Find an example of a lattice such that LLL algorithm can't find the shortest vector of the lattice, satisfying I want to find an example of a basis of a lattice of dimension $n$ such that LLL algorithm can't find the shortest vector of the lattice, and such that the shortest  vector of this lattice, say $b=c_1b_1+\cdots+c_nb_n$, has small $c_i$, i.e., $c_i \in \{-r,\ldots,r\}$ with small $r$.

Note that $b_i \in \mathbb{Z}^n$ and $c_i \in \mathbb{Z}$.

Such example must satisfy $\log_2(2(r+1)) \cdot n \leq 15$, say $n=5$ and $r=4$.

The idea here is to have a treatable value for $2^{\log_2(2(r+1)) \cdot n}$.

I can't find any lattice satisfying this.

Thanks in advance.
Bruteforce appears to work well enough. The following Sage script finds an instance quickly:

from sage.libs.fplll.fplll import FP_LLL
from sage.libs.fplll.fplll import gen_uniform

n = 5 # dimension
q = 16 # size of matrix entries

while True:
    M = gen_uniform(n, n, q)
    L = M.LLL(delta=0.999)
    S = FP_LLL(L).shortest_vector(algorithm='proved')
    if min([ row.norm() for row in L]) != S.norm():
        break

print M
print L
print min([ row.norm() for row in L])
print S
print S.norm()


As a concrete example, the lattice (column vectors)
$$L = 
\begin{pmatrix}
16252&amp;  5541&amp; 48769&amp;  2593&amp; 22395\\
 2404&amp; 46418&amp;  5335&amp; 59631&amp; 29390\\
41816&amp; 27475&amp; 56363&amp; 50417&amp; 62371\\
27419&amp; 53783&amp;  1412&amp; 42344&amp; 55983\\
62503&amp; 52719&amp; 22160&amp;  3940&amp; 64256\\
\end{pmatrix}
$$
can be reduced with LLL ($\delta = 0.999$) to
$$B=
\begin{pmatrix}
  1855&amp;  31849 &amp; -6182&amp;  -5480&amp;  16007\\
 23160&amp; -24484 &amp;  2259&amp; -11807&amp;  10586\\
 29450&amp;   -465 &amp;  8167&amp;   2997&amp; -19059\\
 10069&amp;  -8429 &amp; 24671&amp; -21117&amp; -18320\\
  4328&amp; -17879 &amp; 30280&amp;  29190&amp;  24708\\
\end{pmatrix}
$$
Yet, none of those vectors is the shortest. The shortest vector is $$(-19381, -7964, 16504, -24114, 739),$$ which can be written as $2L_0 - L_1 - L_2 + 2L_3 - L_4$, where $L_i$ is the $i$th row of $L$. So we have $n = 5$ and $r = 2$.