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A065359
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Alternating bit sum for n: replace 2^k with (-1)^k in binary expansion of n.
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41
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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, 3, 4, 2, 3, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2
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OFFSET
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0,6
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COMMENTS
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Notation: (2)[n](-1)
a(n) == 0 (mod 3) iff n == 0 (mod 3).
a(n) == 0 (mod 6) iff (n == 0 (mod 3) and n/3 not in A036556).
a(n) == 3 (mod 6) iff (n == 0 (mod 3) and n/3 in A036556). (End)
First occurrence of k and -k: 0, 1, 2, 5, 10, 21, 42, 85, ..., (A000975); i.e., first 0 occurs for 0, first 1 occurs for 1, first -1 occurs at 2, first 2 occurs for 5, etc.;
a(n)=-3 only if n mod 3 = 0,
a(n)=-2 only if n mod 3 = 1,
a(n)=-1 only if n mod 3 = 2,
a(n)= 0 only if n mod 3 = 0,
a(n)= 1 only if n mod 3 = 1,
a(n)= 2 only if n mod 3 = 2,
a(n)= 3 only if n mod 3 = 0, ..., . (End)
In the Koch curve, number the segments starting with n=0 for the first segment. The net direction (i.e., the sum of the preceding turns) of segment n is a(n)*60 degrees. This is since in the curve each base-4 digit 0,1,2,3 of n is a sub-curve directed respectively 0, +60, -60, 0 degrees, which is the net 0,+1,-1,0 of two bits in the sum here. - Kevin Ryde, Jan 24 2020
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LINKS
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FORMULA
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G.f.: (1/(1-x)) * Sum_{k>=0} (-1)^k*x^2^k/(1+x^2^k). - Ralf Stephan, Mar 07 2003
a(0) = 0, a(2n) = -a(n), a(2n+1) = 1-a(n). - Ralf Stephan, Mar 07 2003
G.f. A(x) satisfies: A(x) = x / (1 - x^2) - (1 + x) * A(x^2). - Ilya Gutkovskiy, Jul 28 2021
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EXAMPLE
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Alternating bit sum for 11 = 1011 in binary is 1 - 1 + 0 - 1 = -1, so a(11) = -1.
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MAPLE
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A065359 := proc(n) local dgs ; dgs := convert(n, base, 2) ; add( -op(i, dgs)*(-1)^i, i=1..nops(dgs)) ; end proc: # R. J. Mathar, Feb 04 2011
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MATHEMATICA
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f[0]=0; f[n_] := Plus @@ (-(-1)^Range[ Floor[ Log2@ n + 1]] Reverse@ IntegerDigits[n, 2]); Array[ f, 107, 0]
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PROG
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(PARI)
SumAD(x)= { local(a=1, s=0); while (x>9, s+=a*(x-10*(x\10)); x\=10; a=-a); return(s + a*x) }
baseE(x, b)= { local(d, e=0, f=1); while (x>0, d=x-b*(x\b); x\=b; e+=d*f; f*=10); return(e) }
{ for (n=0, 1000, b=baseE(n, 2); write("b065359.txt", n, " ", SumAD(b)) ) } \\ Harry J. Smith, Oct 17 2009
(PARI) for(n=0, 106, s=0; u=1; for(k=0, #binary(n)-1, s+=bittest(n, k)*u; u=-u); print1(s, ", ")) /* Washington Bomfim, Jan 18 2011 */
(PARI) a(n) = my(b=binary(n)); b*[(-1)^k|k<-[-#b+1..0]]~; \\ Ruud H.G. van Tol, Oct 16 2023
(Haskell)
a065359 0 = 0
a065359 n = - a065359 n' + m where (n', m) = divMod n 2
(Python)
def a(n):
return sum((-1)**k for k, bi in enumerate(bin(n)[2:][::-1]) if bi=='1')
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CROSSREFS
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KEYWORD
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base,easy,sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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