A053644 -id:A053644 - cp's OEIS frontend

A003188 Decimal equivalent of Gray code for n.

0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8, 24, 25, 27, 26, 30, 31, 29, 28, 20, 21, 23, 22, 18, 19, 17, 16, 48, 49, 51, 50, 54, 55, 53, 52, 60, 61, 63, 62, 58, 59, 57, 56, 40, 41, 43, 42, 46, 47, 45, 44, 36, 37, 39, 38, 34, 35, 33, 32, 96, 97, 99, 98, 102, 103, 101

Offset: 0

N. J. A. Sloane

Inverse of sequence A006068 considered as a permutation of the nonnegative integers, i.e., A006068(A003188(n)) = n = A003188(A006068(n)). - Howard A. Landman, Sep 25 2001
Restricts to a permutation of each {2^(i - 1) .. 2^i - 1}. - Jason Kimberley, Apr 02 2012
a(n) mod 2 = floor(((n + 1) mod 4) / 2), see also A021913. - Reinhard Zumkeller, Apr 28 2012
Invented by Emile Baudot (1845-1903), originally called a "cyclic-permuted" code. Gray codes are named after Frank Gray, who patented their use for shaft encoders in 1953. [F. Gray, "Pulse Code Communication", U.S. Patent 2,632,058, March 17, 1953.] - Robert G. Wilson v, Jun 22 2014
For n >= 2, let G_n be the graph whose vertices are labeled as 0,1,...,2^n-1, and two vertices are adjacent if and only if their binary expansions differ in exactly one bit, then a(0),a(1),...,a(2^n-1),a(0) is a Hamilton cycle in G_n. - Jianing Song, Jun 01 2022

			For n = 13, the binary reflected Gray code representation of n is '1011' and 1011_2 = 11_10. So, a(13) = 11. - _Indranil Ghosh_, Jan 23 2017

M. Gardner, Mathematical Games, Sci. Amer. Vol. 227 (No. 2, Feb. 1972), p. 107.
M. Gardner, Knotted Doughnuts and Other Mathematical Entertainments. Freeman, NY, 1986, p. 15.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Indranil Ghosh, Table of n, a(n) for n = 0..32767 (first 1001 terms from N. J. A. Sloane)
M. W. Bunder, K. P. Tognetti, and G. E. Wheeler, On binary reflected Gray codes and functions, Discr. Math., 308 (2008), 1690-1700.
Hsien-Kuei Hwang, S. Janson, and T.-H. 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.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, 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.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 39.
P. Mathonet, M. Rigo, M. Stipulanti and N. Zénaïdi, On digital sequences associated with Pascal's triangle, arXiv:2201.06636 [math.NT], 2022.
J. A. Oteo and J. Ros, A Fractal Set from the Binary Reflected Gray Code, J. Phys. A: Math Gen. 38 (2005) 8935-8949.
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003.
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)]
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Paul Tarau, Isomorphic Data Encodings and their Generalization to Hylomorphisms on Hereditarily Finite Data Types
Pierluigi Vellucci and Alberto Maria Bersani, Ordering of nested square roots of 2 according to Gray code, arXiv:1604.00222 [math.NT], 2016.
Index entries for sequences that are permutations of the natural numbers

a(2*A003714(n)) = 3*A003714(n) for all n. - Antti Karttunen, Apr 26 1999
Cf. A014550 (in binary), A055975 (first differences), A048724 (even bisection), A065621 (odd bisection).
Cf. A006068, A038554, A048641, A048642.

C
```
int a(int n) { return n ^ (n>>1); }
```

import Data.Bits (xor, shiftR)
a003188 n = n `xor` (shiftR n 1) :: Integer
-- Reinhard Zumkeller, May 26 2013, Apr 28 2012

// A recursive algorithm
N := 10; s := [[]];
for n in [1..N] do
for j in [#s..1 by -1] do
   Append(~s,Append(s[j],1));
   Append(~s[j],0);
end for;
end for;
[SequenceToInteger(b,2):b in s]; // Jason Kimberley, Apr 02 2012

// A direct algorithm
I2B := func< i | [b eq 1: b in IntegerToSequence(i,2)]>;
B2I := func< s | SequenceToInteger([b select 1 else 0:b in s],2)>;
[B2I(Xor(I2B(i),I2B(i div 2)cat[false])):i in [1..127]]; //Jason Kimberley, Apr 02 2012

with(combinat); graycode(6); # to produce first 64 terms
printf(cat(` %.6d`$64), op(map(convert, graycode(6), binary))); lprint(); # to produce binary strings
# alternative:
read("transforms"):
A003188 := proc(n)
    XORnos(n,floor(n/2)) ;
end proc: # R. J. Mathar, Mar 09 2015
# another Maple program:
a:= n-> Bits[Xor](n, iquo(n, 2)):
seq(a(n), n=0..70);  # Alois P. Heinz, Aug 16 2020

f[n_] := BitXor[n, Floor[n/2]]; Array[f, 70, 0] (* Robert G. Wilson v, Jun 09 2010 *)

PARI
```
a(n)=bitxor(n,n>>1);
```

a(n)=sum(k=1,n,(-1)^((k/2^valuation(k,2)-1)/2)*2^valuation(k,2))

def A003188(n):
    return int(bin(n^(n//2))[2:],2) # Indranil Ghosh, Jan 23 2017

def A003188(n): return n^ n>>1 # Chai Wah Wu, Jun 29 2022

maxn <- 63 # by choice
a <- 1
for(n in 1:maxn){ a[2*n  ] <- 2*a[n] + (n%%2 != 0)
                  a[2*n+1] <- 2*a[n] + (n%%2 == 0)}
(a <- c(0,a))
# Yosu Yurramendi, Apr 10 2020
(C#)
static uint a(this uint n) => (n >> 1) ^ n; // Frank Hollstein, Mar 12 2021

a(n) = 2*a(floor(n/2)) + A021913(n - 1). - Henry Bottomley, Apr 05 2001
a(n) = n XOR floor(n/2), where XOR is the binary exclusive OR operator. - Paul D. Hanna, Jun 04 2002
G.f.: (1/(1-x)) * Sum_{k>=0} 2^k*x^2^k/(1 + x^2^(k+1)). - Ralf Stephan, May 06 2003
a(0) = 0, a(2n) = 2a(n) + [n odd], a(2n + 1) = 2a(n) + [n even]. - Ralf Stephan, Oct 20 2003
a(0) = 0, a(n) = 2 a(floor(n/2)) + mod(floor((n + 1)/2), 2).
a(n) = Sum_{k=1..n} 2^A007814(k) * (-1)^((k/2^A007814(k) - 1)/2). - Ralf Stephan, Oct 29 2003
a(0) = 0, a(n + 1) = a(n) XOR 2^A007814(n) - Jaume Simon Gispert (jaume(AT)nuem.com), Sep 11 2004
Inverse of sequence A006068. - Philippe Deléham, Apr 29 2005
a(n) = a(n-1) XOR A006519(n). - Franklin T. Adams-Watters, Jul 18 2011
From Mikhail Kurkov, Sep 27 2023: (Start)
a(2^m + k) = a(2^m - k - 1) + 2^m for 0 <= k < 2^m, m >= 0.
a(n) = a(A053645(A054429(n))) + A053644(n) for n > 0.
a(n) = A063946(a(A053645(n)) + A053644(n)) for n > 0. (End)

A054429 Simple self-inverse permutation of natural numbers: List each block of 2^n numbers (from 2^n to 2^(n+1) - 1) in reverse order.

