%I #12 Aug 29 2014 19:19:56
%S 0,0,0,1,0,1,1,1,2,1,0,2,1,2,2,3,1,2,1,1,3,2,2,3,3,4,2,3,1,2,0,2,4,3,
%T 3,4,2,4,5,3,1,4,2,2,3,1,2,3,5,4,4,5,3,3,5,6,4,2,1,5,2,3,3,4,2,3,1,4,
%U 6,5,1,5,3,6,4,4,6,7,2,5,3,2,2,6,3,4,4,5,3,3,4,2,5,7,6,2,3,6,4,7,4,5,2,5,7,8,3,3,1,6,4,3,2,3,7,4,5,5,6,4
%N a(1)=0, a(p_n) = a(n), a(c_n) = 1 + a(n), where p_n = n-th prime = A000040(n), c_n = n-th composite number = A002808(n); Also one less than the binary weight of terms of A135141.
%C Consider the following algorithm:
%C Start:
%C If n is 1, we have finished,
%C Otherwise:
%C If n is a prime, replace it with its index among the primes, n <- A000720(n), and go back to the start.
%C Otherwise, if n is a composite, replace it with its index among the composites, n <- A065855(n), and go back to the start.
%C At some point, the process is guaranteed to reach the number 1 at which point we stop.
%C a(n) tells how many times a composite number was encountered in the process, before 1 was reached. This count includes also +1 for the cases where the initial n was composite at the beginning.
%H Antti Karttunen, <a href="/A246369/b246369.txt">Table of n, a(n) for n = 1..32998</a>
%F a(1) = 1, and for n >= 1, if A010051(n) = 1 [that is, when n is prime], a(n) = a(A000720(n)), otherwise a(n) = 1 + a(A065855(n)). [A000720(n) and A065855(n) tell the number of primes, and respectively, composites <= n].
%F a(n) = A000120(A135141(n)) - 1. [a(n) is also one less than the Hamming weight (number of 1-bits) of the n-th term of A135141].
%F a(n) = A080791(A246377(n)). [Respectively, the number of 0-bits for 0/1-swapped version of that sequence].
%F a(n) = A246348(n) - A246370(n) - 1.
%e Consider n=30. It is the 19th composite number in A002808: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, ...
%e Thus we consider next n=19, which is the 8th prime in A000040: 2, 3, 5, 7, 11, 13, 17, 19, ...
%e So we proceed with n=8, which is the 3rd composite number, and then with n=3, which is the 2nd prime, and then with n=2 which is the 1st prime, and we have finished.
%e All in all, it took us 5 steps (A246348(30) = 6 = 5+1) to reach 1, and on the journey, we encountered two composites, 30 and 8, thus a(30) = 2.
%o (Scheme, two versions, the first being a direct recurrence employing memoizing definec-macro from _Antti Karttunen_'s IntSeq-library):
%o (definec (A246369 n) (cond ((= 1 n) 0) ((= 1 (A010051 n)) (A246369 (A000720 n))) (else (+ 1 (A246369 (A065855 n))))))
%o (define (A246369 n) (-1+ (A000120 (A135141 n))))
%Y Cf. A000040, A002808, A000120, A000720, A065855, A135141, A246348, A246370, A246377.
%K nonn
%O 1,9
%A _Antti Karttunen_, Aug 27 2014