\documentclass[a4]{article}
\usepackage{amsmath}
\begin{document}
Consider a task of multiplying two bignums, U and V, split into N equal parts (which may or may not coincide with the bignum limbs), each part of $\omega$ bits. Then, the equation that describes schoolbook multiplication is:
\begin{equation}
  \label{eq:schoolbook-multiplication}
  R_{2N} = U_N*V_N = \left( \sum_{i=0}^{N-1}u_i 2^{\omega i} \right) * \left( \sum_{i=0}^{N-1}v_i 2^{\omega i} \right) = \sum_{i=0}^{N-1}\sum_{j=0}^{N-1}u_i v_j 2^{\omega(i+j)} = \sum_{k=0}^{2N-1}r_k 2^{\omega k}
\end{equation}
Here, $N$ is the number of limbs.

Now, lets have a look at the N-way-split Karatsuba:
\begin{equation}
  \label{eq:karatsuba-nway}
  U_N*V_N = \sum_{i=0}^{N-1}\sum_{j=0}^{N-1}u_i v_i 2^{\omega(i+j)} - \sum_{i=1}^{N-1}\sum_{j=0}^{i-1}(u_j-u_i)(v_j-u_i)2^{\omega(i+j)}
\end{equation}

Does it really work? Let's have a look. Let's expand the expression in the second sum in the equation:
\begin{equation}
  \label{eq:second-sum}
  \begin{aligned} 
    \sum_{i=1}^{N-1}\sum_{j=0}^{i-1}(u_j-u_i)(v_j-u_i)2^{\omega(i+j)} &= \sum_{i=1}^{N-1}\sum_{j=0}^{i-1}(u_j v_j - u_i v_j - u_j v_i + u_i v_i)2^{\omega(i+j)}\\
    &= \sum_{i=1}^{N-1}\sum_{j=0}^{i-1}(u_j v_j + u_i v_i)2^{\omega(i+j)} - \sum_{i=1}^{N-1}\sum_{j=0}^{i-1}(u_i v_j + u_j v_i)2^{\omega(i+j)}
\end{aligned}   
\end{equation}


Each of these sums can be further simplified. I'll use the second one as an example:
\begin{equation}
  \label{eq:second-sum-part}
  \sum_{i=1}^{N-1}\sum_{j=0}^{i-1}(u_i v_j + u_j v_i)2^{\omega(i+j)} = \sum_{i=1}^{N-1}\sum_{j=0}^{i-1}u_i v_j2^{\omega(i+j)} + \sum_{i=1}^{N-1}\sum_{j=0}^{i-1}u_j v_i2^{\omega(i+j)}
\end{equation}
Let's change the summation order in the second sum, and then rename the indices:
$$\sum_{i=1}^{N-1}\sum_{j=0}^{i-1}u_j v_i 2^{\omega(i+j)} = \sum_{j=0}^{N-2}\sum_{i=j+1}^{N-1}u_j v_i 2^{\omega(i+j)} = \sum_{i=0}^{N-2}\sum_{j=i+1}^{N-1}u_i v_j2^{\omega(i+j)}$$

Now, plug this sum back into equation (\ref{eq:second-sum-part}), we can see that we can unite the domains of both sums:
$$\sum_{i=1}^{N-1}\sum_{j=0}^{i-1}u_i v_j2^{\omega(i+j)} + \sum_{i=0}^{N-2}\sum_{j=i+1}^{N-1}u_i v_j2^{\omega(i+j)} = \sum_{i=0}^{N-1}\sum_{j=0, j \neq i}^{N-1}u_i v_j2^{\omega(i+j)}$$

Exactly same transformations can be done for the first sum in equation (\ref{eq:second-sum}):
$$\sum_{i=1}^{N-1}\sum_{j=0}^{i-1}(u_j v_j + u_i v_i)2^{\omega(i+j)} = \sum_{i=0}^{N-1}\sum_{j=0, j \neq i}^{N-1}u_i v_i2^{\omega(i+j)}$$

Let's return to equations~(\ref{eq:karatsuba-nway}) and (\ref{eq:schoolbook-multiplication}):
\begin{equation}
  \label{eq:karatsuba-explained}
  \begin{aligned}
    \sum_{i=0}^{N-1}\sum_{j=0}^{N-1}u_i v_j 2^{\omega(i+j)} &= \sum_{i=0}^{N-1}\sum_{j=0}^{N-1}u_i v_i 2^{\omega(i+j)} - \sum_{i=1}^{N-1}\sum_{j=0}^{i-1}(u_j-u_i)(v_j-u_i)2^{\omega(i+j)}\\
    &=\sum_{i=0}^{N-1}\sum_{j=0}^{N-1}u_i v_i 2^{\omega(i+j)} - \sum_{i=0}^{N-1}\sum_{j=0, j \neq i}^{N-1}u_i v_i2^{\omega(i+j)} + \sum_{i=0}^{N-1}\sum_{j = 0, j\neq i}^{N-1}u_i v_j2^{\omega(i+j)}\\
    &=\sum_{i=0}^{N-1}\sum_{j=0, j = i}^{N-1}u_i v_i2^{\omega(i+j)} + \sum_{i=0}^{N-1}\sum_{j = 0, j\neq i}^{N-1}u_i v_j2^{\omega(i+j)}\\
    &=\sum_{i=0}^{N-1}\sum_{j=0}^{N-1}u_i v_j 2^{\omega(i+j)}
  \end{aligned}
\end{equation}
Q.E.D.

Implementing this equation directly as two sums over one working register would be impractical, so equation (\ref{eq:karatsuba-nway}) can be further rewritten as:
\begin{equation}
  \label{eq:karatsuba-optimized}
  \begin{aligned} 
    U_N\cdot V_N &= W_0 + \sum_{k=1}^{2N-3}\left(W_k - \sum_{i=L_i}^{\lceil k/2 \rceil-1}(u_i-u_{k-1})(v_i-v_{k-1})\right)2^{\omega k} + W_{2N-2}2^{\omega(2N-2)} \\
    W_0 &=u_0 v_0\\
    W_i &=W_{i-1} + u_i v_i, i= \overline{1,N-1} \\
    W_i &=W_{i-1} - u_{i-N} v_{i-N}, i= \overline{N,2N-3} \\
    W_{2N-2} &= u_{N-1} v_{N-1}\\
    L_i &=0, i= \overline{1,N-1} \\
    L_i &=k-N+1, i= \overline{N,2N-3}
\end{aligned}   
\end{equation}
\end{document}