Original entry on oeis.org

1, 3, 2, 7, 6, 5, 4, 15, 14, 13, 12, 11, 10, 9, 8, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 127, 126, 125, 124, 123, 122, 121

Offset: 1

Antti Karttunen

a(n) gives the position of the inverse of the n-th term in the full Stern-Brocot tree: A007305(a(n)+2) = A047679(n) and A047679(a(n)) = A007305(n+2). - Reinhard Zumkeller, Dec 22 2008
From Gary W. Adamson, Jun 21 2012: (Start)
The mapping and conversion rules are as follows:
By rows, we have ...
   1;
   3,  2;
   7,  6,  5,  4;
  15, 14, 13, 12, 11, 10, 9, 8;
  ... onto which we are to map one-half of the Stern-Brocot infinite Farey Tree:
  1/2
  1/3, 2/3
  1/4, 2/5, 3/5, 3/4
  1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5
  ...
The conversion rules are: Convert the decimal to binary, adding a duplicate of the rightmost binary term to its right. For example, 10 = 1010, which becomes 10100. Then, from the left, record the number of runs = [1,1,1,2], the continued fraction representation of 5/8. Check: 10 decimal corresponds to 5/8 as shown in the overlaid mapping. Take decimal 9 = 1001 which becomes 10011, with a continued fraction representation of [1,2,2] = 5/7. Check: 9 decimal corresponds to 5/7 in the Farey Tree map. (End)
From Indranil Ghosh, Jan 19 2017: (Start)
a(n) is the value generated when n is converted into its Elias gamma code, the 1's and 0's are interchanged and the resultant is converted back to its decimal value for all values of n > 1. For n = 1, A054429(n) = 1 but after converting 1 to Elias gamma code, interchanging the 1's and 0's and converting it back to decimal, the result produced is 0.
For example, let n = 10. The Elias gamma code for 10 is '1110010'. After interchanging the 1's and 0's it becomes "0001101" and 1101_2 = 13_10. So a(10) = 13. (End)
From Yosu Yurramendi, Mar 09 2017 (similar to Zumkeller's comment): (Start)
A002487(a(n)) = A002487(n+1), A002487(a(n)+1) = A002487(n), n > 0.
A162909(a(n)) = A162910(n), A162910(a(n)) = A162909(n), n > 0.
A162911(a(n)) = A162912(n), A162912(a(n)) = A162911(n), n > 0.
A071766(a(n)) = A245326(n), A245326(a(n)) = A071766(n), n > 0.
A229742(a(n)) = A245325(n), A245325(a(n)) = A229742(n), n > 0.
A020651(a(n)) = A245327(n), A245327(a(n)) = A020651(n), n > 0.
A020650(a(n)) = A245328(n), A245328(a(n)) = A020650(n), n > 0. (End)
From Yosu Yurramendi, Mar 29 2017: (Start)
A063946(a(n)) = a(A063946(n)) = A117120(n), n > 0.
A065190(a(n)) = a(A065190(n)) = A092569(n), n > 0.
A258746(a(n)) = a(A258746(n)) = A165199(n), n > 0.
A258996(a(n)) = a(A258996(n)), n > 0.
A117120(a(n)) = a(A117120(n)), n > 0.
A092569(a(n)) = a(A092569(n)), n > 0. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Index entries for sequences that are permutations of the natural numbers
Index entries for sequences related to binary expansion of n

See also A054424, A054430.
{A000027, A054429, A059893, A059894} form a 4-group.
Cf. A053644, A115303, A115304, A115305, A115306, A115307, A115308, A115309, A106649.
This is Guy Steele's sequence GS(6, 5) (see A135416).

a054429 n = a054429_list !! (n-1)
a054429_list = f [1..] where
   f xs@(x:_) = reverse us ++ f vs where (us, vs) = splitAt x xs
-- Reinhard Zumkeller, Jun 01 2015, Feb 21 2014

A054429 := n -> 3*2^ilog2(n) - n - 1:
seq(A054429(n), n = 1..70); # [Updated by Peter Luschny, Apr 24 2024]

Flatten[Table[Range[2^(n+1)-1,2^n,-1],{n,0,6}]] (* Harvey P. Dale, Dec 17 2013 *)

A054429(n)= 3<<#binary(n\2)-n-1 \\ M. F. Hasler, Aug 18 2014

from itertools import count, islice
def A054429_gen(): # generator of terms
    return (m for n in count(0) for m in range((1<A054429_list = list(islice(A054429_gen(),30)) # Chai Wah Wu, Jul 27 2023

maxblock <- 10 # by choice
a <- NULL
for(m in 0:maxblock) a <- c(a, rev(2^m:(2^(m+1)-1)))
a
# Yosu Yurramendi, Mar 10 2017

a(n) = ReflectBinTreePermutation(n).
a(n) = if n=1 then 1 else 2*a(floor(n/2)) + 1 - n mod 2. - Reinhard Zumkeller, Feb 18 2003
G.f.: 1/(1-x) * ((x-2x^2)/(1-x) + Sum_{k>=0} 3*2^k*x^2^k). - Ralf Stephan, Sep 15 2003
A000120(a(n)) = A000120(A059894(n)) = A023416(n) + 1. - Ralf Stephan, Oct 05 2003
A115310(n, 1) = a(n). - Reinhard Zumkeller, Jan 20 2006
a(1) = 1, a(2^(m+1) + k) = a(2^m+k) + 2^(m+1),
a(2^(m+1) + 2^m+k) = a(2^m+k) + 2^m, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Apr 06 2017
a(n) = A117120(A063946(n)) = A063946(A117120(n)) = A092569(A065190(n)) = A065190(A092569(n)), n > 0. - Yosu Yurramendi, Apr 10 2017
a(n) = 3*A053644(n) - n - 1. - Alan Michael Gómez Calderón, Feb 28 2025

A006068 a(n) is Gray-coded into n.

Original entry on oeis.org

0, 1, 3, 2, 7, 6, 4, 5, 15, 14, 12, 13, 8, 9, 11, 10, 31, 30, 28, 29, 24, 25, 27, 26, 16, 17, 19, 18, 23, 22, 20, 21, 63, 62, 60, 61, 56, 57, 59, 58, 48, 49, 51, 50, 55, 54, 52, 53, 32, 33, 35, 34, 39, 38, 36, 37, 47, 46, 44, 45, 40, 41, 43, 42, 127, 126, 124, 125, 120, 121

Offset: 0

N. J. A. Sloane

Equivalently, if binary expansion of n has m bits (say), compute derivative of n (A038554), getting sequence n' of length m-1; sort on n'.
Inverse of sequence A003188 considered as a permutation of the nonnegative integers, i.e., a(A003188(n)) = n = A003188(a(n)). - Howard A. Landman, Sep 25 2001
The sequence exhibits glide reflections that grow fractally. These show up well on the scatterplot, also audibly using the "listen" link. - Peter Munn, Aug 18 2019
Each bit at bit-index k (counted from the right hand end, with the least significant bit having bit-index 0) in the binary representation of a(n) is the parity of the number of 1's among the bits of the binary representation of n that have a bit-index of k or higher. - Frederik P.J. Vandecasteele, May 26 2025

			The first few values of n' are -,-,1,0,10,11,01,00,100,101,111,110,010,011,001,000,... (for n=0..15) and to put these in lexicographic order we must take n in the order 0,1,3,2,7,6,4,5,15,14,12,13,8,9,11,10,...

