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 A038554 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). 19
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS 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. a(A000225(n)) = 0; a(A062289(n)) > 0; a(A038558(n)) = n. - Reinhard Zumkeller, Mar 06 2013 REFERENCES 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 LINKS T. D. Noe, Table of n, a(n) for n=0..4096 C. Kimberling, Fractal sequences J. W. Layman, View the fractal-like graph of a(n) vs. n R. Stephan, Some divide-and-conquer sequences ... R. Stephan, Table of generating functions FORMULA 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) = 2a(n) + [n odd], a(2n+1) = 2a(n) + [n>0 even]. - Ralf Stephan, Oct 20 2003 a(0) = a(1) = 0, a(4n) = 2a(2n), a(4n+2) = 2a(2n+1)+1, a(4n+1) = 2a(2n)+1, a(4n+3) = 2a(2n+1). Proof by Nikolaus Meyberg following a conjecture by Ralf Stephan. EXAMPLE If n=18=10010, derivative is (1+0)(0+0)(0+1)(1+0) = 1011, so a(18)=11. MAPLE 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) ]; MATHEMATICA a = a = 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 *) PROG (Haskell) import Data.Bits (xor) a038554 n = foldr (\d v -> v * 2 + d) 0 \$ zipWith xor bs \$ tail bs    where bs = a030308_row n -- Reinhard Zumkeller, May 26 2013, Mar 06 2013 (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 CROSSREFS Cf. A038570, A038571. See A003415 for another definition of the derivative of a number. Cf. A038556 (rotates n instead of shifting) Cf. A000035. Cf. A030308. Sequence in context: A167666 A115352 A275808 * A100329 A193535 A332645 Adjacent sequences:  A038551 A038552 A038553 * A038555 A038556 A038557 KEYWORD nonn,nice,easy AUTHOR EXTENSIONS More terms from Erich Friedman STATUS approved

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Last modified April 3 19:43 EDT 2020. Contains 333198 sequences. (Running on oeis4.)