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A324100 A000120-deficiency of n permuted by A156552: a(n) = A192895(A156552(n)).

%I A324100 #9 Feb 19 2019 18:18:58
%S A324100 -1,0,-1,1,-1,2,-2,2,1,3,-2,4,-1,2,1,5,-2,6,-2,6,4,7,-3,5,4,2,3,8,3,9,
%T A324100 -4,2,5,5,1,10,-1,12,2,11,-2,12,-2,2,19,13,-4,8,0,12,-2,14,-3,11,-3,
%U A324100 14,12,15,-3,16,7,2,5,5,4,17,4,2,-2,18,1,19,6,2,5,8,4,20,-4,10,8,21,8,20,24,42,15,22,7,16,6,28,45,20,0,23,1,12,1
%N A324100 A000120-deficiency of n permuted by A156552: a(n) = A192895(A156552(n)).
%H A324100 Antti Karttunen, <a href="/A324100/b324100.txt">Table of n, a(n) for n = 2..4473</a>
%H A324100 <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%H A324100 <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%F A324100 a(n) = A192895(A156552(n)).
%o A324100 (PARI)
%o A324100 A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
%o A324100 A156552(n) = if(1==n, 0, if(!(n%2), 1+(2*A156552(n/2)), 2*A156552(A064989(n))));
%o A324100 A192895(n) = sumdiv(n, d, hammingweight(d)*(-1)^(d==n));
%o A324100 A324100(n) = A192895(A156552(n));
%Y A324100 Cf. A000120, A156552, A192895, A324113 (sign of each term).
%Y A324100 Cf. A324101, A324102 (positions of nonnegative and negative terms).
%Y A324100 Cf. also A324114, A324194.
%K A324100 sign
%O A324100 2,6
%A A324100 _Antti Karttunen_, Feb 19 2019