%I #12 Dec 18 2014 02:07:44

%S 0,0,0,0,0,0,0,0,0,1,0,0,0,2,0,0,0,0,0,1,0,2,0,0,0,2,0,2,0,1,0,0,1,2,

%T 0,0,0,3,1,1,0,2,0,2,0,3,0,0,0,1,1,2,0,0,0,2,1,3,0,1,0,3,0,0,0,2,0,2,

%U 2,2,0,0,0,3,0,3,0,2,0,1,0,4,0,2,0,4,2,2,0,1,0,3,2,4,0,0,0,2,1,1,0,2,0,2,0,4,0,0,0,2,2,2,0,3,0,3,1,4,0,1

%N a(n) = Number of iterations of A003961 starting from n which are needed before the result is one of the numbers in A251726. a(1) = 0 by convention.

%H Antti Karttunen, <a href="/A252459/b252459.txt">Table of n, a(n) for n = 1..10001</a>

%F a(1) = 0 and for n > 1, if A252372(n) = 1 then a(n) = 0, otherwise 1 + a(A003961(n)).

%F Other identities. For all n >= 1:

%F a(n) = a(A066048(n)). [The result depends only on the smallest and the largest prime factor of n.]

%e a(9) = 0, because 9 is already in A251726.

%e For n = 10, as 10 is in A251727, but A003961(10) = A251727(prime(1) * prime(3)) = prime(2) * prime(4) = 3*7 = 21 is in A251726, thus a(10) = 1.

%e For n = 14, as 14 is in A251727, and A003961(14) = 33 (prime(1) * prime(4) -> prime(2) * prime(5)) is also in A251727, and only at the second iteration, A003961(33) = 65 (prime(2) * prime(5) -> prime(3) * prime(6)) the result is in A251726, thus a(14) = 2.

%o (Scheme, with memoization-macro definec)

%o (definec (A252459 n) (cond ((= 1 n) 0) ((not (zero? (A252372 n))) 0) (else (+ 1 (A252459 (A003961 n))))))

%Y Cf. A003961, A066048, A251726 (gives the positions of zeros after a(1)=0), A252372.

%Y Cf. also A246271, A246272.

%K nonn

%O 1,14

%A _Antti Karttunen_, Dec 17 2014