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Alternating sum of the binary expansion of n (row n of A030190). Replace 2^k with (-1)^(A070939(n)-k) in the binary expansion of n (compare to the definition of A065359).
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%I #17 Jul 20 2021 03:30:45

%S 0,1,1,0,1,2,0,1,1,0,2,1,0,-1,1,0,1,2,0,1,2,3,1,2,0,1,-1,0,1,2,0,1,1,

%T 0,2,1,0,-1,1,0,2,1,3,2,1,0,2,1,0,-1,1,0,-1,-2,0,-1,1,0,2,1,0,-1,1,0,

%U 1,2,0,1,2,3,1,2,0,1,-1,0,1,2,0,1,2,3,1,2

%N Alternating sum of the binary expansion of n (row n of A030190). Replace 2^k with (-1)^(A070939(n)-k) in the binary expansion of n (compare to the definition of A065359).

%C The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

%F a(n) = (-1)^(A070939(n)-1)*A065359(n).

%e The binary expansion of 53 is (1,1,0,1,0,1), so a(53) = 1 - 1 + 0 - 1 + 0 - 1 = -2.

%t ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];

%t Table[ats[IntegerDigits[n,2]],{n,0,100}]

%o (PARI) a(n) = subst(Pol(Vecrev(binary(n))), x, -1); \\ _Michel Marcus_, Jul 19 2021

%o (Python)

%o def a(n): return sum((-1)**k for k, bi in enumerate(bin(n)[2:]) if bi=='1')

%o print([a(n) for n in range(84)]) # _Michael S. Branicky_, Jul 19 2021

%Y Binary expansions of each nonnegative integer are the rows of A030190.

%Y The positions of 0's are A039004.

%Y The version for prime factors is A071321 (reverse: A071322).

%Y Positions of first appearances are A086893.

%Y The version for standard compositions is A124754 (reverse: A344618).

%Y The version for prime multiplicities is A316523.

%Y The version for prime indices is A316524 (reverse: A344616).

%Y A003714 lists numbers with no successive binary indices.

%Y A070939 gives the length of an integer's binary expansion.

%Y A103919 counts partitions by sum and alternating sum.

%Y A328594 lists numbers whose binary expansion is aperiodic.

%Y A328595 lists numbers whose reversed binary expansion is a necklace.

%Y Cf. A000037, A000070, A000120, A027187, A028260, A065359, A069010, A116406, A121016, A191232, A245563, A344609, A344619.

%K sign

%O 0,6

%A _Gus Wiseman_, Jul 14 2021