Simplify $\sum_{k=0}$ $n−2k \choose k$

Question

Simplify $$\sum_{k=0} \binom{n−2k} k$$ Or alternatively, show that it can't be simplified any more.

I tried Hockey stick identity but I had issues with the fact that the top had a -2k term. I then tried using $\binom {n+1}{k+1}$ = $\binom{n}{k} + \binom{n}{k+1}$, but again the -2k term means you can't add consecutive terms. Some values I calculated are

3 for n=4

4 for n=5

6 for n =6

9 for n=7.

There doesn't seem to be a pattern.

Answer 1

We have that $\binom{n-2k}{k}=\binom{n-2k-1}{k-1}+\binom{n-2k-1}{k}$. Hence, $$\sum_{k=0} \binom{n-2k}{k}=\sum_{k=0}\binom{(n-1)-2k}{k-1}+\sum_{k=0} \binom{(n-1)-2k}{k}$$ If $a_n=\sum_{k=0} \binom{n-2k}{k}$, then we have $$a_n=a_{n-1}+\sum_{k=0} \binom{(n-1)-2k}{k-1}$$ $$a_n=a_{n-1}+\sum_{k=0} \binom{(n-1)-2(k-1)-2}{k-1}$$ $$a_n=a_{n-1}+\sum_{k=0} \binom{(n-3)-2(k-1)}{k-1}$$ Shifting the indices of this summation gives us $$a_n=a_{n-1}+\sum_{k=0} \binom{(n-3)-2k}{k}$$ $$a_n=a_{n-1}+a_{n-3}$$ The characteristic equation of this linear recurrence is $$x^3-x^2-1=0$$ Wolfram Alpha says that the roots of this equation are very complex. One could possibly solve for the explicit equation for the recurrence using the roots $r_1,r_2,r_3$. However, it would be tedious to find the values of constants $u,v,w$ that satisfy the initial conditions which yield $$a_n=ur_1^n+vr_2^n+wr_3^n$$

Answer 2

This is only an alternative to find the recurrence relation above. The expression $\binom{n-k}{k}$ gives the number of solutions to the equation $n=\sum{x_{i}}\phantom{x}|\phantom{x}x_{i}\in\{1,3\}$ when there are $k$ threes. Therefore the following expression gives the number of possibilities to express $n$ as summation of ones and threes.

$$ a_{n}=\sum{\binom{n-2k}{k}} $$

There are only two possibilities for $x_{1}$. If $x_{1}=1$ then the other $x_{i}$ are ones and threes that sum up to $n-1$. If $x_{1}=3$ then the other $x_{i}$ are ones and threes that sum up to $n-3$. Hence the following recurrence relation:

$$ a_{n}=a_{n-1}+a_{n-3} $$

Answer 3

Maybe expressing this sum using the Generalized Hypergeometric function may form a starting point for further simplifications:

$\binom{0}{\frac{n}{2}} \, _4F_3\left(\frac{1}{2},1,1,-\frac{n}{2};\frac{1}{3}-\frac{n}{6},\frac{2}{3}-\frac{n}{6},1-\frac{n}{6};-\frac{4}{27}\right)$

Hopefully there exist some identities or theorems (such as those from Saalschütz, Dixon or Dougall for example) that enable further simplifications.

Answer 4

Recursive Solution

Considering an old answer to a related problem, I tried $$ \begin{align} a_n &=\sum_k\binom{n-2k}{k}[3k\le n]\tag{1a}\\[3pt] &=\sum_k\binom{n-2k-1}{k}[3k\le n]+\sum_k\binom{n-2k-1}{k-1}[3k\le n]\tag{1b}\\[3pt] &=\sum_k\binom{n-2k-1}{k}[3k\le n]+\sum_k\binom{n-2k-3}{k}[3k+3\le n]\tag{1c}\\ &=a_{n-1}+\color{#C00}{\sum_k\binom{n-2k-1}{k}\overbrace{([3k\le n]-[3k\le n-1])}^{[n=3k]}}\tag{1d}\\ &\phantom{={}}+a_{n-3}+\color{#090}{\sum_k\binom{n-2k-3}{k}\overbrace{([3k+3\le n]-[3k\le n-3])}^0}\tag{1e}\\[12pt] &=a_{n-1}+a_{n-3}+\color{#C00}{[n=0]}+\color{#090}{0}\tag{1f}\\[21pt] &=a_{n-1}+a_{n-3}+[n=0]\tag{1g} \end{align} $$ Explanation:
$\text{(1b)}$: apply Pascal's Rule
$\text{(1c)}$: substitute $k\mapsto k+1$ in the second sum
$\text{(1d)}$: add and subtract $a_{n-1}$ from the left term of $\text{(1c)}$
$\text{(1e)}$: add and subtract $a_{n-3}$ from the right term of $\text{(1c)}$
$\text{(1f)}$: $\binom{k-1}{k}=[k=0]$
$\text{(1g)}$: simplify

