%N Numbers n such that the final nonzero digit of n! is the same as the last digit of C(2n,n) (in base 10).
%F n such that A008904(n) = C(2n, n) reduced (mod 10)
%e 12! = 479001600 C(24,12)= 2704156 and the last nonzero digit of 12! is the same as the last digit of C(24,12) hence 12 is in the sequence.
%A _Benoit Cloitre_, Aug 18 2002