Is it true that $\sum_{k=0}^m\binom{n-k}k$ outputs the $(n+1)$th Fibonacci number, where $m=\frac{n-1}2$ for odd $n$ and $m=\frac n2$ for even $n$?

Question

Does $$\sum_{k=0}^m\binom{n-k}k=F_{n+1}$$ where $m=\left\{\begin{matrix} \frac{n-1}{2}, \text{for odd} \,n\\ \frac n2, \text{for even} \,n \end{matrix}\right.$ hold for all positive integers $n$?

Attempt: I have not yet found a counterexample, so I will attempt to prove it. $$\text{LHS} =\binom n0 + \binom{n-1}1+\binom{n-2}2+...+\left\{\begin{matrix} \binom{1+(n-1)/2}{(n-1)/2}, \text{for odd} \,n\\ \binom{n/2}{n/2}, \text{for even} \,n \end{matrix}\right.$$ Now using the identity that $\binom nk + \binom n{k+1}=\binom {n+1}{k+1}$, where $k$ is a positive integer, I find that $$\binom{n-1}1=\binom n1 - 1, \\ \binom {n-2}2=\binom n2-2\binom n1+3,\\ \binom {n-3}{3} =\binom n3 - 3\binom n2 + 6 \binom n1 - 10, \\ ...$$ This pattern suggests that the coefficients of $\binom{n-4}4$ will be square numbers, those of $\binom{n-5}5$ will be pentagonal numbers, etc. However, I cannot see a way to link these results to any Fibonacci identity.

Edit: @Jack D'Aurizio♢ has provided a very succinct proof to this, but is there a more algebraic method to show the equality?

Possible duplicate of Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ — Guy Fsone, Dec 01 '17 at 18:58
@GuyFsone: the point here is not to prove the hockey stick (or stars and bars) identity, but to understand how it is related to Fibonacci numbers. — Jack D'Aurizio, Dec 01 '17 at 19:15
See The Fibonacci numbers occur in the sums of "shallow" diagonals in Pascal's triangle. — Felix Marin, Dec 01 '17 at 23:50
Does this answer your question? Calculating $\sum_{0\le k\le n/2} \binom{n-k}{k}$ — Anne Bauval, Oct 11 '23 at 14:58

score 6 · Accepted Answer · answered Dec 01 '17 at 19:14

6

There is a simple combinatorial interpretation. Let $S_n$ be the set of strings over the alphabet $\Sigma=\{0,1\}$ with length $n$ and no occurrence of the substring $11$. Let $L_n=|S_n|$. We clearly have $L_1=2$ and $L_2=3$, and $L_n=F_{n+2}$ is straightforward to prove by induction, since every element of $S_n$, for $n\geq 3$, is either $0\text{(element of }S_{n-1})$ or $10\text{(element of }S_{n-2})$, so $L_{n+2}=L_{n+1}+L_n$.

On the other hand, we may consider the elements of $S_n$ with exactly $k$ characters $1$.
There are as many elements with such structure as ways of writing $n+2-k$ as the sum of $k+1$ positive natural numbers. Here it is an example for $n=8$ and $k=3$:

$$ 00101001\mapsto \color{grey}{0}00101001\color{grey}{0}\mapsto \color{red}{000}1\color{red}{0}1\color{red}{00}1\color{red}{0}\mapsto3+1+2+1.$$ By stars and bars it follows that: $$ L_n = F_{n+2} = \sum_{k=0}^{n}[x^{n+2-k}]\left(\frac{x}{1-x}\right)^{k+1}=\sum_{k=0}^{n}\binom{n+1-k}{k} $$ and by reindexing we get $F_{n+1}=\sum_{k=0}^{n}\binom{n-k}{k}$ as wanted.

