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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
1  
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
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You'll see next to each answer a little "check mark". If an answer satisfies you, you click in that check mark. This signifies that you have accepted that answer. Your "accept rate" is the number of questions where you have accepted an answer, divided by the number of questions you have asked. Probably all of this is explained in the faq. –  Gerry Myerson Aug 1 '11 at 12:41
    
Thank you Gerry –  Adi Dani Aug 1 '11 at 15:39

1 Answer 1

up vote 2 down vote accepted

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

$$\sum_{j=0}^{n-1}p(\overline {s}n-\overline{s}j+l-si,N_{s-1})=$$

$$=\sum_{j\in I_n} p(\overline {s-1}\left(\frac{\overline {s}}{\overline {s-1}}n-\frac{\overline {s}}{\overline {s-1}} j+\frac{l-si-r}{\overline{s-1}}\right)+r,N_{s-1})=$$

$$=A_{n}^{s-1}\left(\frac{l-si-r}{\overline{s-1}},r\right)$$

I miss this observation.

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