TITLE: Exercise on Fibonacci Numbers
QUESTION [6 upvotes]: Consider, for positive $n$, the identity
$$ x^n = \sum_{k=1}^{n} F_{n,k} \, x^{\left(1+ \left\lceil \frac{k-1}{2} \right\rceil\right)} (x-1) ^{\left\lfloor \frac{k-1}{2}\right\rfloor}$$
and prove that the sum $F_n= F_{n,1} + F_{n,2} + \cdots + F_{n,n}$ is the $n$-th Fibonacci number satisfying the relations $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n>2$.
Anyone knows how to solve it?
Thank you very much!!

REPLY [4 votes]: This is likely not the most direct proof, but should shed some light on the question.  The table of the $F_{n,k}$ numbers can be generated by this simple MATLAB program
function table = fibtable(nrows)
  table = zeros(nrows);
  for n = 1:nrows
    M = fibident(n);
    row = M \ [zeros(n-1,1); 1];
    table(n,1:n) = row.';
  end
end

function M = fibident(n)
  M = zeros(n);
  for k=1:n
    a = 1 + ceil((k-1)/2);
    b = floor((k-1)/2);
    for d = 0:b
      M(a+d, k) = (-1).^(b-d) * nchoosek(b,d);
    end
  end
end

The program is based on the observation, also made in the comments to the question, that the identity should be interpreted as the identity of two polynomials of real variable.  For 18 rows, we get
$$\scriptsize
\begin{matrix}
  1 \\
  0 & 1 \\
  0 & 1 & 1 \\
  0 & 1 & 1 & 1 \\
  0 & 1 & 1 & 2 & 1 \\
  0 & 1 & 1 & 3 & 2 & 1 \\
  0 & 1 & 1 & 4 & 3 & 3 & 1 \\
  0 & 1 & 1 & 5 & 4 & 6 & 3 & 1 \\
  0 & 1 & 1 & 6 & 5 & 10 & 6 & 4 & 1 \\
  0 & 1 & 1 & 7 & 6 & 15 & 10 & 10 & 4 & 1 \\
  0 & 1 & 1 & 8 & 7 & 21 & 15 & 20 & 10 & 5 & 1 \\
 0 & 1 & 1 & 9 & 8 & 28 & 21 & 35 & 20 & 15 & 5 & 1 \\
  0 & 1 & 1 & 10 & 9 & 36 & 28 & 56 & 35 & 35 & 15 & 6 & 1 \\
  0 & 1 & 1 & 11 & 10 & 45 & 36 & 84 & 56 & 70 & 35 & 21 & 6 & 1 \\
  0 & 1 & 1 & 12 & 11 & 55 & 45 & 120 & 84 & 126 & 70 & 56 & 21 & 7 & 1 \\
  0 & 1 & 1 & 13 & 12 & 66 & 55 & 165 & 120 & 210 & 126 & 126 & 56 & 28 & 7 & 1 \\
  0 & 1 & 1 & 14 & 13 & 78 & 66 & 220 & 165 & 330 & 210 & 252 & 126 & 84 & 28 & 8 & 1 \\
  0 & 1 & 1 & 15 & 14 & 91 & 78 & 286 & 220 & 495 & 330 & 462 & 252 & 210 & 84 & 36 & 8 & 1
\end{matrix}
$$
It then becomes apparent that the columns of the table (except the first) are  the diagonals of Pascal's triangle, which are related to the Fibonacci numbers (See, for example, here.)  In particular,
$$ F_{n,k} = \binom{n-\left\lceil \frac{k}{2}\right\rceil - 1}{n-k} \enspace,$$
which holds for $1 \leq k \leq n$ if we stipulate that $\binom{a}{b} = 0$ when $b < 0$ and $\binom{a}{0} = 1$ also for negative $a$.

Let $S_n = \sum_{1 \leq k \leq n} F_{n,k}.$  It's easy to verify that $S_n = F_n$ for $1 \leq n \leq 3$.  We show that $S_{n+2} = S_{n} + S_{n+1}$ for $n > 1$, hence proving that $S_n = F_n$ for all positive $n$.  
Taking into account that $F_{n,1} = 0$ and $F_{n,2} = F_{n,n} = 1$ for all $n > 1$, we line up the appropriate members of the two sums and then apply the basic identity,
$$ \binom{a}{b} + \binom{a}{b+1} = \binom{a+1}{b+1} \enspace,$$
which holds for $0 \leq b \leq a$.  In detail,
$$\begin{align}
S_n + S_{n+1} &= \sum_{1 \leq k \leq n-1} \binom{n - \left\lceil \frac{k}{2} \right\rceil - 1}{n-k} + \sum_{1 \leq k \leq n+1} \binom{n + 1 - \left\lceil \frac{k}{2} \right\rceil - 1}{n+1-k} + 1 \\
&= 0 + 1 + \sum_{3 \leq k \leq n+1} \binom{n + 1 - \left\lceil \frac{k}{2} \right\rceil - 1}{n+2-k} + \sum_{3 \leq k \leq n+1} \binom{n + 1 - \left\lceil \frac{k}{2} \right\rceil - 1}{n+1-k} + 1 \\
&= 0 + 1 + \sum_{3 \leq k \leq n+1} \left[ \binom{n + 1 - \left\lceil \frac{k}{2} \right\rceil - 1}{n+2-k} + \binom{n + 1 - \left\lceil \frac{k}{2} \right\rceil - 1}{n+1-k}\right] + 1 \\
&= F_{n+2,1} + F_{n+2,2} + \sum_{3 \leq k \leq n+1} \binom{n + 2 - \left\lceil \frac{k}{2} \right\rceil - 1}{n+2-k} + F_{n+2,n+2} \\
&= \sum_{1 \leq k \leq n+2} \binom{n + 2 - \left\lceil \frac{k}{2} \right\rceil - 1}{n+2-k} \\
&= S_{n+2} \enspace.
\end{align}$$

