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A038554
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Derivative of n: write n in binary, replace each pair of adjacent bits with their mod 2 sum (a(0)=a(1)=0 by convention). Also n XOR (n shift 1).
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30
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0, 0, 1, 0, 2, 3, 1, 0, 4, 5, 7, 6, 2, 3, 1, 0, 8, 9, 11, 10, 14, 15, 13, 12, 4, 5, 7, 6, 2, 3, 1, 0, 16, 17, 19, 18, 22, 23, 21, 20, 28, 29, 31, 30, 26, 27, 25, 24, 8, 9, 11, 10, 14, 15, 13, 12, 4, 5, 7, 6, 2, 3, 1, 0, 32, 33, 35, 34, 38, 39, 37, 36, 44, 45, 47, 46, 42, 43, 41, 40, 56, 57
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OFFSET
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0,5
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COMMENTS
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From Antti Karttunen: this is also a version of A003188: a(n) = A003188(n) - 2^floor(log_2(A003188(n))), that is, the corresponding Gray code expansion, but with highest 1-bit turned off. Also a(n) = A003188(n) - 2^floor(log_2(n)).
From John W. Layman: {a(n)} is a self-similar sequence under Kimberling's "upper-trimming" operation.
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REFERENCES
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Hsien-Kuei Hwang, S Janson, TH Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016; http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf. Also Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585
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LINKS
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FORMULA
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If 2*2^k <= n < 3*2^k then a(n) = 2^k + a(2^(k+2)-n-1); if 3*2^k <= n < 4*2^k then a(n) = a(n-2^(k+1)). - Henry Bottomley, May 11 2000
G.f.: (1/(1-x)) * Sum_{k>=0} 2^k*(t^4-t^3+t^2)/(1+t^2), t=x^2^k. - Ralf Stephan, Sep 10 2003
a(0)=0, a(2n) = 2*a(n) + [n odd], a(2n+1) = 2*a(n) + [n>0 even]. - Ralf Stephan, Oct 20 2003
a(0) = a(1) = 0, a(4n) = 2*a(2n), a(4n+2) = 2*a(2n+1)+1, a(4n+1) = 2*a(2n)+1, a(4n+3) = 2*a(2n+1). Proof by Nikolaus Meyberg following a conjecture by Ralf Stephan.
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EXAMPLE
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If n = 18 = 10010_2, derivative is (1+0)(0+0)(0+1)(1+0) = 1011_2, so a(18)=11.
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MAPLE
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A038554 := proc(n) local i, b, ans; ans := 0; b := convert(n, base, 2); for i to nops(b)-1 do ans := ans+((b[ i ]+b[ i+1 ]) mod 2)*2^(i-1); od; RETURN(ans); end; [ seq(A038554(i), i=0..100) ];
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MATHEMATICA
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a[0] = a[1] = 0; a[n_ /; Mod[n, 4] == 0] := a[n] = 2*a[n/2]; a[n_ /; Mod[n, 4] == 1] := a[n] = 2*a[(n-1)/2] + 1; a[n_ /; Mod[n, 4] == 2] := a[n] = 2*a[n/2] + 1; a[n_ /; Mod[n, 4] == 3] := a[n] = 2*a[(n-1)/2]; Table[a[n], {n, 0, 81}] (* Jean-François Alcover, Jul 13 2012, after Ralf Stephan *)
Table[FromDigits[Mod[Total[#], 2]&/@Partition[IntegerDigits[n, 2], 2, 1], 2], {n, 0, 90}] (* Harvey P. Dale, Oct 27 2015 *)
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PROG
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(Haskell)
import Data.Bits (xor)
a038554 n = foldr (\d v -> v * 2 + d) 0 $ zipWith xor bs $ tail bs
where bs = a030308_row n
(PARI) a003188(n)=bitxor(n, n>>1);
a(n)=if(n<2, 0, a003188(n) - 2^logint(a003188(n), 2)); \\ Indranil Ghosh, Apr 26 2017
(Python)
import math
def a003188(n): return n^(n>>1)
def a(n): return 0 if n<2 else a003188(n) - 2**int(math.floor(math.log(a003188(n), 2))) # Indranil Ghosh, Apr 26 2017
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CROSSREFS
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Cf. A038556 (rotates n instead of shifting).
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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