M. Gardner, Mathematical Games, Sci. Amer. Vol. 227 (No. 2, Feb. 1972), p. 107.
M. Gardner, Knotted Doughnuts and Other Mathematical Entertainments. Freeman, NY, 1986, p. 15.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1023
Joerg Arndt, Matters Computational (The Fxtbook), section 1.16 Gray code, C code inverse_gray_code() with "version 3" giving the PARI code here.
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, Dec 08, 2020
Eric Rowland and Reem Yassawi, Profinite automata, arXiv:1403.7659 [math.DS], 2014. See p. 22.
Paul Tarau, Isomorphic Data Encodings and their Generalization to Hylomorphisms on Hereditarily Finite Data Types; see also alternate link.
Index entries for sequences that are permutations of the natural numbers

Cf. A038554, A005811, A003188, A014550, A003100, A265263, A377404.
Cf. A054429, A180200. - Reinhard Zumkeller, Aug 15 2010
Cf. A000079, A055975 (first differences), A209281 (binary weight).
A003987, A010060 are used to express relationship between terms of this sequence.

a006068 n = foldl xor 0 $
                  map (div n) $ takeWhile (<= n) a000079_list :: Integer
-- Reinhard Zumkeller, Apr 28 2012

a:= proc(n) option remember; `if`(n<2, n,
      Bits[Xor](n, a(iquo(n, 2))))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Apr 17 2018

a[n_] := BitXor @@ Table[Floor[n/2^m], {m, 0, Floor[Log[2, n]]}]; a[0]=0; Table[a[n], {n, 0, 69}] (* Jean-François Alcover, Jul 19 2012, after Paul D. Hanna *)
Table[Fold[BitXor, n, Quotient[n, 2^Range[BitLength[n] - 1]]], {n, 0, 70}] (* Jan Mangaldan, Mar 20 2013 *)

{a(n)=local(B=n);for(k=1,floor(log(n+1)/log(2)),B=bitxor(B,n\2^k));B} /* Paul D. Hanna, Jan 18 2012 */

/* the following routine needs only O(log_2(n)) operations */
a(n)= {
    my( s=1, ns );
    while ( 1,
        ns = n >> s;
        if ( 0==ns, break() );
        n = bitxor(n, ns);
        s <<= 1;
    );
    return ( n );
} /* Joerg Arndt, Jul 19 2012 */

def a(n):
    s=1
    while True:
        ns=n>>s
        if ns==0: break
        n=n^ns
        s<<=1
    return n
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 07 2017, after PARI code by Joerg Arndt

nmax <- 63 # by choice
a <- vector()
for(n in 1:nmax){
  ones <- which(as.integer(intToBits(n)) == 1)
  nbit <- as.integer(intToBits(n))[1:tail(ones, n = 1)]
  level <- 0; anbit <- nbit; anbit.s <- nbit
  while(sum(anbit.s) > 0){
    s <- 2^level; if(s > length(anbit.s)) break
    anbit.s <- c(anbit[-(1:s)], rep(0,s))
    anbit <- bitwXor(anbit, anbit.s)
    level <- level + 1
  }
  a <- c(a, sum(anbit*2^(0:(length(anbit) - 1))))
}
(a <- c(0,a))
# Yosu Yurramendi, Oct 12 2021, after PARI code by Joerg Arndt

a(n) = 2*a(ceiling((n+1)/2)) + A010060(n-1). If 3*2^(k-1) < n <= 2^(k+1), a(n) = 2^(k+1) - 1 - a(n-2^k); if 2^(k+1) < n <= 3*2^k, a(n) = a(n-2^k) + 2^k. - Henry Bottomley, Jan 10 2001
a(n) = n XOR [n/2] XOR [n/4] XOR [n/8] ... XOR [n/2^m] where m = [log(n)/log(2)] (for n>0) and [x] is integer floor of x. - Paul D. Hanna, Jun 04 2002
a(n) XOR [a(n)/2] = n. - Paul D. Hanna, Jan 18 2012
A066194(n) = a(n-1) + 1, n>=1. - Philippe Deléham, Apr 29 2005
a(n) = if n<2 then n else 2*m + (n mod 2 + m mod 2) mod 2, with m=a(floor(n/2)). - Reinhard Zumkeller, Aug 10 2010
a(n XOR m) = a(n) XOR a(m), where XOR is the bitwise exclusive-or operator, A003987. - Peter Munn, Dec 14 2019
a(0) = 0. For all n >= 0 if a(n) is even a(2*n) = 2*a(n), a(2*n+1) = 2*a(n)+1, else a(2*n) = 2*a(n)+1, a(2*n+1) = 2*a(n). - Yosu Yurramendi, Oct 12 2021
Conjecture: a(n) = a(A053645(A063946(n))) + A053644(n) for n > 0 with a(0) = 0. - Mikhail Kurkov, Sep 09 2023
a(n) = 2*A265263(n) - 2*A377404(n) - A010060(n). - Alan Michael Gómez Calderón, Jun 26 2025

More terms from Henry Bottomley, Jan 10 2001

A006257 Josephus problem: a(2n) = 2a(n)-1, a(2n+1) = 2a(n)+1.

Original entry on oeis.org

0, 1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29

Offset: 0

N. J. A. Sloane

Write the numbers 1 through n in a circle, start at 1 and cross off every other number until only one number is left.
A version of the children's game "One potato, two potato, ...".
a(n)/A062383(n) = (0, 0.1, 0.01, 0.11, 0.001, ...) enumerates all binary fractions in the unit interval [0, 1). - Fredrik Johansson, Aug 14 2006
Iterating a(n), a(a(n)), ... eventually leads to 2^A000120(n) - 1. - Franklin T. Adams-Watters, Apr 09 2010
By inspection, the solution to the Josephus Problem is a sequence of odd numbers (from 1) starting at each power of 2.  This yields a direct closed form expression (see formula below). - Gregory Pat Scandalis, Oct 15 2013
Also zero together with a triangle read by rows in which row n lists the first 2^(n-1) odd numbers (see A005408), n >= 1. Row lengths give A011782. Right border gives A000225. Row sums give A000302, n >= 1. See example. - Omar E. Pol, Oct 16 2013
For n > 0: a(n) = n + 1 - A080079(n). - Reinhard Zumkeller, Apr 14 2014
In binary, a(n) = ROL(n), where ROL = rotate left = remove the leftmost digit and append it to the right. For example, n = 41 = 101001_2 => a(n) = (0)10011_2 = 19. This also explains FTAW's comment above. - M. F. Hasler, Nov 02 2016
In the under-down Australian card deck separation: top card on bottom of a deck of n cards, next card separated on the table, etc., until one card is left. The position a(n), for n >= 1, from top will be the left over card. See, e.g., the Behrends reference, pp. 156-164. For the down-under case see 2*A053645(n), for n >= 3, n not a power of 2. If n >= 2 is a power of 2 the botton card survives. - Wolfdieter Lang, Jul 28 2020

			From _Omar E. Pol_, Jun 09 2009: (Start)
Written as an irregular triangle the sequence begins:
  0;
  1;
  1,3;
  1,3,5,7;
  1,3,5,7,9,11,13,15;
  1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31;
  1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,
   43,45,47,49,51,53,55,57,59,61,63;
...
(End)
From _Omar E. Pol_, Nov 03 2018: (Start)
An illustration of initial terms, where a(n) is the area (or number of cells) in the n-th region of the structure:
   n   a(n)       Diagram
   0    0     _
   1    1    |_|_ _
   2    1      |_| |
   3    3      |_ _|_ _ _ _
   4    1          |_| | | |
   5    3          |_ _| | |
   6    5          |_ _ _| |
   7    7          |_ _ _ _|
(End)

Erhard Behrends, Der mathematische Zauberstab, Rowolth Taschenbuch Verlag, rororo 62902, 4. Auflage, 2019, pp. 156-164. [English version: The Math Behind the Magic, AMS, 2019.]
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 10.
M. S. Petković, "Josephus problem", Famous Puzzles of Great Mathematicians, page 179, Amer. Math. Soc. (AMS), 2009.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Paul Weisenhorn, Josephus und seine Folgen, MNU, 59(2006), pp. 18-19.

