Partition number problem

Denote by $I_m=\{0,1,2,…m\}$, by $N_s=\{1,2,…,s\}$ , by $\overline s$ least common multiple of elements of set $N_s$ and by $p(k,N_s)$ the number of partitions of natural number $k$ in parts used from set $N_s$

I can prove that

$$p(\overline {s}n+l,N_s)=p(l,N_s)+\sum_{i\in I_{\frac{\overline s}{s}}}\sum_{j\in I_n} p(\overline {s}n+l-\overline {s}j-si,N_{s-1})$$

For $s>0$ define a function

$$A_{n}^{s}(a,r)=\sum_{j\in I_n} p(\overline {s}\left(\frac{\overline {s+1}}{\overline s}n-\frac{\overline {s+1}}{\overline s} j+a\right)+r,N_s),a\in Z,r\in I_{\overline s}$$

$$A_{n}^{s}(a,r)=0,a\notin Z$$

Still there I am all right, but problem for me is how under these conditions get the equation

$$p(\overline {s}n+l,N_s)- p(l,N_s)= \sum_{r\in I_{\overline {s-1}}}\sum_{i\in I_{\frac{\overline s}{s}}}A_{n}^{s-1}\left(\frac{l-si-r}{\overline {s-1}},r\right)$$

I am sure that this formula is correct but some details of proof remain unclear for me.

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The summand in the last equation doesn't involve $j$, yet there is a summation involving the index $j$; the summand also involves both of the indices $i$ and $r$, but there is no summation involving the index $i$; lastly, the variable $l$ doesn't make it from the left side to the right side. You probably transcribed the equation incorrectly. Fix your formulas and then replace $A()$ with its definition (assuming you have that correct), and then work backwards from there. – anon Jul 31 '11 at 11:18
in last formula I correct index sumation instead j must be i – Adi Dani Jul 31 '11 at 14:57
This is your 7th question. You haven't accepted any answers. Why do you ask another question here, when you haven't liked any of the answers to your previous questions? – Gerry Myerson Aug 1 '11 at 2:01
Thank you Gerry – Adi Dani Aug 1 '11 at 15:39
@Gerry If I find the answer of my own problem can I post it here. – Adi Dani Aug 3 '11 at 7:14

Answer to this question is very simple. Take in account observation that for each $i\in I_{\frac{\overline s}{s}}$ exists $r\in I_{\overline{s-1}}$ such that
$$\sum_{j=0}^{n-1}p(\overline {s}n-\overline{s}j+l-si,N_{s-1})=$$
$$=\sum_{j\in I_n} p(\overline {s-1}\left(\frac{\overline {s}}{\overline {s-1}}n-\frac{\overline {s}}{\overline {s-1}} j+\frac{l-si-r}{\overline{s-1}}\right)+r,N_{s-1})=$$
$$=A_{n}^{s-1}\left(\frac{l-si-r}{\overline{s-1}},r\right)$$