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I have the following partitioning condition, $$\text{p(n|parts in N)}=\text{p(n|distinct parts in M)}, \text{for n}\ge1$$

Where N is any set of integers such that no element of N is a power of two times an element of N, and M is the set containing all elements of N together with all their multiples of powers of two.

Suppose I construct a set covered by the theorem $N=(1,3,6)$, for $$ \begin{align} n &= 1 \space 2 \space 3 \space 4 \space 5 \space 6 \\ p(n|\text{parts in (1,3,6)}) &= 1 \space 1 \space 2 \space 2 \space 2 \space 4\end{align}$$

I need your help in understanding how these parts were derived, my initial logic:

$$\begin{align}1 &\to 1 \\ 2 &\to 1+1 \\ 3 &\to 3, 1+1+1 \\ 4 &\to 3+1, 1+1+1+1 \\ 5 &\to 3+1+1, 1+1+1+1+1 \\ 6 &\to 6, 3+3, 3+1+1+1, 1+1+1+1+1+1 \end{align}$$ So essentially, all parts that partition the integer into either 1, 3 or 6. Would this be the correct approach?

Meton
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    This is hard to follow. What is $N$? What is $M$? what is $p(a,|,b)$? If you are quoting from some reference, please link to that reference. I would expect that context resolved all the definitions. – lulu Mar 29 '24 at 12:56
  • If you just want a guess though, then yes. I would guess that $N$ is a set of natural numbers and that you are counting the number of unordered partitions of $n$ using only elements in $N$, and allowing repetition. as you have remarked, that seems consistent with the explicit values you mention. – lulu Mar 29 '24 at 13:00
  • @lulu I have updated – Meton Mar 29 '24 at 13:06
  • Your set $N={1,3,6}$ does not meet the conditions you set, since $6=2\times 3$. – lulu Mar 29 '24 at 13:13
  • Anyway, with the definition the claim seems clear. If you have a partition of the type on the left, then you can group all the $n's$ for $n\in N$ and use the binary expansion to write it uniquely using elements from $M$. – lulu Mar 29 '24 at 13:16
  • @lulu I know that, I was using an example from a book which shows the statement holds true, by a counterexample. I wanted to know whether i had it right in terms of the partitions, so 6 has 4 partitions, and its as given by my examples above, when a partition includes either 1, 3 or 6 – Meton Mar 29 '24 at 13:25
  • Not following, but maybe that doesn't matter. If you are just asking about the definition of the left hand for any given set $N$ then I agree with what you wrote. If you are asking something else, you need to clarify your post. – lulu Mar 29 '24 at 13:27
  • @lulu ok sweet! thanks for confirmation, I am still trying to wrap my head around the counting – Meton Mar 29 '24 at 13:46

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