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Prove that the number of ways to divide $n$ into summands so that no number included in the sum more than $r − 1$ times, equal to the number of ways to divide $n$ into parts that are not divisible by $r$.

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    $\begingroup$ Here is a solution when $r=2$. There are two methods there, bijective and generating function. Both methods generalize to all $r$. $\endgroup$ – Mike Earnest May 8 at 19:10
  • $\begingroup$ Thank you a lot! $\endgroup$ – Arial Pilisov May 8 at 19:11
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The generating function for partitions where there are at most $r-1$ parts of each size is $$ (1+x+x^2+\dots+x^{r-1})(1+x^2+x^4+\dots+x^{2(r-1)})(1+x^3+\dots+x^{3(r-1)})\cdots $$ When expanding this out, the choice of the summand $x^{kj}$ in the factor $1+x^k+\dots+x^{k(r-1)}$ corresponds to having $j$ parts of size $k$ in the partition. We can write this as $$ \frac{1-x^r}{1-x}\cdot \frac{1-x^{2r}}{1-x^2}\cdot \frac{1-x^{3r}}{1-x^3}\cdot \cdots $$ Al factors in the numerator cancel with something in the denominator. What remains is the product of $(1-x^k)^{-1}$ over all $k$ which are not multiples of $r$. This is precisely the generating function for partitions where no parts have a size which is a multiple of $k$; the choice of summand in the factor $(1-x^k)^{-1}=(1+x^k+x^{2k}+\dots)$ determines the number of parts of size $k$.

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  • $\begingroup$ I got it. Thank you! $\endgroup$ – Arial Pilisov May 8 at 19:40

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