A252459 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.

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, 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, 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

Offset: 1

Author

Antti Karttunen, Dec 17 2014

Keywords

• nonn

Examples

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

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10001

Crossrefs

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

Cf. also A246271, A246272.

Formula

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

Other identities. For all n >= 1:

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