A345927 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).
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, 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, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, 2, 3, 1, 2
Offset: 0
Keywords
Examples
The binary expansion of 53 is (1,1,0,1,0,1), so a(53) = 1 - 1 + 0 - 1 + 0 - 1 = -2.
Crossrefs
Binary expansions of each nonnegative integer are the rows of A030190.
The positions of 0's are A039004.
Positions of first appearances are A086893.
The version for prime multiplicities is A316523.
A003714 lists numbers with no successive binary indices.
A070939 gives the length of an integer's binary expansion.
A103919 counts partitions by sum and alternating sum.
A328594 lists numbers whose binary expansion is aperiodic.
A328595 lists numbers whose reversed binary expansion is a necklace.
Programs
-
Mathematica
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}]; Table[ats[IntegerDigits[n,2]],{n,0,100}]
-
PARI
a(n) = subst(Pol(Vecrev(binary(n))), x, -1); \\ Michel Marcus, Jul 19 2021
-
Python
def a(n): return sum((-1)**k for k, bi in enumerate(bin(n)[2:]) if bi=='1') print([a(n) for n in range(84)]) # Michael S. Branicky, Jul 19 2021
Comments