answered Dec 01 '17 at 19:14

Jack D'Aurizio

353,855

I don't really get the interpretation of $S_n$. Say we have the string $0101$. What does that mean? – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Dec 01 '17 at 19:34
@TheSimpliFire: $0101$ it is an element of $S_4$, associated with $k=2$ and $2+1+1$. – Jack D'Aurizio Dec 01 '17 at 19:40
$S_4$ has $8$ elements, namely $$0000,0001,0010,0100,0101,1000,1001,1010$$ associated with $$ 6, 4+1,3+2,2+3,2+1+1,1+4,1+2+1,1+1+2.$$ – Jack D'Aurizio Dec 01 '17 at 19:45
Thank you. But why is $0010$ associated with $3+2$? – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Dec 01 '17 at 19:52
@TheSimpliFire: add an initial and a final zero to get $000100\mapsto 3+2$. – Jack D'Aurizio Dec 01 '17 at 19:58

score 3 · Answer 2 · answered Dec 02 '17 at 13:24

Here is more of an algebraic solution through generating functions. We have

\begin{align} \sum_{n=0}^{\infty}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k}{k}x^n &=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\binom{n}{k}x^{n+k+1}\\ &=\sum_{k=0}^{\infty}x^{2k+1}\sum_{n=k}^{\infty}\binom{n}{k}x^{n-k}\\ &=\sum_{k=0}^{\infty}x^{2k+1}\sum_{n=0}^{\infty}\binom{n+k}{k}x^{n}\\ &=\sum_{k=0}^{\infty}x^{2k+1}\frac{1}{(1-x)^{k+1}}\\ &=\frac{x}{1-x}\sum_{k=0}^{\infty}\left(\frac{x^2}{1-x}\right)^k\\ &=\frac{x}{1-x}\frac{1}{1-\frac{x^2}{1-x}}\\ &=\frac{x}{1-x-x^2} \end{align} where on the last line we arrived at the well known generating function for fibonacci numbers. So by equality of coefficients it follows $$F_{n+1} = \sum_{k=0}^{\lfloor (n)/2\rfloor}\binom{n-k}{k}.$$

clever use of summations and series! – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Dec 02 '17 at 14:05 — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Dec 02 '17 at 14:05

score 1 · Answer 3 · answered Oct 11 '23 at 11:25

A while back, I have had done this proof combinatorically while solving a different probability question on Quora.

Suppose, you toss a fair coin $n$ times. What is the probability that no two consecutive heads appear?

$\underline{\textbf{Notations}}\tag*{}$ For the sake of brevity, I'll denote:

The set of favourable outcomes when a coin is tossed $n$ times (i.e., all the possible arrangements of length $n$ of the letters $H$ and $T$ such that the subsequence $HH$ doesn't occur in them) by $\zeta_n$.
Cardinality of a set by enclosing it in a pair of pipes $|\cdot|$.
Rounding to the nearest integer by a pair of square brackets $[\cdot]$.
Integer part i.e., floor function by $\lfloor\cdot\rfloor$.
Golden ratio $(1+\sqrt 5)/2$ by $\phi$.
A sequence $\{x_0, \ x_1, \ x_2, \ldots\}$ by $\{x_n\}$.
Those elements of $\zeta_n$ which
- end in $H$ by $H_n$.
- end in $T$ by $T_n$.

