A062383 -id:A062383 - cp's OEIS frontend

A339570 Denote the van der Corput sequence of fractions 1/2, 1/4, 3/4, 1/8, 5/8, 3/8, 7/8, 1/16, ... (A030101/A062383) by v(n), n >= 1. Then a(n) = denominator of v(A014486(n)).

4, 16, 16, 64, 64, 64, 64, 64, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024

Offset: 1

N. J. A. Sloane, Dec 09 2020

Comment from N. J. A. Sloane, Dec 11 2020: (Start)
The initial values suggest the conjecture that this sequence consists exactly of Catalan(k) copies of 4^k for k >= 1.
Hugo Pfoertner tested this conjecture with the PARI program given below.
Here is the output from that program:
  [1, 0, 4]
  [2, 4, 16]
  [4, 16, 64]
  [9, 64, 256]
  [23, 256, 1024]
  [65, 1024, 4096]
  [197, 4096, 16384]
  [626, 16384, 65536]
  [2056, 65536, 262144]
  [6918, 262144, 1048576]
  [23714, 1048576, 4194304]
The first column is A014137, the partial sums of the Catalan numbers, which is strong support for the conjecture.
The conjecture has now been proved by Raghavendra Tripathi - see link. (End)

			The van der Corput sequence v(n), n >= 1, is 1/2, 1/4, 3/4, 1/8, 5/8, 3/8, 7/8, 1/16, 9/16, 5/16, 13/16, 3/16, 11/16, ... = A030101/A062383.
Then we construct the sequence b(n) = v(A014486(n)), n >= 1, which is 1/4, 5/16, 3/16, 21/64, 13/64, 19/64, 11/64, 7/64, ...
a(n) is the denominator of b(n), and A072800(n) is the numerator.

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. See Section 2.8.
Raghavendra Tripathi, Proof of conjectured formula

Cf. A000108, A014137, A030101, A062383, A072800.

\\ Program from Hugo Pfoertner for studying the connection with the Catalan numbers mentioned in the Comments.
a30101(n)=fromdigits(Vecrev(binary(n)), 2);
a62383(n)=1<<(log(2*n+1)\log(2));
is_a14486(n)={my(v=binary(n), t=0); for(i=1, #v, t+=if(v[i], 1, -1); if(t<0, return(0))); t==0};
A14486=[];for(k=1,5000000,if(is_a14486(k),A14486=concat(A14486,k)));
aprev=0;for(k=1,#A14486,my(j=A14486[k],a=denominator(a30101(j)/a62383(j)));if(a!=aprev,print([k,aprev,a]);aprev=a));

More terms from Hugo Pfoertner, Dec 09 2020

A153587 a(n) = n mod (A062383(n) - n).

Original entry on oeis.org

0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 4, 1, 0, 1, 0, 0, 0, 2, 4, 6, 8, 10, 2, 5, 0, 4, 2, 2, 0, 2, 0, 0, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 1, 4, 7, 10, 13, 0, 4, 8, 12, 4, 9, 4, 1, 0, 1, 4, 4, 0, 1, 0, 0, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42

Offset: 0

J. Z. Bradley (jzbradley(AT)gmail.com), Dec 29 2008

Michel Marcus, Table of n, a(n) for n = 0..1024

Cf. A062383, A080079.

int A062383(int n) { if(n==0) return 1; return 2*(A062383(n/2)); }
int a(int n) { return n % (A062383(n)-n); }

b(n) = if (n, 2*b(n\2), 1);
a(n) = n % (n - b(n)); \\ Michel Marcus, Jan 28 2018

def A153587(n): return n % ((1 << n.bit_length())-n) # Chai Wah Wu, Jun 30 2022

a(n) = n mod A080079(n), for n > 0. - Michel Marcus, Jan 28 2018

More terms from Michel Marcus, Jan 28 2018

A029837 Binary order of n: log_2(n) rounded up to next integer.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset: 1

