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Number of odd prime factors of n, counted with repetitions.
33

%I #40 Jun 13 2024 10:50:52

%S 0,0,1,0,1,1,1,0,2,1,1,1,1,1,2,0,1,2,1,1,2,1,1,1,2,1,3,1,1,2,1,0,2,1,

%T 2,2,1,1,2,1,1,2,1,1,3,1,1,1,2,2,2,1,1,3,2,1,2,1,1,2,1,1,3,0,2,2,1,1,

%U 2,2,1,2,1,1,3,1,2,2,1,1,4,1,1,2,2,1,2,1,1,3,2,1,2,1,2,1,1,2,3,2,1,2

%N Number of odd prime factors of n, counted with repetitions.

%C Number of parts larger than 1 in the partition with Heinz number n. The Heinz number of an integer partition p = [p_1, p_2, ..., p_r] is defined as Product(p_j-th prime, j=1...r) (concept used by _Alois P. Heinz_ in A215366 as an "encoding" of a partition). Example: a(9) = 2 because the partition with Heinz number 9 (=3*3) is [2,2]. - _Emeric Deutsch_, Oct 02 2015

%C Totally additive because both A001222 and A007814 are. a(2) = 0, and a(p) = 1 for odd primes p, a(m*n) = a(m)+a(n) for m, n > 1. - _Antti Karttunen_, Jul 10 2020

%H Antti Karttunen, <a href="/A087436/b087436.txt">Table of n, a(n) for n = 1..65537</a> (first 1000 terms from G. C. Greubel)

%F a(n) = A001222(n) - A007814(n).

%F a(n) = A001222(A000265(n)). - _Antti Karttunen_, Jul 10 2020

%e a(9) = 2 because 9 = 3*3 has 2 odd prime factors. - _Emeric Deutsch_, Oct 02 2015

%p seq(bigomega(n) - padic[ordp](n, 2), n=1..102); # _Peter Luschny_, Dec 06 2017

%t Join[{0},Table[Length[Select[Flatten[Table[#[[1]],{#[[2]]}]&/@ FactorInteger[ n]],OddQ]],{n,2,110}]] (* _Harvey P. Dale_, Feb 01 2013 *)

%o (PARI) a(n) = bigomega(n) - valuation(n, 2); \\ _Michel Marcus_, Sep 10 2019

%o (PARI) A087436(n) = (bigomega(n>>valuation(n,2))); \\ _Antti Karttunen_, Jul 10 2020

%Y Cf. A000244 (the first occurrence of each n, and also the positions of records).

%Y Cf. A000265, A001222, A005087, A007814, A215366, A336118, A336161.

%K nonn

%O 1,9

%A _Reinhard Zumkeller_, Sep 03 2003