(n+2lambda)C n^(lambda)(x)=2lambda(C n^(lambda+1)(x)-xC (n-1)^(lambda+1)(x))
From specialfunctionswiki
Theorem
The following formula holds: $$(n+2\lambda)C_n^{\lambda}(x) = 2\lambda \left(C_n^{\lambda+1}(x)-xC_{n-1}^{\lambda+1}(x) \right),$$ where $C_n^{\lambda}$ denotes Gegenbauer C.