N. J. A. Sloane

Or, ceiling(log_2(n)).
Worst-case cost of binary search.
Equal to number of binary digits in n unless n is a power of 2 when it is one less.
Thus a(n) gives the length of the binary representation of n - 1 (n >= 2), which is also A070939(n - 1).
Let x(0) = n > 1 and x(k + 1) = x(k) - floor(x(k)/2), then a(n) is the smallest integer such that x(a(n)) = 1. - Benoit Cloitre, Aug 29 2002
Also number of division steps when going from n to 1 by process of adding 1 if odd, or dividing by 2 if even. - Cino Hilliard, Mar 25 2003
Number of ways to write n as (x + 2^y), x >= 0. Number of ways to write n + 1 as 2^x + 3^y (cf. A004050). - Benoit Cloitre, Mar 29 2003
The minimum number of cuts for dividing an object into n (possibly unequal) pieces. - Karl Ove Hufthammer (karl(AT)huftis.org), Mar 29 2010
Partial sums of A209229; number of powers of 2 not greater than n. - Reinhard Zumkeller, Mar 07 2012

			a(1) = 0, since log_2(1) = 0.
a(2) = 1, since log_2(2) = 1.
a(3) = 2, since log_2(3) = 1.58...
a(n) = 7 for n = 65, 66, ..., 127, 128.
G.f. = x^2 + 2*x^3 + 2*x^4 + 3*x^5 + 3*x^6 + 3*x^7 + 3*x^8 + 4*x^9 + ... - _Michael Somos_, Jun 02 2019

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, 1989, p. 70.
G. J. E. Rawlins, Compared to What? An Introduction to the Analysis of Algorithms, W. H. Freeman, 1992; see pp. 108, 118.

T. D. Noe, Table of n, a(n) for n = 1..10000
Cino Hilliard, The x+1 conjecture [broken link]
Lionel Levine, Fractal sequences and restricted Nim, arXiv:math/0409408 [math.CO], 2004.
Eric Weisstein's World of Mathematics, Bit Length.
Index entries for sequences related to binary expansion of n

Cf. A000523, A070939, A000193, A000195, A004233, A113473, A053644.
Partial sums of A036987.
Used for several definitions: A029827, A036378-A036390. Partial sums: A001855.

a029837 n = a029837_list !! (n-1)
a029837_list = scanl1 (+) a209229_list
-- Reinhard Zumkeller, Mar 07 2012
(Common Lisp) (defun A029837 (n) (integer-length (1- n))) ; James Spahlinger, Oct 15 2012

[Ceiling(Log(2, n)): n in [1..100]]; // Vincenzo Librandi, Jun 14 2019

a:= n-> (p-> p+`if`(2^pAlois P. Heinz, Mar 18 2013

a[n_] := Ceiling[Log[2, n]]; Array[a, 105] (* Robert G. Wilson v, Dec 09 2005 *)
Table[IntegerLength[n - 1, 2], {n, 1, 105}] (* Peter Luschny, Dec 02 2017 *)
a[n_] := If[n < 1, 0, BitLength[n - 1]]; (* Michael Somos, Jul 10 2018 *)
Join[{0}, IntegerLength[Range[130], 2]] (* Vincenzo Librandi, Jun 14 2019 *)

{a(n) = if( n<1, 0, ceil(log(n) / log(2)))};

/* Set p = 1, then: */
xpcount(n,p) = for(x=1, n, p1 = x; ct=0; while(p1>1, if(p1%2==0,p1/=2; ct++,p1 = p1*p+1)); print1(ct, ", "))

{a(n) = if( n<2, 0, exponent(n-1)+1)}; /* Michael Somos, Jul 10 2018 */

def A029837(n):
    s = bin(n)[2:]
    return len(s) - (1 if s.count('1') == 1 else 0) # Chai Wah Wu, Jul 09 2020

def A029837(n): return (n-1).bit_length() # Chai Wah Wu, Jun 30 2022

(1 to 80).map(n => Math.ceil(Math.log(n)/Math.log(2)).toInt) // Alonso del Arte, Feb 19 2020