$\underline{\textbf{Combinatorial Perspective}}\tag*{}$

$\zeta_0=\{\emptyset\}$ because if the coin is not tossed, we still have a valid null arrangement. $|\zeta_0|=1\tag{1}$
$\zeta_1=\{H, T\}$ and hence, $|\zeta_1|=2\tag{2}$
$T_n$ and $H_n$ are disjoint and $\zeta_n=T_n\cup H_n$ which implies that: $|\zeta_n|=|T_n|+|H_n|\tag{3}$
$T_n$ is generated by appending a $T$ to the elements of $\zeta_{n-1}$. Thus: $|T_n|=|\zeta_{n-1}|\tag{4}$
$H_n$ is generated by appending an $H$ to the elements of $T_{n-1}$. Thus: $|H_n|=|T_{n-1}|=|\zeta_{n-2}|\tag{5}$ $\because$ $T_{n-1}$ is in turn generated by appending a $T$ to the elements of $\zeta_{n-2}$.
From $(3)$, $(4)$ and $(5)$, $|\zeta_n|=|\zeta_{n-1}|+|\zeta_{n-2}|\tag{6}$ From $(1)$, $(2)$ and $(6)$, we have a recurrence relation, $\displaystyle |\zeta_n|=\begin{cases} 1 & \text{if $n=0$}\\ 2 &\text{if $n=1$}\\ |\zeta_{n-1}|+|\zeta_{n-2}| & \text{if $n>1$}\end{cases}\tag*{}$ This resembles the well-known Fibonacci sequence $\{F_n\}=\{0,1,1,2,3,\ldots\}$ but it is leading by two terms. $\boxed{|\zeta_n|=F_{n+2}}\tag*{}$

$\underline{\textbf{Alternate Perspective}}\tag*{}$ Let's generate $T_n$.

Given any element of $T_n$, every $H$ necessarily occurs in an $HT$ pair. So we can generate $T_n$ by permuting the subsequences $HT$ and $T$.
Say, there are $k$ number of $HT$ pairs and $(n-2k)$ number of free $T$s. (which indeed add up to length $n$ because every $HT$ pair is of length $2$.)
There are $\displaystyle \frac{(k+n-2k)!}{k!\cdot(n-2k)!}=\binom{n-k}{k}\tag*{}$ unique permutations of them.
$k$ i.e, the number of $HT$ pairs can be varied from $0$ to $\lfloor n/2\rfloor$.
Thus we have: $\displaystyle |T_n|=\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n-k}{k}\tag*{}$
From $(04)$, we have that $|T_n|=|\zeta_{n-1}|$. $\displaystyle\therefore |\zeta_{n-1}|=\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n-k}{k}\tag*{}$ This is indeed a valid expression— just as valid as the previously discussed one according to which $|\zeta_n|=F_{n+2}$ i.e., on replacing $n$ by $n-1$, $|\zeta_{n-1}|=F_{n+1}$. Though this alternate perspective is of little use to the main problem, we just ended up revealing an interesting relation between the Fibonacci numbers and the shallow diagonals of Pascal's triangle: $\displaystyle \boxed{F_{n+1}=\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n-k}{k}}\tag*{}$

score 0 · Answer 4 · answered Feb 06 '24 at 08:56

This brings back some memories.

A while ago, my teacher had given the following question as an assignment:

Let $a(n)$ denote the representation of $n$ as a sum of ones and twos. Find a closed form for $n$. (1+1+2 and 1+2+1) are different.

My teacher solved it using recurrence relation to show that $a(n) = F_{n+1}$, as $a(1) = 1, a(2) = 2$ and for $n >2$, $a(n) = a(n-1)+a(n-2)$ which can be seen when we go from $n-1$ to $n$ by concatenating a $1$, or go from $n-2$ to $n$ by concatenating a $2$.

I, however, completely missed the recurrence and finished with a sum which I didn't know how to simplify:
Let $k$ denote the number of twos in the sum. $k$ goes from $0$ till $\lfloor n/2 \rfloor$. Then we have $n-2k$ ones. We can assign the $k$ twos in $k+(n-2k)$ places in $\binom{n-k}{k}$ ways. This gives $$a(n) = \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n-k}{k}$$Thus, $$\boxed{\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n-k}{k} = F_{n+1}}$$

Moreover, my friend found $F_{n+1}$ using generating functions, almost the same method as Sil's answer — D S, Feb 06 '24 at 08:58

Is it true that $\sum_{k=0}^m\binom{n-k}k$ outputs the $(n+1)$th Fibonacci number, where $m=\frac{n-1}2$ for odd $n$ and $m=\frac n2$ for even $n$?

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