$\text{(1a)}$ says that $a_n=0$ for $n\lt0$. $\text{(1g)}$ says that $a_0=1$, and for $n\gt0$, $$ a_n=a_{n-1}+a_{n-3}\tag2 $$ Thus, $$ \begin{array}{c|c}n&0&1&2&3&4&5&6&7&8\\\hline a_n&1&1&1&2&3&4&6&9&13\end{array} $$ Care needs to be taken with the upper limits of the sum, as shown in $\text{(1d)}$ and $\text{(1e)}$, but luckily, the terms cancel for $n\gt0$.

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Solution to the Linear Recurrence Equation

A closed form solution to the constant coefficient linear recurrence equation takes the form $$\newcommand{\csch}{\operatorname{csch}} a_n=c_1r_1^n+c_2r_2^n+c_3r_3^n\tag3 $$ where $r_1,r_2,r_3$ are roots of $x^3-x^2-1=0$, which, if we substitute $x=t+\frac13$, becomes $t^3-\frac13t=\frac{29}{27}$. Multiplying the identity $4\cosh^3(u)-3\cosh(u)=\cosh(3u)$ by $\frac2{27}$ gives $$ \left(\frac23\cosh(u)\right)^3-\frac13\left(\frac23\cosh(u)\right)=\frac2{27}\cosh(3u)\tag4 $$ Applying $(4)$, we get the real root $$ r_1=\frac13+\frac23\cosh\left(\frac13\cosh^{-1}\left(\frac{29}2\right)\right)\tag5 $$ and then, because $r_1+r_2+r_3=r_1r_2r_3=1$, we can compute the complex roots $$ r_2=\frac{1-r_1}2+i\frac{\sqrt{3r_1^2-2r_1-1}}2\tag{6a} $$ and $$ r_3=\frac{1-r_1}2-i\frac{\sqrt{3r_1^2-2r_1-1}}2\tag{6b} $$ and we can compute the coefficients using Cramer's Rule and Vandermonde Determinants $$ c_1=\frac{\det\begin{bmatrix} 1&1&1\\ 1&r_2&r_3\\ 1&r_2^2&r_3^2 \end{bmatrix}} {\det\begin{bmatrix} 1&1&1\\ r_1&r_2&r_3\\ r_1^2&r_2^2&r_3^2 \end{bmatrix}} =\frac{r_1^2+1}{r_1^2+3}\tag{7a} $$ and similarly $$ c_2=\frac{r_2^2+1}{r_2^2+3}\tag{7b} $$ and $$ c_3=\frac{r_3^2+1}{r_3^2+3}\tag{7c} $$ Now we simply need to plug $(5)$, $(6)$, and $(7)$ into $(3)$ to get a closed form for the solution.

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A Simple Closed Form

Note that $|c_2|=|c_3|\approx0.22968031\lt\frac14$. Since $r_1\approx1.46557123\gt1$, and $r_1r_2r_3=1$, we have that $|r_2|=|r_3|\lt1$. Thus, since $\left|c_2r_2^n+c_3r_3^n\right|\lt\frac12$, according to $(3)$, $a_n$ is the closest integer to $c_1r_1^n$: $$ \begin{array}{r|r|l} n&a_n&c_1r_1^n\\\hline 0&1&0.61149199\\ 1&1&0.89618507\\ 2&1&1.31342306\\ 3&2&1.92491505\\ 4&3&2.82110012\\ 5&4&4.13452318\\ 6&6&6.05943824\\ 7&9&8.88053836\\ 8&13&13.0150615\\ \end{array} $$ Therefore, a simple closed form would be $$ a_n=\left\lfloor c_1r_1^n+\tfrac12\right\rfloor\tag8 $$ where, computed from $(5)$, $$ r_1\approx1.4655712318767680267\tag{9a} $$ and, computed from $\text{(7a)}$, $$ c_1\approx0.61149199195081251841\tag{9b} $$