Iain Fox, Table of n, a(n) for n = 0..100000 (terms 0..1000 from T. D. Noe, terms 1001..10000 from Indranil Ghosh).
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197, ex. 34.
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
Paul Barry, Conjectures and results on some generalized Rueppel sequences, arXiv:2107.00442 [math.CO], 2021.
Daniel Erman and Brady Haran, The Josephus Problem, Numberphile video (2016)
Chris Groër, The Mathematics of Survival: From Antiquity to the Playground, Amer. Math. Monthly, 110 (No. 9, 2003), 812-825.
Alasdair MacFhraing, Aireamh Muinntir Fhinn Is Dhubhain, Agus Sgeul Josephuis Is An Da Fhichead Iudhaich, [Gaelic with English summary], Proc. Royal Irish Acad., Vol. LII, Sect. A., No. 7, 1948, 87-93.
Yuri Nikolayevsky and Ioannis Tsartsaflis, Cohomology of N-graded Lie algebras of maximal class over Z_2, arXiv:1512.87676 [math.RA], (2016), pages 2, 6.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Eric Weisstein's World of Mathematics, Josephus Problem
Wikipedia, Josephus problem
Index entries for sequences related to the Josephus Problem

Cf. A005428, A035327, A038572, A049940, A049964, A053644, A053645, A088147, A088442.
Second column, and main diagonal, of triangle A032434.
Cf. A181281 (with s=5), A054995 (with s=3).
Column k=2 of A360099.

Require Import ZArith.
Fixpoint a (n : positive) : Z :=
match n with
| xH => 1
| xI n' => (2*(a n') + 1)%Z
| xO n' => (2*(a n') - 1)%Z
end.
(* Stefan Haan, Aug 27 2023 *)

a006257 n = a006257_list !! n
a006257_list =
   0 : 1 : (map (+ 1) $ zipWith mod (map (+ 1) $ tail a006257_list) [2..])
-- Reinhard Zumkeller, Oct 06 2011

[0] cat [2*(n-2^Floor(Log(2,n)))+1: n in [1..100]]; // Vincenzo Librandi, Jan 14 2016

a(0):=0: for n from 1 to 100 do a(n):=(a(n-1)+1) mod n +1: end do:
seq(a(i),i=0..100); # Paul Weisenhorn, Oct 10 2010; corrected by Robert Israel, Jan 13 2016
A006257 := proc(n)
    convert(n,base,2) ;
    ListTools[Rotate](%,-1) ;
    add( op(i,%)*2^(i-1),i=1..nops(%)) ;
end proc: # R. J. Mathar, May 20 2016
A006257 := n -> 2*n  - Bits:-Iff(n, n):
seq(A006257(n), n=0..78); # Peter Luschny, Sep 24 2019

Table[ FromDigits[ RotateLeft[ IntegerDigits[n, 2]], 2], {n, 0, 80}] (* Robert G. Wilson v, Sep 21 2003 *)
Flatten@Table[Range[1, 2^n - 1, 2], {n, 0, 5}] (* Birkas Gyorgy, Feb 07 2011 *)
m = 5; Range[2^m - 1] + 1 - Flatten@Table[Reverse@Range[2^n], {n, 0, m - 1}] (* Birkas Gyorgy, Feb 07 2011 *)

a(n)=sum(k=1,n,if(bitxor(n,k)Paul D. Hanna

a(n)=if(n, 2*n-2^logint(2*n,2)+1, 0) \\ Charles R Greathouse IV, Oct 29 2016

import math
def A006257(n):
     return 0 if n==0 else 2*(n-2**int(math.log(n,2)))+1 # Indranil Ghosh, Jan 11 2017

def A006257(n): return bool(n&(m:=1<Chai Wah Wu, Jan 22 2023
(C#)
static long cs_A006257(this long n) => n == 0 ? 0 : 1 + (1 + (n - 1).cs_A006257()) % n; // Frank Hollstein, Feb 24 2021

To get a(n), write n in binary, rotate left 1 place.
a(n) = 2*A053645(n) + 1 = 2(n-msb(n))+1. - Marc LeBrun, Jul 11 2001. [Here "msb" = "most significant bit", A053644.]
G.f.: 1 + 2/(1-x) * ((3*x-1)/(2-2*x) - Sum_{k>=1} 2^(k-1)*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = number of positive integers k < n such that n XOR k < n. a(n) = n - A035327(n). - Paul D. Hanna, Jan 21 2006
a(n) = n for n = 2^k - 1. - Zak Seidov, Dec 14 2006
a(n) = n - A035327(n). - K. Spage, Oct 22 2009
a(2^m+k) = 1+2*k; with 0 <= m and 0 <= k < 2^m; n = 2^m+k; m = floor(log_2(n)); k = n-2^m; a(n) = ((a(n-1)+1) mod n) + 1; a(1) = 1. E.g., n=27; m=4; k=11; a(27) = 1 + 2*11 = 23. - Paul Weisenhorn, Oct 10 2010
a(n) = 2*(n - 2^floor(log_2(n))) + 1 (see comment above). - Gregory Pat Scandalis, Oct 15 2013
a(n) = 0 if n = 0 and a(n) = 2*a(floor(n/2)) - (-1)^(n mod 2) if n > 0. - Marek A. Suchenek, Mar 31 2016
G.f. A(x) satisfies: A(x) = 2*A(x^2)*(1 + x) + x/(1 + x). - Ilya Gutkovskiy, Aug 31 2019
For n > 0: a(n) = 2 * A062050(n) - 1. - Frank Hollstein, Oct 25 2021

More terms from Robert G. Wilson v, Sep 21 2003

A059893 Reverse the order of all but the most significant bit in binary expansion of n: if n = 1ab..yz then a(n) = 1zy..ba.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 7, 8, 12, 10, 14, 9, 13, 11, 15, 16, 24, 20, 28, 18, 26, 22, 30, 17, 25, 21, 29, 19, 27, 23, 31, 32, 48, 40, 56, 36, 52, 44, 60, 34, 50, 42, 58, 38, 54, 46, 62, 33, 49, 41, 57, 37, 53, 45, 61, 35, 51, 43, 59, 39, 55, 47, 63, 64, 96, 80, 112, 72, 104, 88, 120

