关于Lucas数立方与二项式数的卷积公式

对于非负整数l，L_l表示第l个Lucas数；$left( {array}{l}ni{array} right) = frac{{n!}}{{i!left( {n - i} right)!}}$为二项式系数；对于非负整数l和k以及正整数n，设l(k, 3, n)是数列$left{ {left( {array}{l}ni{array} right)} right}_{i = 0}^n$和$left{ {L_{k + i}^3} right}_{i = 0}^n$的卷积，即l(k, 3, n)=$left( {array}{l}n0{array} right)L_k^3 + left( {array}{l}n1{array} right)L_{k + 1}^3 + cdots + left( {array}{l}nn{array} right)L_{k + n}^3 = sumlimits_{i = 0}^n {left( {array}{l}ni{array} right)L_{k + i}^3} $。文章证明了k≥n时，l(k, 3, n)=2nL_3k+2n+3(－1)k+nL_k－n; 当k < n时，l(k, 3, n)=2nL_3k+2n+3L_n－k成立。

A Relationship Between Binomial Coefficients and Cubes of Lucas Numbers

For nonnegative integer l, L_l is the lth Lucas Number and $left( {array}{l}ni{array} right) = frac{{n!}}{{i!left( {n - i} right)!}}$ is the binomial coeffient. For any nonnegative integer l, k and positive integer n, l(k, 3, n) denotes the convolution of sequence $left{ {left( {array}{l}ni{array} right)} right}_{i = 0}^n$ and $left{ {L_{k + i}^3} right}_{i = 0}^n$, namely, l(k, 3, n)=$left( {array}{l}n0{array} right)L_k^3 + left( {array}{l}n1{array} right)L_{k + 1}^3 + cdots + left( {array}{l}nn{array} right)L_{k + n}^3 = sumlimits_{i = 0}^n {left( {array}{l}ni{array} right)L_{k + i}^3} $. According to the definition of the Fibonacci sequence and by using the knowledge of elementary number theory, it is proved that l(k, 3, n) is equal to 2nL_3k+2n+3(-1)k+nL_k-n or 2nL_3k+2n+3L_n-k when k≥n or not.