- #1
ramsey2879
- 841
- 3
I derived the following identity after considering my thread "Recursive series equality" but the result is so clean and neat that I post it as a new topic.
Let [tex]S_{0} = 0 \quad S_{1} = 1 \quad S_{n} = b*S_{n-1} - S_{n-2}[/tex]
Then [tex]S_{n}*(S_{n+b} -S_{b-2}) = (S_{n+1}+1)*(S_{n+b-1} -S_{b-1})[/tex]
Has anyone seen this before?
Let [tex]S_{0} = 0 \quad S_{1} = 1 \quad S_{n} = b*S_{n-1} - S_{n-2}[/tex]
Then [tex]S_{n}*(S_{n+b} -S_{b-2}) = (S_{n+1}+1)*(S_{n+b-1} -S_{b-1})[/tex]
Has anyone seen this before?