Offset: 1

Marc LeBrun, Feb 06 2001

A self-inverse permutation of the natural numbers.
a(n)=n if and only if A081242(n) is a palindrome. - Clark Kimberling, Mar 12 2003
a(n) is the position in B of the reversal of the n-th term of B, where B is the left-to-right binary enumeration sequence (A081242 with the empty word attached as first term). - Clark Kimberling, Mar 12 2003
From Antti Karttunen, Oct 28 2001: (Start)
When certain Stern-Brocot tree-related permutations are conjugated with this permutation, they induce a permutation on Z (folded to N), which is an infinite siteswap permutation (see, e.g., figure 7 in the Buhler and Graham paper, which is permutation A065174). We get:
   A065260(n) = a(A057115(a(n))),
   A065266(n) = a(A065264(a(n))),
   A065272(n) = a(A065270(a(n))),
   A065278(n) = a(A065276(a(n))),
   A065284(n) = a(A065282(a(n))),
   A065290(n) = a(A065288(a(n))). (End)
Every nonnegative integer has a unique representation c(1) + c(2)*2 + c(3)*2^2 + c(4)*2^3 + ..., where every c(i) is 0 or 1. Taking tuples of coefficients in lexical order (i.e., 0, 1; 01,11; 001,011,101,111; ...) yields A059893. - Clark Kimberling, Mar 15 2015
From Ed Pegg Jr, Sep 09 2015: (Start)
The reduced rationals can be ordered either as the Calkin-Wilf tree A002487(n)/A002487(n+1) or the Stern-Brocot tree A007305(n+2)/A047679(n). The present sequence gives the order of matching rationals in the other sequence.
For reference, the Calkin-Wilf tree is 1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4, 1/5, 5/4, 4/7, 7/3, 3/8, 8/5, 5/7, 7/2, 2/7, 7/5, 5/8, 8/3, 3/7, 7/4, 4/5, ..., which is A002487(n)/A002487(n+1).
The Stern-Brocot tree is 1, 1/2, 2, 1/3, 2/3, 3/2, 3, 1/4, 2/5, 3/5, 3/4, 4/3, 5/3, 5/2, 4, 1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5, 5/4, 7/5, 8/5, 7/4, 7/3, 8/3, 7/2, ..., which is A007305(n+2)/A047679(n).
There is a great little OEIS-is-useful story here. I had code for the position of fractions in the Calkin-Wilf tree. The best I had for positions of fractions in the Stern-Brocot tree was the paper "Locating terms in the Stern-Brocot tree" by Bruce Bates, Martin Bunder, Keith Tognetti. The method was opaque to me, so I used my Calkin-Wilf code on the Stern-Brocot fractions, and got A059893. And thus the problem was solved. (End)

			a(11) = a(1011) = 1110 = 14.
With empty word e prefixed, A081242 becomes (e,1,2,11,21,12,22,111,211,121,221,112,...); (reversal of term #9) = (term #12); i.e., a(9)=12 and a(12)=9. - _Clark Kimberling_, Mar 12 2003
From _Philippe Deléham_, Jun 02 2015: (Start)
This sequence regarded as a triangle with rows of lengths 1, 2, 4, 8, 16, ...:
   1;
   2,  3;
   4,  6,  5,  7;
   8, 12, 10, 14,  9,  13,  11,  15;
  16, 24, 20, 28, 18,  26,  22,  30,  17,  25,  21,  29,  19,  27,  23,  31;
  32, 48, 40, 56, 36,  52,  44, ...
Row sums = A010036. (End)

Alois P. Heinz, Table of n, a(n) for n = 1..8191 (first 1023 terms from T. D. Noe)
Bruce Bates, Martin Bunder, and Keith Tognetti, Locating terms in the Stern-Brocot tree, European Journal of Combinatorics 31.3 (2010): 1020-1033.
Joe Buhler and R. L. Graham, Juggling Drops and Descents, Amer. Math. Monthly, 101, (no. 6) 1994, 507-519.
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, 2020-2021.
Wikipedia, Calkin-Wilf Tree.
Wikipedia, Stern-Brocot tree.
Index entries for sequences related to Stern's sequences
Index entries for sequences that are permutations of the natural numbers

{A000027, A054429, A059893, A059894} form a 4-group.
The set of permutations {A059893, A080541, A080542} generates an infinite dihedral group.
Cf. A000079, A000523, A007931, A030308.
In other bases: A351702 (balanced ternary), A343150 (Zeckendorf), A343152 (lazy Fibonacci).

a059893 = foldl (\v b -> v * 2 + b) 1 . init . a030308_row
-- Reinhard Zumkeller, May 01 2013
(Scheme, with memoization-macro definec)
(definec (A059893 n) (if (<= n 1) n (let* ((k (- (A000523 n) 1)) (r (A059893 (- n (A000079 k))))) (if (= 2 (floor->exact (/ n (A000079 k)))) (* 2 r) (+ 1 r)))))
;; Antti Karttunen, May 16 2015

# Implements Bottomley's formula
A059893 := proc(n) option remember; local k; if(1 = n) then RETURN(1); fi; k := floor_log_2(n)-1; if(2 = floor(n/(2^k))) then RETURN(2*A059893(n-(2^k))); else RETURN(1+A059893(n-(2^k))); fi; end;
floor_log_2 := proc(n) local nn,i; nn := n; for i from -1 to n do if(0 = nn) then RETURN(i); fi; nn := floor(nn/2); od; end;
# second Maple program:
a:= proc(n) local i, m, r; m, r:= n, 0;
      for i from 0 while m>1 do r:= 2*r +irem(m,2,'m') od;
      r +2^i
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 28 2015

A059893 = Reap[ For[n=1, n <= 100, n++, a=1; b=n; While[b > 1, a = 2*a + 2*FractionalPart[b/2]; b=Floor[b/2]]; Sow[a]]][[2, 1]] (* Jean-François Alcover, Jul 16 2012, after Harry J. Smith *)
ro[n_]:=Module[{idn=IntegerDigits[n,2]},FromDigits[Join[{First[idn]}, Reverse[ Rest[idn]]],2]]; Array[ro,80] (* Harvey P. Dale, Oct 24 2012 *)

a(n) = my(b=binary(n)); fromdigits(concat(b[1], Vecrev(vector(#b-1, k, b[k+1]))), 2); \\ Michel Marcus, Sep 29 2021

def a(n): return int('1' + bin(n)[3:][::-1], 2)
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Mar 21 2017

maxrow <- 6 # by choice
a <- 1
for(m in 0:maxrow) for(k in 0:(2^m-1)) {
  a[2^(m+1)+    k] <- 2*a[2^m+k]
  a[2^(m+1)+2^m+k] <- 2*a[2^m+k] + 1
}
a
# Yosu Yurramendi, Mar 20 2017

maxblock <- 7 # by choice
a <- 1
for(n in 2:2^maxblock){
  ones <- which(as.integer(intToBits(n)) == 1)
  nbit <- as.integer(intToBits(n))[1:tail(ones, n = 1)]
  anbit <- nbit
  anbit[1:(length(anbit) - 1)] <- anbit[rev(1:(length(anbit)-1))]
  a <- c(a, sum(anbit*2^(0:(length(anbit) - 1))))
}
a
# Yosu Yurramendi, Apr 25 2021

a(n) = A030109(n) + A053644(n). If 2*2^k <= n < 3*2^k then a(n) = 2*a(n-2^k); if 3*2^k <= n < 4*2^k then a(n) = 1 + a(n-2^k) starting with a(1)=1. - Henry Bottomley, Sep 13 2001

A053645 Distance to largest power of 2 less than or equal to n; write n in binary, change the first digit to zero, and convert back to decimal.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 3, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

Offset: 1

Henry Bottomley, Mar 22 2000

Triangle read by rows in which row n lists the first 2^n nonnegative integers (A001477), n >= 0. Right border gives A000225. Row sums give A006516. See example. - Omar E. Pol, Oct 17 2013

Without the initial zero also: zeroless numbers in base 3 (A032924: 1, 2, 11, 12, 21, ...), ternary digits decreased by 1 and read as binary. - M. F. Hasler, Jun 22 2020

