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A349345 Sum of A109168 and its Dirichlet inverse, where A109168(n) = (n+A006519(n))/2, and A006519 is the highest power of 2 dividing n.

%I A349345 #8 Dec 09 2021 01:31:25
%S A349345 2,0,0,4,0,8,0,8,4,12,0,8,0,16,12,16,0,12,0,12,16,24,0,16,9,28,12,16,
%T A349345 0,8,0,32,24,36,24,20,0,40,28,24,0,12,0,24,26,48,0,32,16,34,36,28,0,
%U A349345 32,36,32,40,60,0,32,0,64,36,64,42,20,0,36,48,24,0,40,0,76,46,40,48,24,0,48,37,84,0,44,54,88,60,48
%N A349345 Sum of A109168 and its Dirichlet inverse, where A109168(n) = (n+A006519(n))/2, and A006519 is the highest power of 2 dividing n.
%H A349345 Antti Karttunen, <a href="/A349345/b349345.txt">Table of n, a(n) for n = 1..20000</a>
%F A349345 a(n) = A109168(n) + A349344(n).
%F A349345 a(1) = 2, and for n > 1, a(n) = -Sum_{d|n, 1<d<n} A109168(d) * A349344(n/d).
%F A349345 For all n >= 1, a(4*n) = 4*A109168(n). - _Antti Karttunen_, Dec 07 2021
%o A349345 (PARI)
%o A349345 up_to = 20000;
%o A349345 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 A349345 A109168(n) = ((n+bitand(n, -n))\2); \\ From A109168
%o A349345 v349344 = DirInverseCorrect(vector(up_to,n,A109168(n)));
%o A349345 A349344(n) = v349344[n];
%o A349345 A349345(n) = (A109168(n)+A349344(n));
%Y A349345 Cf. A109168, A349344.
%K A349345 nonn,look
%O A349345 1,1
%A A349345 _Antti Karttunen_, Nov 15 2021