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here I asked how to get the coefficients of the expression $$\prod_{p=1}^n (x^p+1)^p$$ and I got that $$ a(n,k)=\sum_{\substack{\sum_{j=1}^n j g_j=k \\ g_j\in\{0,1,..,j\}}}\prod_{j=1}^n \binom{j}{g_j}$$ and then I found formula for A026007

$$ a_n=\sum_{\substack{\sum_{j=1}^n j g_j=n \\ g_j\in\{0,1,..,j\}}}\prod_{j=1}^n \binom{j}{g_j}$$ So my question is there a way to find all $(g_1,g_2,...,g_n)$ such that $$ \sum_{j=1}^n j g_j=n \space \space , \space g_j\in\{0,1,..,j\}$$

for example for $n=1$ we get easily that $g_1=1$

and for $n=2$

$$ g_1+2g_2=2 , g_1 \in \{0,1 \} , g_2 \in \{0,1,2 \}$$

So $(g_1,g_2)=(0,1)$

and for $n=3$

$$ g_1+2g_2+3g_3=3 , g_1\in\{0,1\} , g_2\in\{0,1,2\} ,g_3\in \{0,1,2,3\} $$

So $$ (g_1,g_2,g_3)=\{ (0,0,1) , (1,1,0)\} $$

Faoler
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  • Specify better, please,. Your last equation has as last term $ng_n$ and this plus something should be equal to $n$. ¿How to act if the numbers should be positive as it seems to be the case? – Ataulfo Aug 17 '24 at 19:17
  • The same constraint may be expressible via multiple sums or an $n$th derivative like here – Тyma Gaidash Aug 17 '24 at 19:29
  • @Piquito the numbers can be zero also not only positive So for $g_1+2g_2+3g_3=0$ we can see that $(1,1,0) , (0,0,1)$ is a solution – Faoler Aug 17 '24 at 19:30
  • @Faoler: I mean no reference to $n$ but for $j$ and $g_j$, for example for $n=50$ we must have $P+50g_{50}=50$ where $P\ge0$ . What $P$ is in this example? – Ataulfo Aug 17 '24 at 21:13
  • @Piquito here we can put $P=0 , g_{50}=1$ or $g_{50}=0 , P=50$ that means $g_{50}\in{0,1} $ and also don't forget that $P=\sum_{j=1}^{49} j g_j $ – Faoler Aug 17 '24 at 22:15
  • @Faoler: Yes but you do have this FOR ALL $n$ so what....... I see now your reply that then we have $P$ another way. But again this would be valid for all $n$. – Ataulfo Aug 18 '24 at 02:03

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