Difference between revisions of "L(n+1)L(n-1)-L(n)^2=5(-1)^(n+1)"
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==References== | ==References== | ||
| − | * {{PaperReference|A Primer on the Fibonacci Sequence Part I|1963|S.L. Basin|author2=V.E. Hoggatt, Jr.|prev=F(n+1)F(n-1)-F(n)^2=(-1)^n|next= | + | * {{PaperReference|A Primer on the Fibonacci Sequence Part I|1963|S.L. Basin|author2=V.E. Hoggatt, Jr.|prev=F(n+1)F(n-1)-F(n)^2=(-1)^n|next=L(n)=F(n+1)+F(n-1)}} |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 00:25, 25 May 2017
Theorem
The following formula holds: $$L(n+1)L(n-1)-L(n)^2=5(-1)^{n+1},$$ where $L(n)$ denotes a Lucas number.
