a(n) = 2^L(n+1)*3^L(n), where L(n) is the n-th Lucas number (A000032(n)).
A166471
a(n) = 2^L(n+1)*3^L(n), where L(n) is the n-th Lucas number (A000032(n)).
Terms
- a(0) =18a(1) =24a(2) =432a(3) =10368a(4) =4478976a(5) =46438023168
External references
- oeis: A166471