[THOUGHT] We need to count the number of proper colorings of a cycle of length $L=n+1=11$ with 11 colors such that no two vertices at distance 1 or 2 share the same color.
This can be modeled as a walk on the state graph where a state is an ordered pair of distinct colors $(a,b)$.
From $(a,b)$ we may go to $(b,c)$ iff $c\neq a$ and $c\neq b$.
The number of valid sequences of length $L$ forming a closed loop equals the trace of the adjacency matrix $M$ of this state graph raised to the $L$‑th power: $k_n = \operatorname{tr}(M^{L})$.

We will:

1. Enumerate all ordered pairs of distinct colors from $\{0,\dots,10\}$ (11 colors).
2. Build the $110\times110$ adjacency matrix $M$ where $M[(a,b),(b,c)]=1$ iff $c\neq a,b$.
3. Compute $\operatorname{tr}(M^{11})$.

[/THOUGHT] [PYTHON] import numpy as np

number of colors

C = 11

generate all ordered pairs (a,b) with a != b

pairs = [(a, b) for a in range(C) for b in range(C) if a != b] N = len(pairs) # 110

map pair to index

pair_to_idx = {pair: i for i, pair in enumerate(pairs)}

build adjacency matrix

M = np.zeros((N, N), dtype=int)

for i, (a, b) in enumerate(pairs): for c in range(C): if c != a and c != b: j = pair_to_idx[(b, c)] M[i, j] = 1

compute trace of M^L for L = 11

L = 11 M_L = np.linalg.matrix_power(M, L) trace = int(np.trace(M_L))

trace [/PYTHON]

[THOUGHT] The Python computation confirms that the number of valid sequences is 31384473120. [/THOUGHT] $\boxed{31384473120}$