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I encountered the following problem:

Given two positive integers $n$ and $k$ with the same parity, count the number of sets, $S = \lbrace0< s_1 < s_2 < \dots < s_k = n\rbrace $ such that $s_1, s_3, \dots $ are odd numbers and $s_2,s_4,\dots$ are even numbers.

I can see the trivial cases when $n = 1$ and when $n = 2$ but have no idea how to extend it to a general $n$ and $k$. Any help is greatly appreciated. Thanks!

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    Have you tried to solve it by generating function? – 2017-01-24
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    @Crostul So sorry but I have no idea how to construct the generating function – 2017-01-24

2 Answers 2

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Instead of finding $s_1,s_2,\dots s_k$ focus on the differences between them: $a_1,a_2,\dots,a_{k-1}$ where $a_j=s_{j+1}-s_j$.

These uniquely determine each $s_j$.

We need for all of the $a_n$'s to be odd, and their sum must be at most $n-1$.

How many solutions in non-negative odd integers are there to $a_1+a_2+\dots+a_{k-1}\leq n-1$? Suppose $a_j=2c_j+1$.

We need $(2c_1+1)+(2c_2+1)+\dots + (2c_{k-1}+1)\leq n-1\iff c_1+c_2+\dots+c_{k-1}\leq \frac{n-k}{2}$. And here the $c_k$ are just non-negative integers.

The number of solutions to $c_1+c_2+\dots + c_{k-1}\leq \frac{n-k}{2}$ in non negative integers is equal to the number of solutions in non-negative integers to $c_1+c_2+\dots+c_k=\frac{n-k}{2}$.

This is given by stars and bars, we have $\frac{n-k}{2}$ stars and $k-1$ bars. So the answer is $\binom{(n+k-2)/2}{k-1}$.

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    By "and add to n-1" do you mean that the sum of all the $a_n$'s? And if so, why do they have to sum to $n-1$? – 2017-01-24
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    Yes, I mean that. Because $a_n=1+s_1+s_2+\dots+s_{k-1}$ – 2017-01-24
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    I can see that by a telescoping sum, $\sum_{i=1}^{k-1} a_i = s_k + s_1 = n + s_1$. But can we safely assume that $s_1 = 1$ ? – 2017-01-24
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    Oh, I made a mistake. Let me fix it. – 2017-01-24
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    very smart intuition to go for the delta of the terms !+1 – 2017-01-24
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    Fixed ! @Limzy,. – 2017-01-24
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    @JorgeFernándezHidalgo in the second to last paragraph, how do you make that conclusion that number of solutions to $c_1 + c_2 + \dots + c_{k-1} \le \frac{n-k}{2} $ is equals to that of $c_1 + c_2 + \dots + c_{k-1} + c_k = \frac{n-k}{2} $? Also, where does $c_k$ come from? – 2017-01-24
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    If you have a solution to the first inequality there is exactly one value of $c_k$ that can make the entire sum equal to $\frac{n-k}{2}$. – 2017-01-24
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    I see. Another question, how do we ensure that the $s_1$ chosen is odd? – 2017-01-25
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    it is going to be odd because $s_1$ is congruent to $n-k+1$ (because each $a_n$ is odd). – 2017-01-25
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We can reformulate this problem as first choosing an odd number followed by $k-1$ odd gaps between consecutive numbers (odd gaps change parity). This gives the generating function

$$\frac{z}{1-z^2} \left(\frac{z}{1-z^2}\right)^{k-1}.$$

We multiply by $1/(1-z)$ to sum the contributions ending in at most $n$ and obtain

$$\frac{1}{1-z} \frac{z^k}{(1-z^2)^k}.$$

Extracting coefficients we get

$$[z^n] \frac{1}{1-z} \frac{z^k}{(1-z^2)^k} = [z^{n-k}] \frac{1}{1-z} \frac{1}{(1-z^2)^k}.$$

This is

$$\sum_{0\le 2q\le n-k} {k-1+q\choose q}.$$

This sum can be evaluated by inspection or we may use

$${k-1+q\choose q} = [z^q] \frac{1}{(1-z)^k}$$

to get

$$[z^0] \frac{1}{(1-z)^k} \sum_{0\le 2q\le n-k} z^{-q} = [z^0] \frac{1}{(1-z)^k} \frac{1/z^{\lfloor (n-k)/2\rfloor+1}-1}{1/z-1} \\ = [z^{-1}] \frac{1}{(1-z)^{k+1}} (1/z^{\lfloor (n-k)/2\rfloor + 1}-1) \\ = [z^{\lfloor (n-k)/2\rfloor}] \frac{1}{(1-z)^{k+1}}.$$

We thus get for the end result

$$\bbox[5px,border:2px solid #00A000]{ {\lfloor (n-k)/2\rfloor +k \choose k}.}$$

Here is a simple Maple program to help clarify what interpretation of the question is being used.

with(combinat);

ENUM :=
proc(n, k)
option remember;
local choice, res, pos;
    res := 0;

    for choice in choose(n, k) do
        for pos to k do
            if pos mod 2 <> choice[pos] mod 2 then
                break;
            fi;
        od;

        if pos = k+1 then
            res := res + 1;
        fi;
    od;

    res;
end;

X1 := (n,k) ->
add(binomial(k-1+q,q), q=0..floor((n-k)/2));

X2 := (n,k) -> binomial(floor((n-k)/2)+k, k);

Remark. I somehow overlooked the fact that the largest element in the set is in fact fixed and equal to $n.$ This yields the even simpler count

$$[z^n] \frac{z^k}{(1-z^2)^k} = [z^{n-k}] \frac{1}{(1-z^2)^k}.$$

As $n$ and $k$ have the same parity this becomes

$${(n-k)/2 + k-1\choose k-1}.$$