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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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    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)=\sum_{j=0}^{n-1}p(\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)\,$2011-08-05

1 Answers 1

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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.