a(n) = ceiling(log_2(n)).
a(1) = 0; for n > 1, a(2n) = a(n) + 1, a(2n + 1) = a(n) + 1. Alternatively, a(1) = 0; for n > 1, a(n) = a(ceiling(n/2)) + 1. [corrected by Ilya Gutkovskiy, Mar 21 2020]
a(n) = k such that n^(1/k - 1) > 2 > n^(1/k), or the least value of k for which floor n^(1/k) = 1. a(n) = k for all n such that 2^(k - 1) < n < 2^k. - Amarnath Murthy, May 06 2001
G.f.: x/(1 - x) * Sum_{k >= 0} x^2^k. - Ralf Stephan, Apr 13 2002
A062383(n-1) = 2^a(n). - Johannes W. Meijer, Jul 06 2009
a(n+1) = -Sum_{k = 1..n} mu(2*k)*floor(n/k). - Benoit Cloitre, Oct 21 2009
a(n+1) = A113473(n). - Michael Somos, Jun 02 2019

Additional comments from Daniele Parisse

More terms from Michael Somos, Aug 02 2002

A030101 a(n) is the number produced when n is converted to binary digits, the binary digits are reversed and then converted back into a decimal number.

Original entry on oeis.org

0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 13, 3, 11, 7, 15, 1, 17, 9, 25, 5, 21, 13, 29, 3, 19, 11, 27, 7, 23, 15, 31, 1, 33, 17, 49, 9, 41, 25, 57, 5, 37, 21, 53, 13, 45, 29, 61, 3, 35, 19, 51, 11, 43, 27, 59, 7, 39, 23, 55, 15, 47, 31, 63, 1, 65, 33, 97, 17, 81, 49, 113, 9, 73, 41, 105, 25, 89, 57

Offset: 0

David W. Wilson

As with decimal reversal, initial zeros are ignored; otherwise, the reverse of 1 would be 1000000... ad infinitum.
Numerators of the binary van der Corput sequence. - Eric Rowland, Feb 12 2008
It seems that in most cases A030101(x) = A000265(x) and that if A030101(x) <> A000265(x), the next time A030101(y) = A000265(x), A030101(x) = A000265(y). Also, it seems that if a pair of values exist at one index, they will exist at any index where one of them exist. It also seems like the greater of the pair always shows up on A000265 first. - Dylan Hamilton, Aug 04 2010
The number of occasions A030101(n) = A000265(n) before n = 2^k is A053599(k) + 1. For n = 0..2^19, the sequences match less than 1% of the time. - Andrew Woods, May 19 2012
For n > 0: a(a(n)) = n if and only if n is odd; a(A006995(n)) = A006995(n). - Juli Mallett, Nov 11 2010, corrected: Reinhard Zumkeller, Oct 21 2011
n is binary palindromic if and only if a(n) = n. - Reinhard Zumkeller, corrected: Jan 17 2012, thanks to Hieronymus Fischer, who pointed this out; Oct 21 2011
Given any n > 1, the set of numbers A030109(i) = (A030101(i) - 1)/2 for indexes i ranging from 2^n to 2^(n + 1) - 1 is a permutation of the set of consecutive integers {0, 1, 2, ..., 2^n - 1}. This is important in the standard FFT algorithms (starting or ending bit-reversal permutation). - Stanislav Sykora, Mar 15 2012
Row n of A030308 gives the binary digits of a(n), prepended with zero at even positions. - Reinhard Zumkeller, Jun 17 2012
The binary van der Corput sequence is the infinite sequence of fractions { A030101(n)/A062383(n), n = 0, 1, 2, 3, ... }, and begins 0, 1/2, 1/4, 3/4, 1/8, 5/8, 3/8, 7/8, 1/16, 9/16, 5/16, 13/16, 3/16, 11/16, 7/16, 15/16, 1/32, 17/32, 9/32, 25/32, 5/32, 21/32, 13/32, 29/32, 3/32, 19/32, 11/32, 27/32, 7/32, 23/32, 15/32, 31/32, 1/64, 33/64, 17/64, 49/64, ... - N. J. A. Sloane, Dec 01 2019
Record highs occur at n = A209492(m) (for n>=1) with values a(n) = A224195(m) (for n>=3). - Bill McEachen, Aug 02 2023

			a(100) = 19 because 100 (base 10) = 1100100 (base 2) and R(1100100 (base 2)) = 10011 (base 2) = 19 (base 10).

