f ( n ) = ∑ i = 0 [ n 3 ] ∑ j = 0 [ n − 3 i 2 ] n − i n − 2 i − j ( n − 2 i − j i ) ( n − 3 i − j j ) , n > 2 , n ∈ N {\displaystyle f(n)=\sum _{i=0}^{\left[{\frac {n}{3}}\right]}\sum _{j=0}^{\left[{\frac {n-3i}{2}}\right]}{\frac {n-i}{n-2i-j}}{\binom {n-2i-j}{i}}{\binom {n-3i-j}{j}},n>2,n\in \mathbb {N} }
f ( n ) = ∑ k = 0 [ n 4 ] ∑ i = 0 [ n − 4 k 3 ] ∑ j = 0 [ n − 4 k − 3 i 2 ] n − 2 k − i n − 3 k − 2 i − j ( n − 3 k − 2 i − j k ) ( n − 4 k − 2 i − j i ) ( n − 4 k − 3 i − j j ) , n > 3 , n ∈ N {\displaystyle f(n)=\sum _{k=0}^{\left[{\frac {n}{4}}\right]}\sum _{i=0}^{\left[{\frac {n-4k}{3}}\right]}\sum _{j=0}^{\left[{\frac {n-4k-3i}{2}}\right]}{\frac {n-2k-i}{n-3k-2i-j}}{\binom {n-3k-2i-j}{k}}{\binom {n-4k-2i-j}{i}}{\binom {n-4k-3i-j}{j}},n>3,n\in \mathbb {N} }
<<School:李煌数学研究院
![{\displaystyle f(n)=\sum _{i=0}^{\left[{\frac {n}{3}}\right]}\sum _{j=0}^{\left[{\frac {n-3i}{2}}\right]}{\frac {n-i}{n-2i-j}}{\binom {n-2i-j}{i}}{\binom {n-3i-j}{j}},n>2,n\in \mathbb {N} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/51b8804a16a424cdeb4d7c3cda908ac3eca77a45)
![{\displaystyle f(n)=\sum _{k=0}^{\left[{\frac {n}{4}}\right]}\sum _{i=0}^{\left[{\frac {n-4k}{3}}\right]}\sum _{j=0}^{\left[{\frac {n-4k-3i}{2}}\right]}{\frac {n-2k-i}{n-3k-2i-j}}{\binom {n-3k-2i-j}{k}}{\binom {n-4k-2i-j}{i}}{\binom {n-4k-3i-j}{j}},n>3,n\in \mathbb {N} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/14c2b54c608a90e1bc5b9f0c25fac920b3681930)
