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Following on from this question:

How many distinct ways can a sequence of $n$ $1$s be partitioned into triples or singles, in which $\{1,1,1\}=\{3\}$ is considered a triple and $\{1\}$ is considered a single?

For example $\{1,1,1,1,1\}$ can be partitioned into:

$\{3,1,1\}$

$\{1,3,1\}$

$\{1,3,1\}$

But no result containing $\{1,1,1\}$ should be enumerated since this should be counted as a triple.

So for $n=5$, the answer is 3 ways.

I think the answer to this question equals the number of Dyck words which give unique results when exponentiating powers of $2$... as asked in this question, since groups of 3 powers of 2 always give the same result; 16 irrespective of order, but the orders of carrying out individual exponents around them make distinct results.

Number of solutions for small $N$: $$\begin{pmatrix}n & N\\1&1\\2&1\\3&1\\4&2\\5&3\\6&3\\7&4\end{pmatrix}$$

Parcly Taxel
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Robert Frost
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1 Answers1

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Let's see… using Millikan's approach on the other question, a partition of $n$ is one of the following:

  • partition of $n-3$ + 3
  • partition of $n-4$ + 31
  • partition of $n-5$ + 311

Therefore $f(n)=f(n-3)+f(n-4)+f(n-5)$ with $f(1,2,3,4,5)=1,1,1,2,3$. This is OEIS A017818.

Parcly Taxel
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