Hlawka E. The theory of uniform distribution. Academic Publishers, Berkhamsted, 1984. See pp. 93, 94 for the van der Corput sequence. - N. J. A. Sloane, Dec 01 2019

T. D. Noe, Table of n, a(n) for n = 0..10000
Sean Anderson, Bit Twiddling Hacks, for fixed-width reversals.
Joerg Arndt, Matters Computational (The Fxtbook), section 1.14 Reversing the bits of a word, page 33.
Harold L. Dorwart, Seventeenth annual USA Mathematical Olympiads, Math. Mag., 62 (1989), 210-214 (#3).
Bernd Fischer and Lothar Reichel, Newton interpolation in Fejér and Chebyshev points, Mathematics of Computation 53.187 (1989): 265-278. See pp. 266, 267 for the van der Corput sequence. - _N. J. A. Sloane_, Dec 01 2019
Michael Gilleland, Some Self-Similar Integer Sequences
E. Hlawka, The theory of uniform distribution. Academic Publishers, Berkhamsted, 1984. See pp. 93, 94 for the van der Corput sequence. - _N. J. A. Sloane_, Dec 01 2019
Project Euler, Problem 463: A weird recurrence relation
Wikipedia, van der Corput sequence.
Index entries for sequences related to binary expansion of n

Cf. A030102 - A030109, A036044, A004086, A005408.
Cf. A055944 (reverse and add), A178225, A273258.
Cf. A056539, A057889 (bijective variants), A224195, A209492.

a030101 = f 0 where
   f y 0 = y
   f y x = f (2 * y + b) x'  where (x', b) = divMod x 2
-- Reinhard Zumkeller, Mar 18 2014, Oct 21 2011

([: #. [: |. #:)"0 NB. Stephen Makdisi, May 07 2018

A030101:=func; // Jason Kimberley, Sep 19 2011

A030101 := proc(n)
    convert(n,base,2) ;
    ListTools[Reverse](%) ;
    add(op(i,%)*2^(i-1),i=1..nops(%)) ;
end proc: # R. J. Mathar, Mar 10 2015
# second Maple program:
a:= proc(n) local m, r; m:=n; r:=0;
      while m>0 do r:=r*2+irem(m, 2, 'm') od; r
    end:
seq(a(n), n=0..80);  # Alois P. Heinz, Nov 17 2015

Table[FromDigits[Reverse[IntegerDigits[i, 2]], 2], {i, 0, 80}]
bitRev[n_] := Switch[Mod[n, 4], 0, bitRev[n/2], 1, 2 bitRev[(n + 1)/2] - bitRev[(n - 1)/4], 2, bitRev[n/2], 3, 3 bitRev[(n - 1)/2] - 2 bitRev[(n - 3)/4]]; bitRev[0] = 0; bitRev[1] = 1; bitRev[3] = 3; Array[bitRev, 80, 0] (* Robert G. Wilson v, Mar 18 2014 *)

a(n)=if(n<1,0,subst(Polrev(binary(n)),x,2))

a(n) = fromdigits(Vecrev(binary(n)), 2); \\ Michel Marcus, Nov 10 2017

def a(n): return int(bin(n)[2:][::-1], 2) # Indranil Ghosh, Apr 24 2017

def A030101(n): return Integer(bin(n).lstrip("0b")[::-1],2) if n!=0 else 0
[A030101(n) for n in (0..78)]  # Peter Luschny, Aug 09 2012