			From _Omar E. Pol_, Oct 17 2013: (Start)
Written as an irregular triangle the sequence begins:
  0;
  0,1;
  0,1,2,3;
  0,1,2,3,4,5,6,7;
  0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15;
  ...
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, preprint, Theoretical Computer Sci., 98 (1992), 163-197.
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197 (see Ex. 24).
Index entries for sequences related to binary expansion of n

Cf. A000225, A000523, A002262, A004760, A006257, A006516, A030308, A036987, A053644, A062050, A083741, A160588.

a053645 1 = 0
a053645 n = 2 * a053645 n' + b  where (n', b) = divMod n 2
-- Reinhard Zumkeller, Aug 28 2014
a053645_list = concatMap (0 `enumFromTo`) a000225_list
-- Reinhard Zumkeller, Feb 04 2013, Mar 23 2012

[n - 2^Ilog2(n): n in [1..70]]; // Vincenzo Librandi, Jul 18 2019

seq(n - 2^ilog2(n), n=1..1000); # Robert Israel, Dec 23 2015

Table[n - 2^Floor[Log2[n]], {n, 100}] (* IWABUCHI Yu(u)ki, May 25 2017 *)
Table[FromDigits[Rest[IntegerDigits[n, 2]], 2], {n, 100}] (* IWABUCHI Yu(u)ki, May 25 2017 *)

a(n)=n-2^(#binary(n)-1) \\ Charles R Greathouse IV, Sep 02 2015

def a(n): return n - 2**(n.bit_length()-1)
print([a(n) for n in range(1, 85)]) # Michael S. Branicky, Jul 03 2021

def A053645(n): return n&(1<Chai Wah Wu, Jan 22 2023

a(n) = n - 2^A000523(n).
G.f.: 1/(1-x) * ((2x-1)/(1-x) + Sum_{k>=1} 2^(k-1)*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = (A006257(n)-1)/2. - N. J. A. Sloane, May 16 2003
a(1) = 0, a(2n) = 2a(n), a(2n+1) = 2a(n) + 1. - N. J. A. Sloane, Sep 13 2003
a(n) = A062050(n) - 1. - N. J. A. Sloane, Jun 12 2004
a(A004760(n+1)) = n. - Reinhard Zumkeller, May 20 2009
a(n) = f(n-1,1) with f(n,m) = if n < m then n else f(n-m,2*m). - Reinhard Zumkeller, May 20 2009
Conjecture: a(n) = (1 - A036987(n-1))*(1 + a(n-1)) for n > 1 with a(1) = 0. - Mikhail Kurkov, Jul 16 2019

A055938 Integers not generated by b(n) = b(floor(n/2)) + n (complement of A005187).

Original entry on oeis.org

2, 5, 6, 9, 12, 13, 14, 17, 20, 21, 24, 27, 28, 29, 30, 33, 36, 37, 40, 43, 44, 45, 48, 51, 52, 55, 58, 59, 60, 61, 62, 65, 68, 69, 72, 75, 76, 77, 80, 83, 84, 87, 90, 91, 92, 93, 96, 99, 100, 103, 106, 107, 108, 111, 114, 115, 118, 121, 122, 123, 124, 125, 126, 129

Offset: 1

Alford Arnold, Jul 21 2000

Note that the lengths of the consecutive runs in a(n) form sequence A001511.
Integers that are not a sum of distinct integers of the form 2^k-1. - Vladeta Jovovic, Jan 24 2003
Also n! never ends in this many 0's in base 2 - Carl R. White, Jan 21 2008
A079559(a(n)) = 0. - Reinhard Zumkeller, Mar 18 2009
These numbers are dead-end points when trying to apply the iterated process depicted in A071542 in reverse, i.e. these are positive integers i such that there does not exist k with A000120(i+k)=k. See also comments at A179016. - Antti Karttunen, Oct 26 2012
Conjecture: a(n)=b(n) defined as b(1)=2, for n>1, b(n+1)=b(n)+1 if n is already in the sequence, b(n+1)=b(n)+3 otherwise. If so, then see Cloitre comment in A080578. - Ralf Stephan, Dec 27 2013
Numbers n for which A257265(m) = 0. - Reinhard Zumkeller, May 06 2015. Typo corrected by Antti Karttunen, Aug 08 2015
Numbers which have a 2 in their skew-binary representation (cf. A169683). - Allan C. Wechsler, Feb 28 2025

			Since A005187 begins 0 1 3 4 7 8 10 11 15 16 18 19 22 23 25 26 31... this sequence begins 2 5 6 9 12 13 14 17 20 21

Complement of A005187. Setwise difference of A213713 and A213717.
Row 1 of arrays A257264, A256997 and also of A255557 (when prepended with 1). Equally: column 1 of A256995 and A255555.
Cf. also arrays A254105, A254107 and permutations A233276, A233278.
Left inverses: A234017, A256992.
Cf. A001511, A046699, A079559, A080578, A086343, A227359, A227408, A234016.
Gives positions of zeros in A213714, A213723, A213724, A213731, A257265, positions of ones in A213725-A213727 and A256989, positions of nonzeros in A254110.
Cf. also A010061 (integers that are not a sum of distinct integers of the form 2^k+1).
Analogous sequence for factorial base number system: A219658, for Fibonacci number system: A219638, for base-3: A096346. Cf. also A136767-A136774.
Cf. A257508, A257509, A257126.

a055938 n = a055938_list !! (n-1)
a055938_list = concat $
   zipWith (\u v -> [u+1..v-1]) a005187_list $ tail a005187_list
-- Reinhard Zumkeller, Nov 07 2011

a[0] = 0; a[1] = 1; a[n_Integer] := a[Floor[n/2]] + n; b = {}; Do[ b = Append[b, a[n]], {n, 0, 105}]; c =Table[n, {n, 0, 200}]; Complement[c, b]
(* Second program: *)
t = Table[IntegerExponent[(2n)!, 2], {n, 0, 100}]; Complement[Range[t // Last], t] (* Jean-François Alcover, Nov 15 2016 *)

L=listcreate();for(n=1,1000,for(k=2*n-hammingweight(n)+1,2*n+1-hammingweight(n+1),listput(L,k)));Vec(L) \\ Ralf Stephan, Dec 27 2013

def a053644(n): return 0 if n==0 else 2**(len(bin(n)[2:]) - 1)
def a043545(n):
    x=bin(n)[2:]
    return int(max(x)) - int(min(x))
def a079559(n): return 1 if n==0 else a043545(n + 1)*a079559(n + 1 - a053644(n + 1))
print([n for n in range(1, 201) if a079559(n)==0]) # Indranil Ghosh, Jun 11 2017, after the comment by Reinhard Zumkeller

;; utilizing COMPLEMENT-macro from Antti Karttunen's IntSeq-library)
(define A055938 (COMPLEMENT 1 A005187))
;; Antti Karttunen, Aug 08 2015

a(n) = A080578(n+1) - 2 = A080468(n+1) + 2*n (conjectured). - Ralf Stephan, Dec 27 2013
From Antti Karttunen, Aug 08 2015: (Start)
Other identities. For all n >= 1:
A234017(a(n)) = n.
A256992(a(n)) = n.
A257126(n) = a(n) - A005187(n).
(End)

More terms from Robert G. Wilson v, Jul 24 2000

A035327 Write n in binary, interchange 0's and 1's, convert back to decimal.

