%I #12 Oct 01 2021 22:09:03
%S 1,-1,-2,-6,-2,2,-4,0,-15,2,-2,12,-4,4,4,0,-2,15,-4,12,8,2,-6,0,-35,4,
%T 0,24,-2,-4,-6,0,4,2,8,90,-4,4,8,0,-2,-8,-4,12,30,6,-6,0,-77,35,4,24,
%U -6,0,4,0,8,2,-2,-24,-6,6,60,0,8,-4,-4,12,12,-8,-2,0,-6,4,70,24,8,-8,-4,0,0,2,-6,-48,4,4,4,0,-8
%N Dirichlet inverse of A347236.
%C Multiplicative because A347236 is.
%C It seems that A046099 gives the positions of zeros.
%C This follows from the formula for a(p^e). - _Sebastian Karlsson_, Sep 01 2021
%H Antti Karttunen, <a href="/A347238/b347238.txt">Table of n, a(n) for n = 1..20000</a>
%H <a href="/index/Pri#gaps">Index entries for primes, gaps between</a>
%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%F a(1) = 1; a(n) = -Sum_{d|n, d < n} A347236(n/d) * a(d).
%F a(n) = A347239(n) - A347236(n).
%F For all n >= 1, a(A000040(n)) = -A001223(n).
%F Multiplicative with a(p^e) = p - A151800(p) if e = 1, -p*A151800(p) if e = 2 and 0 if e > 2. - _Sebastian Karlsson_, Sep 01 2021
%o (PARI)
%o up_to = 16384;
%o DirInverseCorrect(v) = { my(u=vector(#v)); u[1] = (1/v[1]); for(n=2, #v, u[n] = (-u[1]*sumdiv(n, d, if(d<n, v[n/d]*u[d], 0)))); (u) }; \\ Compute the Dirichlet inverse of the sequence given in input vector v.
%o A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
%o A061019(n) = (((-1)^bigomega(n))*n);
%o A347236(n) = sumdiv(n,d,A061019(d)*A003961(n/d));
%o v347238 = DirInverseCorrect(vector(up_to,n,A347236(n)));
%o A347238(n) = v347238[n];
%Y Cf. A000040, A001223, A003961, A046099, A061019, A347236, A347239.
%Y Cf. A151800.
%K sign,mult
%O 1,3
%A _Antti Karttunen_, Aug 24 2021