(0 to 127).map(n => Integer.parseInt(Integer.toString(n, 2).reverse, 2)) // Alonso del Arte, Feb 11 2020

a(n) = 0, a(2n) = a(n), a(2n+1) = a(n) + 2^(floor(log_2(n)) + 1). For n > 0, a(n) = 2*A030109(n) - 1. - Ralf Stephan, Sep 15 2003
a(n) = b(n, 0) with b(n, r) = r if n = 0, otherwise b(floor(n/2), 2*r + n mod 2). - Reinhard Zumkeller, Mar 03 2010
a(1) = 1, a(3) = 3, a(2n) = a(n), a(4n+1) = 2a(2n+1) - a(n), a(4n+3) = 3a(2n+1) - 2a(n) (as in the Project Euler problem). To prove this, expand the recurrence into binary strings and reversals. - David Applegate, Mar 16 2014, following a posting to the Sequence Fans Mailing List by Martin Møller Skarbiniks Pedersen.
Conjecture: a(n) = 2*w(n) - 2*w(A053645(n)) - 1 for n > 0, where w = A264596. - Velin Yanev, Sep 12 2017

Edits (including correction of an erroneous date pointed out by J. M. Bergot) by Jon E. Schoenfield, Mar 16 2014

Name clarified by Antti Karttunen, Nov 09 2017

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

A053644 Most significant bit of n, msb(n); largest power of 2 less than or equal to n; write n in binary and change all but the first digit to zero.

Original entry on oeis.org

0, 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 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

Offset: 0

Henry Bottomley, Mar 22 2000

Except for the initial term, 2^n appears 2^n times. - Lekraj Beedassy, May 26 2005
a(n) is the smallest k such that row k in triangle A265705 contains n. - Reinhard Zumkeller, Dec 17 2015
a(n) is the sum of totient function over powers of 2 <= n. - Anthony Browne, Jun 17 2016
Given positive n, reverse the bits of n and divide by 2^floor(log_2 n). Numerators are in A030101. Ignoring the initial 0, denominators are in this sequence. - Alonso del Arte, Feb 11 2020

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
N. J. A. Sloane, Transforms
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions

See A000035 for least significant bit(n).
MASKTRANS transform of A055975 (prepended with 0), MASKTRANSi transform of A048678.
Bisection of A065267, A065279, A065291, A072376.
First differences of A063915. Cf. A076877, A073121.
This is Guy Steele's sequence GS(5, 5) (see A135416).
Equals for n >= 1 the first right hand column of A160464. - Johannes W. Meijer, May 24 2009
Diagonal of A088370. - Alois P. Heinz, Oct 28 2011
Cf. A265705, A000010.

a053644 n = if n <= 1 then n else 2 * a053644 (div n 2)
-- Reinhard Zumkeller, Aug 28 2014
a053644_list = 0 : concat (iterate (\zs -> map (* 2) (zs ++ zs)) [1])
-- Reinhard Zumkeller, Dec 08 2012, Oct 21 2011, Oct 17 2010

[0] cat [2^Ilog2(n): n in [1..90]]; // Vincenzo Librandi, Dec 11 2018

a:= n-> 2^ilog2(n):
seq(a(n), n=0..80);  # Alois P. Heinz, Dec 20 2016

A053644[n_] := 2^(Length[ IntegerDigits[n, 2]] - 1); A053644[0] = 0; Table[A053644[n], {n, 0, 74}] (* Jean-François Alcover, Dec 01 2011 *)
nv[n_] := Module[{c = 2^n}, Table[c, {c}]]; Join[{0}, Flatten[Array[nv, 7, 0]]] (* Harvey P. Dale, Jul 17 2012 *)

a(n)=my(k=1);while(k<=n,k<<=1);k>>1 \\ Charles R Greathouse IV, May 27 2011

a(n) = if(!n, 0, 2^exponent(n)) \\ Iain Fox, Dec 10 2018

def a(n): return 0 if n==0 else 2**(len(bin(n)[2:]) - 1) # Indranil Ghosh, May 25 2017

def A053644(n): return 1<Chai Wah Wu, Jul 27 2022

(0 to 127).map(Integer.highestOneBit()) // _Alonso del Arte, Feb 26 2020

a(n) = a(floor(n / 2)) * 2.
a(n) = 2^A000523(n).
From n >= 1 onward, A053644(n) = A062383(n)/2.
a(0) = 0, a(1) = 1 and a(n+1) = a(n)*floor(n/a(n)). - Benoit Cloitre, Aug 17 2002
G.f.: 1/(1 - x) * (x + Sum_{k >= 1} 2^(k - 1)*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = (A003817(n) + 1)/2 = A091940(n) + 1. - Reinhard Zumkeller, Feb 15 2004
a(n) = Sum_{k = 1..n} (floor(2^k/k) - floor((2^k - 1)/k))*A000010(k). - Anthony Browne, Jun 17 2016
a(2^m+k) = 2^m, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Aug 07 2016

A038712 Let k be the exponent of highest power of 2 dividing n (A007814); a(n) = 2^(k+1)-1.

