Let $V(y,b;N,m,r)$ calculate the value of the interest of an initial amount of money, $N$, put into an account on the $b$th month with a monthly payback of $m$ and a yearly interest rate of $r$ at the end of the $y$th year.
Naturally, for $V(0,b;N,m,r)$ we have $N$ since this is just the initial amount and $0$ years have passed. At the end of year one, we have $N+(N-(12-b)m)r=N+(N-12m+bm)r$. This is because the money is put in at the $b$th month and $(12-b)$ months remain in the year. Thus, we have $(12-b)$ monthly paybacks, and that is where we get $(N-(12-b)m)$. Given that $(N-(12-b)m)$ is the yearly balance in the account at the end of the year, it accrues an interest of $(N-(12-b)m)r$. Thus, the total value at the end of the year is that (the interest accrued) added to the original, so we have $N+(N-(12-b)m)r=N+(N-12m+bm)r$.
For the end of the second year, we have $N+(N-12m+bm)r+[N+(N-12m+bm)r-12m]r$. If we let $N_1=N+(N-12m+bm)r$, this makes much more sense: $N+(N-12m+bm)r+[N+(N-12m+bm)r-12m]r=N_1+[N_1-12m]r;$ that is, it's simply the amount of year one, $N_1$, plus the interest accrued after the $12$ monthly paybacks (the interest accrued being the $[N_1-12m]r$).
For the end of the third year, we have $N_1+[N_1-12m]r+(N_1+[N_1-12m]r-12m)r$. In other words, if we let $N_y=V(y,b;N,m,r)$, our recurrence becomes almost obvious: $N_x=N_{x-1}+(N_{x-1}-12m)r$. So, your recurrence is:
$V(y,b;N,m,r)=V(y-1,b;N,m,r)+\left[V(y-1,b;N,m,r)-12m\right]r,$ or, using the $N_y$ notation, $N_x=N_{x-1}+(N_{x-1}-12m)r.$
Addendum: Solving the Recursion
To solve the recursion, I first put it into a nicer form: $N_x=N_{x-1}+rN_{x-1}-12mr=(1+r)N_{x-1}-12mr.$ Now, let $C=-12mr$ and $u=1+r$. We have, then, $N_{x}=(1+r)N_{x-1}-12mr=uN_{x-1}+C.$
As we continue putting the recursion inside itself, we see a pattern:
$\begin{align} N_x&=uN_{x-1}+C\\ &=u(uN_{x-2}+C)+C\\ &=u^2N_{x-2}+uC+C\\ &=u^2(uN_{x-3}+C)+uC+C\\ &=u^3N_{x-3}+u^2C+uC+C\\ &=\dots \end{align} $
In other words, we're noticing that if we continue this until we're at $N_2$, we have (note: we're not going all the way back to $N_1$ or $N_0$ since this pattern begins at $N_2$):
$N_x=u^{x-2}N_{2}+\left(\sum_{0 \le i \le x-3}u^i\right)C.$
The sum has a closed form since it is the canonical geometric sum. Its closed form is simply $\frac{1-u^{x-2}}{1-u}$. Thus, we have: $N_x=u^{x-2}N_2+\left(\frac{1-u^{x-2}}{1-u}\right)C.$
After subbing back in and completing some simplifications, we finally arrive at
$N_x=(1+r)^{x-2}N_2+12((1+r)^{x-2}-1)m.$
Now, recalling that $N_2=N_1+(N_1-12m)r$ and $N_1=N_0+(N_0-bm)r$, we have $N_2=N_0+(N_0-bm)r+(N_0+(N_0-bm)r-12m)r$. This is more compactly stated as $N_2=N_0+\left((2+r)N_0-(b+br+12)m \right)r.$
Subbing $N_2$ in, our final statement is thusly:
$V(y,b;N,m,r)=(1+r)^{x-2}\left[N_0+\left((2+r)N_0-(b+br+12)m \right)r\right]+12((1+r)^{x-2}-1)m.$
Remark
[. . .]it does (Nā6m)r interests at the end of the first year
It would appear that your case is $b=6$. Throw that into the above formula and it simplifies a tad.