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Dirichlet inverse of Kimberling's paraphrases, A003602.
18

%I #14 Nov 14 2021 17:13:53

%S 1,-1,-2,0,-3,2,-4,0,-1,3,-6,0,-7,4,4,0,-9,1,-10,0,5,6,-12,0,-4,7,-2,

%T 0,-15,-4,-16,0,7,9,6,0,-19,10,8,0,-21,-5,-22,0,3,12,-24,0,-9,4,10,0,

%U -27,2,8,0,11,15,-30,0,-31,16,4,0,9,-7,-34,0,13,-6,-36,0,-37,19,8,0,9,-8,-40,0,-4,21,-42,0,11,22

%N Dirichlet inverse of Kimberling's paraphrases, A003602.

%H Antti Karttunen, <a href="/A349134/b349134.txt">Table of n, a(n) for n = 1..16384</a>

%H Antti Karttunen, <a href="/A349134/a349134.txt">Data supplement: n, a(n) computed for n = 1..65537</a>

%F a(1) = 1; a(n) = -Sum_{d|n, d < n} A003602(n/d) * a(d).

%F a(n) = A349135(n) - A003602(n).

%t k[n_] := (n/2^IntegerExponent[n, 2] + 1)/2; a[1] = 1; a[n_] := a[n] = -DivisorSum[n, a[#]*k[n/#] &, # < n &]; Array[a, 100] (* _Amiram Eldar_, Nov 13 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 A003602(n) = (1+(n>>valuation(n,2)))/2;

%o v349134 = DirInverseCorrect(vector(up_to,n,A003602(n)));

%o A349134(n) = v349134[n];

%Y Cf. A003602, A349135, A349136.

%Y Cf. also A323881, A349125.

%K sign

%O 1,3

%A _Antti Karttunen_, Nov 13 2021