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A305614 Expansion of Sum_{p prime} x^p/(1 + x^p).

%I A305614 #20 Aug 01 2018 10:31:59
%S A305614 0,0,1,1,-1,1,0,1,-1,1,0,1,-2,1,0,2,-1,1,0,1,-2,2,0,1,-2,1,0,1,-2,1,
%T A305614 -1,1,-1,2,0,2,-2,1,0,2,-2,1,-1,1,-2,2,0,1,-2,1,0,2,-2,1,0,2,-2,2,0,1,
%U A305614 -3,1,0,2,-1,2,-1,1,-2,2,-1,1,-2,1,0,2,-2,2,-1
%N A305614 Expansion of Sum_{p prime} x^p/(1 + x^p).
%C A305614 a(n) is the number of prime divisors p|n such that n/p is odd, minus the number of prime divisors p|n such that n/p is even.
%H A305614 Robert Israel, <a href="/A305614/b305614.txt">Table of n, a(n) for n = 0..10000</a>
%F A305614 a(n) = -Sum_{p|n prime} (-1)^(n/p).
%F A305614 From _Robert Israel_, Jun 07 2018: (Start)
%F A305614 If n is odd, a(n) = A001221(n).
%F A305614 If n == 2 (mod 4), a(n) = 2 - A001221(n).
%F A305614 If n == 0 (mod 4) and n > 0, a(n) = -A001221(n). (End)
%F A305614 L.g.f.: log(Product_{k>=1} (1 + x^prime(k))^(1/prime(k))) = Sum_{n>=1} a(n)*x^n/n. - _Ilya Gutkovskiy_, Jul 30 2018
%e A305614 The prime divisors of 12 are 2, 3. We see that 12/2 = 6, 12/3 = 4. None of those are odd, but both of them are even, so a(12) = -2.
%e A305614 The prime divisors of 30 are {2,3,5} with quotients {15,10,6}. One of these is odd and two are even, so a(30) = 1 - 2 = -1.
%p A305614 a:= n-> -add((-1)^(n/i[1]), i=ifactors(n)[2]):
%p A305614 seq(a(n), n=0..100);  # _Alois P. Heinz_, Jun 07 2018
%p A305614 # Alternative
%p A305614 N:= 1000: # to get a(0)..a(N)
%p A305614 V:= Vector(N):
%p A305614 p:= 1:
%p A305614 do
%p A305614   p:= nextprime(p);
%p A305614   if p > N then break fi;
%p A305614   R:= [seq(i,i=p..N,p)];
%p A305614   W:= <seq((-1)^(n+1),n=1..nops(R))>;
%p A305614   V[R]:= V[R]+W;
%p A305614 od:
%p A305614 [0,seq(V[i],i=1..N)]; # _Robert Israel_, Jun 07 2018
%t A305614 Table[Sum[If[PrimeQ[d], (-1)^(n/d - 1), 0], {d, Divisors[n]}], {n, 30}]
%Y A305614 Cf. A000005, A000607, A001221, A008683, A048165, A083399, A088705, A106404, A205745, A280954.
%K A305614 sign
%O A305614 0,13
%A A305614 _Gus Wiseman_, Jun 06 2018