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Inverse Moebius transform of bigomega(n).
5

%I #19 May 19 2017 14:37:37

%S 0,1,1,3,1,4,1,6,3,4,1,9,1,4,4,10,1,9,1,9,4,4,1,16,3,4,6,9,1,12,1,15,

%T 4,4,4,18,1,4,4,16,1,12,1,9,9,4,1,25,3,9,4,9,1,16,4,16,4,4,1,24,1,4,9,

%U 21,4,12,1,9,4,12,1,30,1,4,9,9,4,12,1,25,10,4,1,24,4,4,4,16,1,24,4,9,4,4

%N Inverse Moebius transform of bigomega(n).

%C a(n) is the total number of prime factors (counted with multiplicity) over all the divisors of n. - _Geoffrey Critzer_, Feb 03 2015

%H Antti Karttunen, <a href="/A069264/b069264.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = tau(n)*bigomega(n)/2. - _Vladeta Jovovic_, Jan 25 2004

%F G.f.: Sum_{k>=1} bigomega(k)*x^k/(1 - x^k). - _Ilya Gutkovskiy_, Feb 19 2017

%e a(12)=9 because the divisors of 12 are: 1,2,3,4,6,12 and the number (with multiplicity) of prime factors of these divisors is: 0+1+1+2+2+3=9. - _Geoffrey Critzer_, Feb 03 2015

%t Table[Sum[PrimeOmega[d], {d, Divisors[n]}], {n, 1, 94}] (* _Geoffrey Critzer_, Feb 03 2015 *)

%o (PARI) for(n=1,120,print1(sumdiv(n,d,bigomega(d)),","))

%Y Cf. A000005, A001222.

%K easy,nonn

%O 1,4

%A _Benoit Cloitre_, Apr 19 2002