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What is a closed-form or recursive expression for the number of compositions of an integer $N$ that contain a specific integer $X$? Assume $0at least one instance of $X$, not the number of instances of $X$ in all compositions of $N$.

An equivalent way of thinking about this is to consider the number of binary sequences of length $N-1$ that contain a run of length $X-1$.

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    What are the compositions of an integer? – 2017-01-06
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    https://oeis.org/wiki/Integer_compositions – 2017-01-06
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    Is it not just the number of compositions of $N - X$? – 2017-01-06
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    @pjs36 If I have understood correctly (not guaranteed), a composition is an ordered partition. Your claim would be correct for general partitions, but not for ordered ones. – 2017-01-06
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    The compositions of an integer are just the ways it can be written as a sum of other integers less than or equal to itself. Every integer $N$ has $2^{N-1}$ compositions. For example, $1+1+2$, $1+2+1$, $3+1$ and $4$ are some of the compositions of $4$. – 2017-01-06
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    Have you looked at [this book](https://www.crcpress.com/Combinatorics-of-Compositions-and-Words/Heubach-Mansour/p/book/9781420072679)? I haven't, but it looks promising. – 2017-01-06

2 Answers 2

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It will be easier to compute the number of compositions of $n$ that don't contain a specific integer $k$. Then since there are $2^{n-1}$ compositions of $n$, so subtract.

To do this, we'll use generating functions. The possible parts of a composition which avoids $k$ have generating function $$ F(z) = (z + z^2 + z^3 + z^4 + \cdots) - z^k = {z \over 1-z} - z^k = {z - z^k + z^{k+1} \over 1-z}$$ and so the generating function which counts $k$-avoiding compositions is $$ G(z) = {1 \over 1 - F(z)} = {1 \over 1 - {z - z^k + z^{k+1} \over 1-z}} = {1-z \over 1-2z+z^k - z^{k+1}}. $$

Let $g(n)$ denote the number of $k$-free compositions of $n$. Then from standard results in the theory of generating functions, you have the recurrence

$$ g(n) = 2 g(n-1) - g(n-k) + g(n-k-1) $$

which you can basically read off the denominator of $G(z)$. The initial conditions are:

  • $g(\ell) = 2^{\ell-1}$ for $l = 1, 2, \cdots, k-1$ (all compositions of numbers up to $k-1$ are $k$-avoiding)
  • $g(k) = 2^{k-1} - 1$ (all compositions of $k$ are $k$-avoiding except for the one consisting of $k$ alone)
  • $g(k+1) = 2^k - 2$ (all compositions of $k+1$ are $k$-avoiding except for the two with parts $k$ and $1$ in either order).

In the particular case $k = 1$, note this reduces to $g(n) = g(n-1) + g(n-2)$, which is the recurrence for the Fibonacci numbers. So the Fibonacci numbers count the $1$-avoiding compositions.

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    I knew something seemed wrong. Editing accordingly. – 2017-01-06
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    Apologies for the delay...I like this approach a lot and have been trying to understand the theory behind generating functions so I can prove it to myself, although I can see that it works...it's very elegant. Thank you – 2017-01-10
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    Depending on your level of existing knowledge, some good books on this are "generatingfunctionology" by Wilf (less advanced) and "Analytic Combinatorics" by Flajolet and Sedgewick (more advanced). Both are available online, legally, for free. – 2017-01-10
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Michael Lugo’s recurrence is probably more useful, but we can also get the desired number in the form of a summation. We want the number of compositions of $n$ that contain at least one part $m$; we can get this with an inclusion-exclusion argument.

Fix $p$, and consider compositions of $n$ with $p$ parts. For $k\in[p]$ let $C_k$ be the set of compositions of $n$ whose $k$-th part is $m$. If $\varnothing\ne I\subseteq[p]$, then

$$\left|\,\bigcap_{k\in I}C_k\,\right|=\binom{n-m|I|-1}{p-|I|-1}\;,$$

the number of compositions of $n-m|I|$ into $p-|I|$ parts, provided that neither the upper nor the lower number in the binomial coefficient is negative. For convenience I’ll adopt the non-standard convention that $\binom{n}k=0$ not only if $k<0$, but also if $n<0$.

Thus, the number of compositions of $n$ with $p$ parts, at least one of which is $m$, is

$$\begin{align*} \left|\,\bigcup_{k\in[p]}C_k\,\right|&=\sum_{\varnothing\ne I\subseteq[p]}(-1)^{|I|+1}\binom{n-m|I|-1}{p-|I|-1}\\ &=\sum_{k=1}^p(-1)^{k+1}\binom{p}k\binom{n-km-1}{p-k-1}\;. \end{align*}$$

Summing over $p$ then gives the desired number,

$$\sum_p\sum_{k=1}^p(-1)^{k+1}\binom{p}k\binom{n-km-1}{p-k-1}\;.\tag{1}$$

For example, take $n=5$ and $m=2$. There are $9$ compositions of $5$ containing at least one $2$:

$$\begin{align*} &2+2+1\\ &2+1+2\\ &1+2+2\\ &2+3\\ &3+2\\ &2+1+1+1\\ &1+2+1+1\\ &1+1+2+1\\ &1+1+1+2 \end{align*}$$

And formula $(1)$ yields

$$\begin{align*} \sum_{p\ge 1}\sum_{k=1}^p\binom{p}k\binom{4-2k}{p-k-1}&=\binom11\binom2{-1}+\left(\binom21\binom20-\binom22\binom0{-1}\right)+\\ &\qquad+\left(\binom31\binom21-\binom32\binom00+\binom33\binom{-2}{-1}\right)+\\ &\qquad+\left(\binom41\binom22-\binom42\binom01+\binom43\binom{-2}0-\binom44\binom{-4}{-1}\right)\\ &=0+(2-0)+(6-3+0)+(4-0+0-0)\\ &=9\;, \end{align*}$$

as it should.