footnote added

src/sections/01_introduction/footnote.tex

@@ -0,0 +1,5 @@
+One may assume that it is possible to reach the form $n^{2m+1} = \sum_{k=1}^{n} \mathbf{A}_{m,0} k^0 (n-k)^0 + \mathbf{A}_{m,1}(n-k)^1
++ \cdots + \mathbf{A}_{m,m} k^m (n-k)^m$ simply taking finite differences of the polynomial $n^{2m+1}$ up to order of $2m+1$ 
+and interpolating it backwards similarly as shown in~\eqref{eq:cubes_interpolation}.
+However, my observations do not provide any evidence of such assumption.
+Interestingly enough is that we could have been arrived to the pure differential approach of the relation~\eqref{eq:odd_power_conjecture} then.

src/sections/01_introduction/introduction.tex

@@ -17,16 +17,21 @@
     \end{center}
     \caption{Table of finite differences of the polynomial $n^3$.} \label{tab:table}
 \end{table}
-We can observe easily that finite differences of the polynomial $n^3$ may be expressed according
+We can observe easily that finite differences
+\footnote{\input{sections/01_introduction/footnote}}
+of the polynomial $n^3$ may be expressed according
 to the following relation, via rearrangement of the terms
-\begin{align*}
-    \Delta(0^3) &= 1+6 \cdot 0 \\
-    \Delta(1^3) &= 1+6\cdot0+6\cdot1 \\
-    \Delta(2^3) &= 1+6\cdot0+6\cdot1+6\cdot2 \\
-    \Delta(3^3) &= 1+6\cdot0+6\cdot1+6\cdot2+6\cdot3 \\
-    &\; \; \vdots \\
-    \Delta(n^3) &= 1+6\cdot0+6\cdot1+6\cdot2+6\cdot3+\cdots+6\cdot n
-\end{align*}
+\begin{align}
+    \label{eq:cubes_interpolation}
+    \begin{split}
+        \Delta(0^3) &= 1+6 \cdot 0 \\
+        \Delta(1^3) &= 1+6\cdot0+6\cdot1 \\
+        \Delta(2^3) &= 1+6\cdot0+6\cdot1+6\cdot2 \\
+        \Delta(3^3) &= 1+6\cdot0+6\cdot1+6\cdot2+6\cdot3 \\
+        &\; \; \vdots \\
+        \Delta(n^3) &= 1+6\cdot0+6\cdot1+6\cdot2+6\cdot3+\cdots+6\cdot n
+    \end{split}
+\end{align}
 Furthermore, the polynomial $n^3$ is identical to
 \begin{align*}
     n^3 &= [1+6\cdot0]+[1+6\cdot0+6\cdot1]+[1+6\cdot0+6\cdot1+6\cdot2]+\cdots \\
@@ -52,8 +57,9 @@
 Therefore, let be a conjecture
 \begin{conj}
     For every $n\geq 1, \; n,m\in\mathbb{N}$ there are coefficients $\coeffA{m}{0}, \coeffA{m}{1}, \ldots, \coeffA{m}{m}$ such that
-    \begin{equation*}
+    \begin{equation}
+        \label{eq:odd_power_conjecture}
         n^{2m+1} = \sum_{k=1}^{n} \coeffA{m}{0} k^0 (n-k)^0 + \coeffA{m}{1} (n-k)^1
         + \cdots + \coeffA{m}{m} k^m (n-k)^m
-    \end{equation*}
+    \end{equation}
 \end{conj}