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I have the follow two questions :

How many sequences $a_1,a_2,...,a_n$ in length $k$ so $a_i \in \{1,2,3,4....,n\}$ satisfy :

1) $a_1<a_2<....<a_k$ while $(a_{i+1} \neq a_i+1)$

2) $a_1 \leq a_2 \leq .... \leq a_k$ while $(a_{i+1} \neq a_i+1)$

I did some thinking about these questions, if I understood it correctly $a_1,...a_n$ are numbers therefore the question is to choose $k$ numbers without sequential numbers yet find it hard to solve, I thought of maybe using inclusion exclusion principle since I don't find a way to count this.

Any ideas? Thanks.

user84413
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JaVaPG
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    Relate this problem involving $0<a_1<a_2<\dots<a_k<n+1$ to the sequence $b_1=(a_1-0),b_2=(a_2-a_1),b_3=(a_3-a_2),\dots,b_k=(a_k-a_{k-1}),b_{k+1}=(n-a_k)$. What can you say about $b_1+b_2+\dots+b_{k+1}$? Are each $b_i$ bounded in a particular way? Does this match a known problem type to you? – JMoravitz May 31 '16 at 17:09
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    Not sure I understand the "while" condition tacked on these - it seems like the inequalities would take of this without further specification. Possible typo? – Joffan May 31 '16 at 17:24
  • @Joffan The while condition means that there are not sequential numbers for example 1,5,6,8 is not valid since 5,6 – JaVaPG May 31 '16 at 17:35
  • @JMoravitz I say that the following need to apply $1 \leq i \leq k+1,b_{i+1}-b{i}>1$ I also think that $b_1+...b_n=n$ – JaVaPG May 31 '16 at 17:39
  • @JaVaPG so you meant to write "while $a_i \neq a_{i-1}+1$" - ok – Joffan May 31 '16 at 17:45

1 Answers1

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From the original length $k$ sequence $a_1,\dots,a_k$, we form a bijection to the related length $k+1$ sequence $b_1,\dots,b_{k+1}$ where

$b_i = \begin{cases} a_1&\text{if}~i=1\\ a_i-a_{i-1}&\text{if}~i\in\{2,3,\dots,k\}\\ n-a_k&\text{if}~i=k+1\end{cases}$

Notice that $b_1+b_2+b_3+\dots+b_{k+1} = a_1+(a_2-a_1)+(a_3-a_2)+\dots+n-a_k = n$ as the series telescopes.

Given the conditions that $1\leq a_1\leq a_2\leq a_3\leq\dots \leq n$, this implies the conditions:

$\begin{cases}b_1+b_2+\dots+b_{k+1} = n\\ 1\leq b_1\\0\leq b_2\\ \vdots\\ 0\leq b_{k+1}\end{cases}$

Alternatively, given the conditions that $0<a_1<a_2<a_3<\dots a_k< n+1$ this implies the conditions:

$\begin{cases} b_1+b_2+\dots+b_{k+1} = n\\ 1\leq b_1\\ 1\leq b_2\\ \vdots\\ 1\leq b_k\\0\leq b_{k+1}\end{cases}$

If you add the additional condition that $a_i\neq a_{i-1}+1$ for any $i$ (note: your typo $a_i\neq a_{i+1}+1$ is always true since $a_i\leq a_{i+1}<a_{i+1}+1$ in both scenarios), then this is equivalent to adding the restriction that $b_i\neq 1$ for any $i\in\{2,3,\dots,k\}$

In the case with strict inequalities, this is easily managed by making the change of variable $c_1=b_1-1$, $c_i = b_i-2$ for each $i\in\{2,3,\dots,k\}$. Approach as usual with stars and bars.

In the case with $a_1\leq a_2\leq\dots$, then you can approach as in my answer for Number of ways to write $n$ as sum of $k$ non-negative integers without 1, iterating over how many and which of the $b_i$ are zero.

As an aside and to make for a more complete answer, the question of finding how many integer solutions there are to the system:

$\begin{cases} x_1+x_2+\dots+x_k=n\\ 0\leq x_i~\text{for each}~i\end{cases}$

can be found via stars-and-bars and has answer $\binom{n+k-1}{k-1}=\binom{n+k-1}{n} = \left(\!\!\binom{n}{k}\!\!\right)$

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JMoravitz
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