TITLE: Neighbourhood of a word and Levenshtein distance
QUESTION [5 upvotes]: The Levenshtein distance or Edit distance $$ lev(U,V) $$ between two strings $U$ and $V$ over a finite alphabet $\Sigma$ of size $ \left| \Sigma \right| = \sigma ,$ is the minimal number of insertions, deletions  and replacements  to make the strings equal.
For $k \in \mathbb{N} $ and $U \in  \Sigma^*$ we define :
$$ N_k(U) = \{ V \in  \Sigma^* : lev(U,V)  \leq k \}  $$
For example if $\Sigma = \{A,B,L\}$ : 
\begin{eqnarray*}
 N_1(AAA) &=& \{ AAA, AA, AAB, AAL, ABA, ALA, BAA, LAA, AAAA, BAAA,  \\ &&LAAA, ABAA, ALAA, AABA, AALA, AAAB, AAAL \}  \\
N_1(LAB) &=& \{LAB, LA, AB, LB, AAB, BAB, LBB, LLB, LAA, LAL, ALAB, \\
&&BLAB, LLAB, LAAB, LBAB, LALB, LABA, LABB, LABL\} 
\end{eqnarray*}
My goal is to compute for $\Sigma, n $ and $k$ fixed  :
$$ \max_{U \in  \Sigma^* , ~ \left| U \right|  = n}  \left|N_k(U) \right| $$
or an upper bound $f(\Sigma,k,n)$.
I conjecture that the maximum is reached for the words of the form :
$$ c_1c_2 \dots c_{\sigma}c_1c_2 \dots c_{\sigma}\dots $$
where $\Sigma = \{  c_1,c_2, \dots,c_{\sigma}  \}. $
Does anyone have an idea of how to compute $ \max_{U \in  \Sigma^* , ~ \left| U \right|  = n}  \left|N_k(U) \right| $ or to prove this conjecture?
A special case has already been studied in the section 6 of the paper What's Behind Blast by Gene Myers.

REPLY [2 votes]: A recent paper by Hélène Touzet discusses how to compute the size of the neighborhood of a given word via the Universal Levenshtein automaton.
"On the Levenshtein Automaton and the Size of the Neighbourhood of a Word" (unfortunately not freely available online yet).
Maybe it is possible to exploit the structure of the universal DFA or to at least do some practical experiments regarding your conjecture.