a(n) = 2^(2^(n-1))*b(n) where b(1) = 1/2 and b(n+1) = b(n) - b(n)^2.
A076628
a(n) = 2^(2^(n-1))*b(n) where b(1) = 1/2 and b(n+1) = b(n) - b(n)^2.
Terms
- a(0) =1a(1) =1a(2) =3a(3) =39a(4) =8463a(5) =483008799
External references
- oeis: A076628
A076628
a(n) = 2^(2^(n-1))*b(n) where b(1) = 1/2 and b(n+1) = b(n) - b(n)^2.