Two $k$-ary Fibonacci recurrences are $a_k(n) = a_k(n-1) + k \cdot a_k(n-2)$ and $b_k(n) = k \cdot b_k(n-1) + b_k(n-2)$. We provide a simple proof that $a_k(n)$ is the number of $k$-regular words over $[n] = \{1,2,\ldots,n\}$ that avoid patterns $\{121, 123, 132, 213\}$ when using base cases $a_k(0) = a_k(1) = 1$ for any $k \geq 1$. This was previously proven by Kuba and Panholzer in the context of Wilf-equivalence for restricted Stirling permutations, and it creates Simion and Schmidt's classic result on the Fibonacci sequence when $k=1$, and the Jacobsthal sequence when $k=2$. We complement this theorem by proving that $b_k(n)$ is the number of $k$-regular words over $[n]$ that avoid $\{122, 213\}$ with $b_k(0) = b_k(1) = 1$ for any~$k \geq 2$. Finally, we conjecture that $|Av^{2}_{n}(\underline{121}, 123, 132, 213)| = a_1(n)^2$ for $n \geq 0$. That is, vincularizing the Stirling pattern in Kuba and Panholzer's Jacobsthal result gives the Fibonacci-squared numbers.

翻译：两个$k$元斐波那契递推关系分别为$a_k(n) = a_k(n-1) + k \cdot a_k(n-2)$和$b_k(n) = k \cdot b_k(n-1) + b_k(n-2)$。我们给出一个简洁证明：当初始条件为$a_k(0) = a_k(1) = 1$且任意$k \geq 1$时，$a_k(n)$表示定义于$[n] = \{1,2,\ldots,n\}$上、规避模式集合$\{121, 123, 132, 213\}$的$k$-正则词的个数。该结论此前由Kuba和Panholzer在受限Stirling排列的Wilf等价性背景下证明，并推导出Simion和Schmidt关于经典斐波那契数列($k=1$时)以及Jacobsthal数列($k=2$时)的著名结论。我们补充该定理：证明当$b_k(0) = b_k(1) = 1$且任意$k \geq 2$时，$b_k(n)$表示定义于$[n]$上、规避模式$\{122, 213\}$的$k$-正则词的个数。最后，我们猜想对于$n \geq 0$，有$|Av^{2}_{n}(\underline{121}, 123, 132, 213)| = a_1(n)^2$，即对Kuba和Panholzer Jacobsthal结果中的Stirling模式添加附标化约束，将得到斐波那契平方数。