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A139351
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Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives e(n).
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18
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0, 1, 0, 1, 1, 2, 1, 2, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 1, 2, 2, 3, 2, 3, 0, 1, 0, 1, 1, 2, 1, 2, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 1, 2, 2, 3, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 2, 3, 3, 4, 3, 4, 1, 2, 1
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OFFSET
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0,6
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COMMENTS
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e(n)+o(n) = A000120(n), the binary weight of n.
a(n) is also number of 1's and 3's in 4-ary representation of n. - Frank Ruskey, May 02 2009
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LINKS
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FORMULA
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G.f.: (1/(1-z))*Sum_{m>=0} (z^(4^m)/(1+z^(4^m))). - Frank Ruskey, May 03 2009
Recurrence relation: a(0)=0, a(4m) = a(4m+2) = a(m), a(4m+1) = a(4m+3) = 1+a(m). - Frank Ruskey, May 11 2009
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EXAMPLE
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For n = 43 = 2^0 + 2^1 + 2^3 + 2^5, e(43)=1, o(43)=3.
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MAPLE
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local a, bdgs, r;
a := 0 ;
bdgs := convert(n, base, 2) ;
for r from 1 to nops(bdgs) by 2 do
if op(r, bdgs) = 1 then
a := a+1 ;
end if;
end do:
a;
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MATHEMATICA
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terms = 99; s = (1/(1-z))*Sum[z^(4^m)/(1+z^(4^m)), {m, 0, Log[4, terms] // Ceiling}] + O[z]^terms; CoefficientList[s, z] (* Jean-François Alcover, Jul 21 2017 *)
a[0] = 0; a[n_] := a[n] = a[Floor[n/4]] + If[OddQ[Mod[n, 4]], 1, 0]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)
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PROG
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(Fortran) See Sloane link.
(Haskell)
import Data.List (unfoldr)
a139351 = sum . map (`mod` 2) .
unfoldr (\x -> if x == 0 then Nothing else Just (x, x `div` 4)
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CROSSREFS
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Cf. A000120, A030308, A059841, A139352, A139353, A139354, A139355, A039004, A139370, A139371, A139372, A139373.
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KEYWORD
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nonn,base,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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