

A091304


a(n) = Omega(2n1) (number of prime factors of the nth odd number, counted with multiplicity).


6



0, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 3, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 2, 1, 2, 2, 1, 1, 3, 2, 1, 2, 1, 1, 3, 2, 1, 4, 1, 2, 2, 1, 2, 2, 2, 1, 3, 1, 1, 3, 1, 1, 2, 1, 2, 3, 2, 2, 2, 3, 1, 2, 1, 2, 4, 1, 1, 2, 2, 2, 3, 1, 1, 3, 2, 1, 2, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 1, 2, 2, 2, 4, 1, 1, 3, 1, 1, 2, 2, 2, 3, 2
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OFFSET

1,5


COMMENTS

Omega(n) of the odd integers follows a pattern similar to A001222, with 4 maxima instead of 2  i.e. between 2^n and (2^(n+1)  1) there are two numbers with exactly n factors (2^n and 2^(n1) * 3) while the odd integers have 4 maxima (3^n, 3^(n1) * 5, 3^(n1) * 7, 5^2*3^(n2)) between 3^n and 3^(n+1)  1.


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..10000


FORMULA

a(n) = Omega(2n1). [Odd bisection of A001222.]
From Antti Karttunen, May 31 2017: (Start)
For n >= 1, a(n) = A000120(A244153(n)).
For n >= 2, a(n) = 1+A285716(n).
(End)


EXAMPLE

Omega(1) = 0, Omega(9) = 2 (3 * 3 = 9), Omega (243) = 5 (3 * 3 * 3 * 3 * 3 = 243), Omega(51) = 2 (3 * 17 = 51).
For n = 92, A001222(2*92  1) = A001222(183) = 2 as 183 = 3*61, thus a(92) = 2.  Antti Karttunen, May 31 2017


PROG

(PARI) a(n) = bigomega(2*n1) \\ Michel Marcus, Jul 26 2013, edited to reflect the changed starting offset by Antti Karttunen, May 31 2017


CROSSREFS

One more than A285716 (after the initial term).
Cf. A001222, A000120, A092523, A099774, A099990, A244153, A278223, A286582.
Cf. A006254 (positions of ones).
Sequence in context: A232550 A091853 A193773 * A049847 A255274 A025431
Adjacent sequences: A091301 A091302 A091303 * A091305 A091306 A091307


KEYWORD

easy,nonn


AUTHOR

Andrew Plewe, Feb 20 2004


EXTENSIONS

Starting offset changed to 1 and the definition modified respectively. Also values of the initial term and of term a(92) (= 2, previously a(91) = 1) corrected by Antti Karttunen, May 31 2017


STATUS

approved