Inspection of the matrix above suggested a formula for $F_{n,k}$.  It remains to prove that the suggestion was correct.  We'll start by revisiting the program that produces the table of the $F_{n,k}$ coefficients.
If we look at a few matrices produced by fibident, we'll see that they have a very interesting structure:
$$\small
M_5 = \begin{bmatrix}
  1 & 0 & 0 & 0 & 0 \\
  0 & 1 & -1 & 0 & 0 \\
  0 & 0 & 1 & -1 & 1 \\
  0 & 0 & 0 & 1 & -2 \\
  0 & 0 & 0 & 0 & 1
\end{bmatrix}
~~~~~~
M_7 = \begin{bmatrix}
  1 & 0 & 0 & 0 & 0 & 0 & 0 \\
  0 & 1 & -1 & 0 & 0 & 0 & 0 \\
  0 & 0 & 1 & -1 & 1 & 0 & 0 \\
  0 & 0 & 0 & 1 & -2 & 1 & -1 \\
  0 & 0 & 0 & 0 & 1 & -2 & 3 \\
  0 & 0 & 0 & 0 & 0 & 1 & -3 \\
  0 & 0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix} \enspace.
$$
Each matrix is a submatrix of all larger matrices and the matrices are triangular, with a diagonal of all ones (which implies they are unimodular).  Therefore we only need to generate one matrix and invert it to get (the transpose of) the table.  (The right hand sides of all our systems of linear equations are the columns of the identity matrix.)  Our revised code does just that.
function [tab, M] = fibmatrices(n)
  M = fibident(n);
  tab = inv(M).';
end

This revision also suggests that we can prove the correctness of our guess for the coefficients by proving that the transpose of the matrix defined by our $F_{n,k}$'s is the inverse of the $M$ matrix above.  We write the expression of the generic element $P_{i,j}$ of the product of the two matrices and, after a nontrivial amount of algebra, we find that $P_{i,i} = 1$ and $P_{i,j} = 0$ whenever $i \neq j$.
In some more detail, we start by observing that the columns of $M$ contain the coefficients of $(x-1)^{\left\lfloor \frac{k-1}{2} \right\rfloor}$; hence
$$ M_{n,k} = (-1)^{k-n} \binom{\left\lfloor \frac{k-1}{2} \right\rfloor}{n - \left\lceil \frac{k+1}{2} \right\rceil}  = (-1)^{k-n} \binom{\left\lfloor \frac{k-1}{2} \right\rfloor}{k-n} \enspace. $$
We then write, for $i,j>1$,
$$ P_{i,j} = \sum_{i \leq k \leq j} F_{k,i} M_{k,j} = \sum_{i \leq k \leq j} (-1)^{j-k} \binom{k - \left\lceil \frac{i}{2} \right\rceil}{k-i} \binom{\left\lfloor \frac{j-1}{2} \right\rfloor}{j-k} \enspace. $$
In preparation for further manipulation, we write,
$$ P_{i,j} = (-1)^j \sum_{i \leq k \leq j} (-1)^k \binom{\left\lfloor \frac{j-1}{2} \right\rfloor}{k- \left\lceil \frac{j-1}{2} \right\rceil - 1} \binom{k - \left\lceil \frac{i}{2} \right\rceil - 1}{\left\lfloor \frac{i}{2} \right\rfloor - 1} \enspace. $$
Now we spit cases; for $i>j$ the summation is empty and $P_{i,j}=0$.  For $i=j$ the summation contains one term,
$$ P_{i,i}  = \binom{\left\lfloor \frac{i-1}{2} \right\rfloor}{\left\lfloor \frac{i-1}{2} \right\rfloor} \binom{\left\lfloor \frac{i}{2} \right\rfloor - 1}{\left\lfloor \frac{i}{2} \right\rfloor - 1} = 1 \enspace. $$
Finally, for $j>i$, if $k < \left\lceil \frac{i}{2} \right\rceil + 1$, it is also $k < \left\lceil \frac{j-1}{2} \right\rceil + 1$.  Hence, we can safely extend the summation to all values of $k$.  This allows us to apply a variation of Vadermonde's convolution ((5.24) in Concrete Mathematics, second edition), which gives,
$$ P_{i,j} = \binom{- \left\lceil \frac{i}{2} \right\rceil + \left\lceil\frac{j-1}{2} \right\rfloor}{\left\lfloor \frac{i}{2} \right\rfloor - \left\lfloor \frac{j-1}{2} \right\rfloor - 1} \enspace. $$
For $j>i$, the upper index is non-negative, and the lower index is always greater than the upper index.  Hence $P_{i,j} = 0$.