0,5

FromAntti Karttunen:this is also a version ofA003188: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)).

FromJohn 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

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

T. D. Noe,Table of n, a(n) for n = 0..4096

Clark Kimberling,Fractal sequences.

Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang,Finding structure in sequences of real numbers via graph theory: a problem list,arXiv:2012.04625 [math.CO], 2020-2021.

J. W. Layman,View the fractal-like graph of a(n) vs. n.

Ralf Stephan,Some divide-and-conquer sequences....

Ralf Stephan,Table of generating functions.

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 byRalf Stephan.

If n = 18 = 10010_2, derivative is (1+0)(0+0)(0+1)(1+0) = 1011_2, so a(18)=11.

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) ];

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, afterRalf Stephan*)

Table[FromDigits[Mod[Total[#], 2]&/@Partition[IntegerDigits[n, 2], 2, 1], 2], {n, 0, 90}] (*Harvey P. Dale,Oct 27 2015 *)

(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

Cf.A038570,A038571.SeeA003415for another definition of the derivative of a number.

Cf.A038556(rotates n instead of shifting).

Cf.A000035.

Cf.A030308.

Sequence in context:A167666A115352A275808*A362160A356120A100329

Adjacent sequences:A038551A038552A038553*A038555A038556A038557

nonn,base,look,nice,easy

More terms fromErich Friedman

approved