Original entry on oeis.org

1, 0, 1, 0, 3, 2, 1, 0, 7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46

Offset: 0

N. J. A. Sloane

For n>0: largest m<=n such that no carry occurs when adding m to n in binary arithmetic: A003817(n+1) = a(n) + n = a(n) XOR n. - Reinhard Zumkeller, Nov 14 2009
a(0) could be considered to be 0 (it was set so from 2004 to 2008) if the binary representation of zero was chosen to be the empty string. - Jason Kimberley, Sep 19 2011
For n > 0: A240857(n,a(n)) = 0. - Reinhard Zumkeller, Apr 14 2014
This is a base-2 analog of A048379. Another variant, without converting back to decimal, is given in A256078. - M. F. Hasler, Mar 22 2015
For n >= 2, a(n) is the least nonnegative k that must be added to n+1 to make a power of 2. Hence in a single-elimination tennis tournament with n entrants, a(n-1) is the number of players given a bye in round one, so that the number of players remaining at the start of round two is a power of 2. For example, if 39 players register, a(38)=25 players receive a round-one bye leaving 14 to play, so that round two will have 25+(14/2)=32 players. - Mathew Englander, Jan 20 2024

			8 = 1000 -> 0111 = 111 = 7.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Index entries for sequences related to binary expansion of n
Index entries for sequences related to the Josephus Problem

a(n) = A003817(n) - n, for n>0.
Cf. A080079, A087734, A167831, A167877, A007088, A061601, A171960, A010078, A000225, A006257 (Josephus problem).
Cf. A240857.
Cf. A048379, A256078.

a035327 n = if n <= 1 then 1 - n else 2 * a035327 n' + 1 - b
            where (n',b) = divMod n 2
-- Reinhard Zumkeller, Feb 21 2014

using IntegerSequences
A035327List(len) = [Bits("NAND", n, n) for n in 0:len]
println(A035327List(100))  # Peter Luschny, Sep 25 2021

A035327:=func; // Jason Kimberley, Sep 19 2011

seq(2^(1 + ilog2(max(n, 1))) - 1 - n, n = 0..81); # Emeric Deutsch, Oct 19 2008
A035327 := n -> `if`(n=0, 1, Bits:-Nand(n, n)):
seq(A035327(n), n=0..81); # Peter Luschny, Sep 23 2019

Table[BaseForm[FromDigits[(IntegerDigits[i, 2]/.{0->1, 1->0}), 2], 10], {i, 0, 90}]
Table[BitXor[n, 2^IntegerPart[Log[2, n] + 1] - 1], {n, 100}] (* Alonso del Arte, Jan 14 2006 *)
Join[{1},Table[2^BitLength[n]-n-1,{n,100}]] (* Paolo Xausa, Oct 13 2023 *)
Table[FromDigits[IntegerDigits[n,2]/.{0->1,1->0},2],{n,0,90}] (* Harvey P. Dale, May 03 2025 *)

a(n)=sum(k=1,n,if(bitxor(n,k)>n,1,0)) \\ Paul D. Hanna, Jan 21 2006

a(n) = bitxor(n, 2^(1+logint(max(n,1), 2))-1) \\ Rémy Sigrist, Jan 04 2019

a(n)=if(n, bitneg(n, exponent(n)+1), 1) \\ Charles R Greathouse IV, Apr 13 2020

def a(n): return int(''.join('1' if i == '0' else '0' for i in bin(n)[2:]), 2) # Indranil Ghosh, Apr 29 2017

def a(n): return 1 if n == 0 else n^((1 << n.bit_length()) - 1)
print([a(n) for n in range(100)]) # Michael S. Branicky, Sep 28 2021

def A035327(n): return (~n)^(-1<Chai Wah Wu, Dec 20 2022

def a(n):
    if n == 0:
        return 1
    return sum([(1 - b) << s for (s, b) in enumerate(n.bits())])
[a(n) for n in srange(82)]  # Peter Luschny, Aug 31 2019

a(n) = 2^k - n - 1, where 2^(k-1) <= n < 2^k.
a(n+1) = (a(n)+n) mod (n+1); a(0) = 1. - Reinhard Zumkeller, Jul 22 2002
G.f.: 1 + 1/(1-x)*Sum_{k>=0} 2^k*x^2^(k+1)/(1+x^2^k). - Ralf Stephan, May 06 2003
a(0) = 0, a(2n+1) = 2*a(n), a(2n) = 2*a(n) + 1. - Philippe Deléham, Feb 29 2004
a(n) = number of positive integers k < n such that n XOR k > n. a(n) = n - A006257(n). - Paul D. Hanna, Jan 21 2006
a(n) = 2^{1+floor(log[2](n))}-n-1 for n>=1; a(0)=1. - Emeric Deutsch, Oct 19 2008
a(n) = if n<2 then 1 - n else 2*a(floor(n/2)) + 1 - n mod 2. - Reinhard Zumkeller, Jan 20 2010
a(n) = abs(2*A053644(n) - n - 1). - Mathew Englander, Jan 22 2024

More terms from Vit Planocka (planocka(AT)mistral.cz), Feb 01 2003
a(0) corrected by Paolo P. Lava, Oct 22 2007
Definition completed by M. F. Hasler, Mar 22 2015

A056971 Number of (binary) heaps on n elements.

Original entry on oeis.org

1, 1, 1, 2, 3, 8, 20, 80, 210, 896, 3360, 19200, 79200, 506880, 2745600, 21964800, 108108000, 820019200, 5227622400, 48881664000, 319258368000, 3143467008000, 25540669440000, 299677188096000, 2261626278912000, 25732281217843200, 241240136417280000

Offset: 0

Eric W. Weisstein

A sequence {a_i}{i=1..N} forms a (binary) heap if it satisfies a_i<a{2i} and a_i

Proof of recurrence: a_1 must take the greatest of the n values. Deleting a_1 gives 2 heaps of size b+r1, b+r2. - Sascha Kurz, Mar 24 2002

Note that A132862(n)*a(n) = n!. - Alois P. Heinz, Nov 22 2007

			There is 1 heap if n is in {0,1,2}, 2 heaps for n=3, 3 heaps for n=4 and so on.
a(5) = 8 (min-heaps): 12345, 12354, 12435, 12453, 12534, 12543, 13245, 13254.

Alois P. Heinz, Table of n, a(n) for n = 0..500
Sean Cleary, M. Fischer, R. C. Griffiths, and R. Sainudiin, Some distributions on finite rooted binary trees, UCDMS Research Report NO. UCDMS2015/2, School of Mathematics and Statistics, University of Canterbury, Christchurch, NZ, 2015.
D. Levin, L. Pudwell, M. Riehl, and A. Sandberg, Pattern Avoidance on k-ary Heaps, Slides of Talk, 2014.
Eric Weisstein's World of Mathematics, Heap
Wikipedia, Binary heap

Cf. A053644, A056972, A132862, A373452 (allows repeated elements).
Column k=2 of A273693.
Column k=0 of A306343 and of A306393.
Main diagonal of A373451.

a[0] := 1: a[1] := 1:
for n from 2 to 50 do
h := ilog2(n+1)-1:
b := 2^h-1: r := n-1-2*b: r1 := r-floor(r/2^h)*(r-2^h): r2 := r-r1:
a[n] := binomial(n-1, b+r1)*a[b+r1]*a[b+r2]:end do:
q := seq(a[j], j=0..50);
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 1, (g-> (f-> a(f)*
      binomial(n-1, f)*a(n-1-f))(min(g-1, n-g/2)))(2^ilog2(n)))
    end:
seq(a(n), n=0..28);  # Alois P. Heinz, Feb 14 2019

a[0] = 1; a[1] = 1; For[n = 2, n <= 50, n++, h = Floor[Log[2, n + 1]] - 1; b = 2^h - 1; r = n - 1 - 2*b; r1 = r - Floor[r/2^h]*(r - 2^h); r2 = r - r1; a[n] = Binomial[n - 1, b + r1]*a[b + r1]*a[b + r2]]; Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Oct 22 2012, translated from Maple program *)

from functools import lru_cache
from math import comb
@lru_cache(maxsize=None)
def A056971(n):
    if n<=1: return 1
    h = (n+1).bit_length()-2
    b = (1<A056971(b+r1)*A056971(b+r2) # Chai Wah Wu, May 06 2024

See recurrence in Maple and Mma programs.

