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A139352
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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 o(n).
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16
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0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 3, 3, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 3, 3, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2
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OFFSET
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0,11
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COMMENTS
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e(n) + o(n) = A000120(n), the binary weight of n.
a(n) is also the number of 2's and 3's in the 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^(2*4^m)/(1+(2*4^m))). - Frank Ruskey, May 03 2009
Recurrence relation: a(0)=0, a(4m) = a(4m+1) = a(m), a(4m+2) = 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. [Typo fixed by Reinhard Zumkeller, Apr 22 2011]
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MAPLE
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local a, bdgs, r;
a := 0 ;
bdgs := convert(n, base, 2) ;
for r from 2 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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a[n_] := Count[Position[Reverse@IntegerDigits[n, 2], 1]-1, {_?OddQ}];
a[0] = 0; a[n_] := a[n] = a[Floor[n/4]] + If[Mod[n, 4] > 1, 1, 0]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)
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PROG
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See link in A139351 for Fortran program.
(Haskell)
import Data.List (unfoldr)
a139352 = sum . map ((`div` 2) . (`mod` 4)) .
unfoldr (\x -> if x == 0 then Nothing else Just (x, x `div` 4))
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CROSSREFS
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Cf. A000035, A000120, A030308, A039004, A139351, A139353, A139354, A139355, 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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STATUS
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approved
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