_{n+1}$ is not sitting to the right of his wife, because $M_1$ is in that seat, but he could be sitting to the left of $F_{n+1}$. And if it’s any of the men $M_2,M_3,\ldots,M_n$ who is sitting next to his wife, he could be sitting on either side of her. Thus, there are $2n-1$ possibilities:
In each of these cases we can again remove the first couple, leaving
$$X_1F_2X_2\ldots F_nX_nF_{n+1}\;,$$
where $X_1\ne M_{n+1}$. This is an arrangement of $n$ couples in which exactly one man is sitting next to his wife. If $M_2$ was sitting to the left of $F_2$ originally, the new arrangement is
$$M_2F_2X_2\ldots F_nX_nF_{n+1}\;,$$
and $M_2$ is the only man sitting next to his wife: $M_1$ was the only other man sitting next to his wife originally, he is now gone, and since we’re assuming at this point that $X_1\ne M_{n+1}$, the departure of the first couple did not cause any other man to be seated next to his wife.
If $M_2$ was sitting to the right of $F_2$ originally, he still, is and we have the arrangement
$$X_1F_2M_2\ldots F_nX_nF_{n+1}\;,$$
in which he is the only man sitting next to his wife.
And so on through all $2n-1$ possibilities: in each case we end up with an arrangement of $n$ couples in which exactly one man is sitting next to his wife, i.e., a B-arrangement. By definition there are $B_n$ such arrangements, and there are $2n-1$ ways to choose which man is sitting next to his wife and on which side of her, so the case $X_1\ne M_{n+1}$ contributes $(2n-1)B_n$ C-arrangements of $n+1$ couples. And every C-arrangement of $n+1$ couples in which $X_1\ne M_{n+1}$ is uniquely derivable from one of these B-arrangements of $n$ couples.
Thus,
$$C_{n+1}=C_n+(2n-1)B_n\;.$$
By the way, there’s a small error on p.494 of Lucas’s *Théorie des nombres’ from your earlier question about this result: in the little table of initial values below formula $(5)$ the correct value for $\nu_4$ is $5$, not $1$. The exposition to which you’ve linked here avoids the type that Lucas was counting with the numbers $\nu_n$, folding it into the count of A-arrangements from B-arrangements, so it has only $A_n$ (Lucas’s $\lambda_n$), $B_n$ (Lucas’s $\mu_n$), and $C_n$ (Lucas’s $\rho_n$).