(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: $$(n+2\lambda)C_n^{\lambda}(x) = 2\lambda(C_n^{\lambda+1}(x)-xC_{n-1}^{\lambda+1}(x)),$$ where $C_n^{\lambda}$ denotes Gegenbauer C.