More terms from Sascha Kurz, Mar 24 2002

Offset and some terms corrected by Alois P. Heinz, Nov 21 2007

A062383 a(0) = 1: for n>0, a(n) = 2^floor(log_2(n)+1) or a(n) = 2*a(floor(n/2)).

Original entry on oeis.org

1, 2, 4, 4, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 128, 128, 128, 128, 128, 128

Offset: 0

Antti Karttunen, Jun 19 2001

Informally, write down 1 followed by 2^k 2^(k-1) times, for k = 1,2,3,4,... These are the denominators of the binary van der Corput sequence (see A030101 for the numerators). - N. J. A. Sloane, Dec 01 2019
a(n) is the denominator of the form 2^k needed to make the ratio (2n-1)/2^k lie in the interval [1-2], i.e. such ratios are 1/1, 3/2, 5/4, 7/4, 9/8, 11/8, 13/8, 15/8, 17/16, 19/16, 21/16, ... where the numerators are A005408 (The odd numbers).
Let A_n be the upper triangular matrix in the group GL(n,2) that has zero entries below the diagonal and 1 elsewhere. For example for n=4 the matrix is / 1,1,1,1 / 0,1,1,1 / 0,0,1,1 / 0,0,0,1 /. The order of this matrix as an element of GL(n,2) is a(n-1). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 14 2001
A006257(n)/a(n) = (0, 0.1, 0.01, 0.11, 0.001, ...) enumerates all binary fractions in the unit interval [0, 1). - Fredrik Johansson, Aug 14 2006
a(n) = maximum of row n+1 in A240769. - Reinhard Zumkeller, Apr 13 2014
This is the discriminator sequence for the odious numbers. - N. J. A. Sloane, May 10 2016
From Jianing Song, Jul 05 2025: (Start)
a(n) is the period of {binomial(N,n) mod 2: N in Z}. For the general result, see A349593.
Since the modulus (2) is a prime, the remainder of binomial(N,n) is given by Lucas's theorem. (End)

Harry J. Smith, Table of n, a(n) for n = 0..1000
L. K. Arnold, S. J. Benkoski and B. J. McCabe, The discriminator (a simple application of Bertrand's postulate). Amer. Math. Monthly 92 (1985), 275-277.
Sajed Haque, Chapter 2.6.1 of Discriminators of Integer Sequences, 2017, See p. 33.
S. Haque and J. Shallit, Discriminators and k-regular sequences, arXiv:1605.00092 [cs.DM], 2016.
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, Dec 08, 2020
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Wikipedia, Lucas's theorem
Index to divisibility sequences

Apart from the initial term, equals 2 * A053644. MASKTRANSi(A062383) seems to give a signed form of A038712. (See identities at A053644). floor_log_2 given in A054429.
Equals A003817(n)+1. Cf. A002884.
Bisection of A065285. Cf. A076877.
Equals for n>=1 the r(n) sequence of A160464. - Johannes W. Meijer, May 24 2009
Equals the r(n) sequence of A162440 for n>=1. - Johannes W. Meijer, Jul 06 2009
Cf. A030101, A264618, A264619.
Discriminator of the odious numbers (A000069). - Jeffrey Shallit, May 08 2016
Column 2 of A349593. A064235 (if offset 0), A385552, A385553, and A385554 are respectively columns 3, 5, 6, and 10.

import Data.List (transpose)
a062383 n = a062383_list !! n
a062383_list = 1 : zs where
   zs = 2 : (map (* 2) $ concat $ transpose [zs, zs])
-- Reinhard Zumkeller, Aug 27 2014, Mar 13 2014

[2^Floor(Log(2,2*n+1)): n in [0..70]]; // Bruno Berselli, Mar 04 2016

[seq(2^(floor_log_2(j)+1),j=0..127)]; or [seq(coerce1st_octave((2*j)+1),j=0..127)]; or [seq(a(j),j=0..127)];
coerce1st_octave := proc(r) option remember; if(r < 1) then coerce1st_octave(2*r); else if(r >= 2) then coerce1st_octave(r/2); else (r); fi; fi; end;
A062383 := proc(n)
    option remember;
    if n = 0 then
        1 ;
    else
        2*procname(floor(n/2));
    end if;
end proc:
A062383 := n -> 1 + Bits:-Iff(n, n):
seq(A062383(n), n=0..69); # Peter Luschny, Sep 23 2019

a[n_] := a[n] = 2 a[n/2 // Floor]; a[0] = 1; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Mar 04 2016 *)
Table[2^Floor[Log2[n] + 1], {n, 0, 20}] (* Eric W. Weisstein, Nov 17 2017 *)
2^Floor[Log2[Range[0, 20]] + 1] (* Eric W. Weisstein, Nov 17 2017 *)
2^BitLength[Range[0, 100]] (* Paolo Xausa, Jan 29 2025 *)

{ a=1; for (n=0, 1000, write("b062383.txt", n, " ", a*=ceil((n + 1)/a)) ) } \\ Harry J. Smith, Aug 06 2009

a(n)=1<<(log(2*n+1)\log(2)) \\ Charles R Greathouse IV, Dec 08 2011

def A062383(n): return 1 << n.bit_length() # Chai Wah Wu, Jun 30 2022

a(1) = 1 and a(n+1) = a(n)*ceiling(n/a(n)). - Benoit Cloitre, Aug 17 2002
G.f.: 1/(1-x) * (1 + Sum_{k>=0} 2^k*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = A142151(2*n)/2 + 1. - Reinhard Zumkeller, Jul 15 2008
log(a(n))/log(2) = A029837(n+1). - Johannes W. Meijer, Jul 06 2009
a(n+1) = a(n) + A099894(n). - Reinhard Zumkeller, Aug 06 2009
a(n) = A264619(n) - A264618(n). - Reinhard Zumkeller, Dec 01 2015
a(n) is the smallest power of 2 > n. - Chai Wah Wu, Nov 04 2016
a(n) = 2^ceiling(log_2(n+1)). - M. F. Hasler, Sep 20 2017

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A003188 Decimal equivalent of Gray code for n.

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A054429 Simple self-inverse permutation of natural numbers: List each block of 2^n numbers (from 2^n to 2^(n+1) - 1) in reverse order.

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A006068 a(n) is Gray-coded into n.

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A006257 Josephus problem: a(2*n) = 2*a(n)-1, a(2*n+1) = 2*a(n)+1.

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A059893 Reverse the order of all but the most significant bit in binary expansion of n: if n = 1ab..yz then a(n) = 1zy..ba.

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A006257 Josephus problem: a(2n) = 2a(n)-1, a(2n+1) = 2a(n)+1.