

A246369


a(1)=0, a(p_n) = a(n), a(c_n) = 1 + a(n), where p_n = nth prime = A000040(n), c_n = nth composite number = A002808(n); Also one less than the binary weight of terms of A135141.


5



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, 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, 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
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OFFSET

1,9


COMMENTS

Consider the following algorithm:
Start:
If n is 1, we have finished,
Otherwise:
If n is a prime, replace it with its index among the primes, n < A000720(n), and go back to the start.
Otherwise, if n is a composite, replace it with its index among the composites, n < A065855(n), and go back to the start.
At some point, the process is guaranteed to reach the number 1 at which point we stop.
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.


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..32998


FORMULA

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].
a(n) = A000120(A135141(n))  1. [a(n) is also one less than the Hamming weight (number of 1bits) of the nth term of A135141].
a(n) = A080791(A246377(n)). [Respectively, the number of 0bits for 0/1swapped version of that sequence].
a(n) = A246348(n)  A246370(n)  1.


EXAMPLE

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, ...
Thus we consider next n=19, which is the 8th prime in A000040: 2, 3, 5, 7, 11, 13, 17, 19, ...
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.
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.


PROG

(Scheme, two versions, the first being a direct recurrence employing memoizing definecmacro from Antti Karttunen's IntSeqlibrary):
(definec (A246369 n) (cond ((= 1 n) 0) ((= 1 (A010051 n)) (A246369 (A000720 n))) (else (+ 1 (A246369 (A065855 n))))))
(define (A246369 n) (1+ (A000120 (A135141 n))))


CROSSREFS

Cf. A000040, A002808, A000120, A000720, A065855, A135141, A246348, A246370, A246377.
Sequence in context: A287360 A035443 A180430 * A036261 A140575 A091917
Adjacent sequences: A246366 A246367 A246368 * A246370 A246371 A246372


KEYWORD

nonn


AUTHOR

Antti Karttunen, Aug 27 2014


STATUS

approved



