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%I #28 Jul 26 2023 21:02:14
%S 1,1,-1,1,1,-1,1,1,-1,1,-1,-1,1,1,-1,1,1,-1,1,1,-1,-1,1,-1,1,1,-1,1,1,
%T -1,1,1,-1,1,-1,-1,1,1,-1,1,-1,-1,1,-1,-1,1,-1,-1,1,1,-1,1,1,-1,1,1,
%U -1,1,-1,-1,1,1,-1,1,1,-1,1,1,-1,-1,1,-1,1,1,-1,1,1,-1,1,1,-1,-1,1,-1,-1,1
%N a(n) = (-1)^A055941(n).
%C This sequence and A246017 are the subject of the Lafrance et al. (2014) paper.
%H Philip Lafrance, Narad Rampersad, Randy Yee, <a href="http://arxiv.org/abs/1408.2277">Some properties of a Rudin-Shapiro-like sequence</a>, arXiv:1408.2277 [math.CO], 2014.
%t b[n_] := b[n] = If[n == 0, 0, If[EvenQ[n], b[n/2] + DigitCount[n/2, 2, 1], b[(n-1)/2] + 1]];
%t a55941[n_] := b[n] - DigitCount[n, 2, 1];
%t a[n_] := (-1)^a55941[n];
%t Table[a[n], {n, 0, 100}] (* _Jean-François Alcover_, Sep 23 2018 *)
%o (PARI) a055941(n) = {my(b=binary(n)); nb = 0; for (i=1, #b-1, if (b[i], nb += sum(j=i+1, #b, !b[j]));); nb;}
%o a(n) = (-1)^a055941(n);
%o (Python)
%o def A246016(n):
%o a, b = 0, 0
%o for i, j in enumerate(bin(n)[:1:-1]):
%o if int(j):
%o a ^= (i&1)^b
%o b ^= 1
%o return -1 if a else 1 # _Chai Wah Wu_, Jul 26 2023
%Y Cf. A055941, A161511, A246017 (partial sums).
%K sign
%O 0
%A _Michel Marcus_ and _N. J. A. Sloane_, Aug 13 2014
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