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?