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

1,14

Links

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

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.]

Example

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.

Prog

(Scheme, with memoization-macro definec)
(definec (A252459 n) (cond ((= 1 n) 0) ((not (zero? (A252372 n))) 0) (else (+ 1 (A252459 (A003961 n))))))

Crossrefs

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

Cf. also A246271, A246272.

Sequence in context: A248394 A368843 A127268 * A083918 A083895 A093488

Adjacent sequences: A252456 A252457 A252458 * A252460 A252461 A252462

Keyword

nonn

Author

Antti Karttunen, Dec 17 2014

Status

approved