A formula for the solution of a kind of recurrence relation with two parameters. (Chinese. English summary) Zbl 0706.11009
Summary: An explicit formula for the solution of a recurrence relation with two parameters
\[ u_{i,j} = \sum^{p-2}_{\lambda =0}a_{(p-1)- \lambda}u_{(i-1-\lambda),j}+a_0u_{i-n,j-1}\quad (i\ge p-1+nj,\quad j=0,1,\ldots) \]
\[ u_{\lambda,0} = c_{\lambda}\quad (\lambda =0,1,\ldots,p- 2),\quad u_{i,j} = 0\quad (j<0, \text{ or } j\geq 1\text{ and } i<p - 1+nj,\quad i=0,1, \ldots) \]
(where \(p\ge 2\), \(n\ge 0\), \(a_k\) \((k=0,1,\ldots,p-1)\) and \(c_{\lambda}\) \((\lambda = 0,1,\ldots,p-2)\) are arbitrary constants) is given.
\[ u_{i,j} = \sum^{p-2}_{\lambda =0}a_{(p-1)- \lambda}u_{(i-1-\lambda),j}+a_0u_{i-n,j-1}\quad (i\ge p-1+nj,\quad j=0,1,\ldots) \]
\[ u_{\lambda,0} = c_{\lambda}\quad (\lambda =0,1,\ldots,p- 2),\quad u_{i,j} = 0\quad (j<0, \text{ or } j\geq 1\text{ and } i<p - 1+nj,\quad i=0,1, \ldots) \]
(where \(p\ge 2\), \(n\ge 0\), \(a_k\) \((k=0,1,\ldots,p-1)\) and \(c_{\lambda}\) \((\lambda = 0,1,\ldots,p-2)\) are arbitrary constants) is given.
MSC:
| 11B37 | Recurrences |
