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By definition, a(n)>0 if and only if n is a member of A034878. If n>2, then a(n!)>max(a(n),a(n!-1)), as (n!)!=n!*(n!-1)!. Similarly, a(A001013(n))>0 for n>2. Clearly a(n)=0 if n is a prime A000040. So a(n+1)=1 if n=2^p-1 is a Mersenne prime A000668, as (n+1)!=(2!)^p*n! and n is prime. - Jonathan Sondow, Dec 15 2004
R. K. Guy, Unsolved Problems in Number Theory, B23.