Original entry on oeis.org

1, 3, 1, 7, 1, 3, 1, 15, 1, 3, 1, 7, 1, 3, 1, 31, 1, 3, 1, 7, 1, 3, 1, 15, 1, 3, 1, 7, 1, 3, 1, 63, 1, 3, 1, 7, 1, 3, 1, 15, 1, 3, 1, 7, 1, 3, 1, 31, 1, 3, 1, 7, 1, 3, 1, 15, 1, 3, 1, 7, 1, 3, 1, 127, 1, 3, 1, 7, 1, 3, 1, 15, 1, 3, 1, 7, 1, 3, 1, 31, 1, 3, 1, 7, 1, 3, 1, 15, 1, 3, 1, 7, 1, 3, 1, 63, 1, 3

Offset: 1

Henry Bottomley, May 02 2000

n XOR n-1, i.e., nim-sum of a pair of consecutive numbers.
Denominator of quotient sigma(2*n)/sigma(n). - Labos Elemer, Nov 04 2003
a(n) = the Towers of Hanoi disc moved at the n-th move, using standard moves with discs labeled (1, 3, 7, 15, ...) starting from top (smallest = 1). - Gary W. Adamson, Oct 26 2009
Equals row sums of triangle A168312. - Gary W. Adamson, Nov 22 2009
In the binary expansion of n, delete everything left of the rightmost 1 bit, and set all bits to the right of it. - Ralf Stephan, Aug 22 2013
Every finite sequence of positive integers summing to n may be termwise dominated by a subsequence of the first n values in this sequence [see Bannister et al., 2013]. - David Eppstein, Aug 31 2013
Sum of powers of 2 dividing n. - Omar E. Pol, Aug 18 2019
Given the binary expansion of (n-1) as {b[k-1], b[k-2], ..., b[2], b[1], b[0]}, then the binary expansion of a(n) is {bitand(b[k-1], b[k-2], ..., b[2], b[1], b[0]), bitand(b[k-2], ..., b[2], b[1], b[0]), ..., bitand(b[2], b[1], b[0]), bitand(b[1], b[0]), b[0], 1}. Recursively stated - 0th bit (L.S.B) of a(n), a(n)[0] = 1, a(n)[i] = bitand(a(n)[i-1], (n-1)[i-1]), where n[i] = i-th bit in the binary expansion of n. - Chinmaya Dash, Jun 27 2020

			a(6) = 3 because 110 XOR 101 = 11 base 2 = 3.
From _Omar E. Pol_, Aug 18 2019: (Start)
Illustration of initial terms:
a(n) is also the area of the n-th region of an infinite diagram of compositions (ordered partitions) of the positive integers, where the length of the n-th horizontal line segment is equal to A001511(n) and the length of the n-th vertical line segment is equal to A006519(n), as shown below (first eight regions):
-----------------------------
n    a(n)    Diagram
-----------------------------
.            _ _ _ _
1     1     |_| | | |
2     3     |_ _| | |
3     1     |_|   | |
4     7     |_ _ _| |
5     1     |_| |   |
6     3     |_ _|   |
7     1     |_|     |
8    15     |_ _ _ _|
.
The above diagram represents the eight compositions of 4: [1,1,1,1],[2,1,1],[1,2,1],[3,1],[1,1,2],[2,2],[1,3],[4].
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
M. J. Bannister, Z. Cheng, W. E. Devanny, and D. Eppstein, Superpatterns and universal point sets, 21st Int. Symp. Graph Drawing, 2013, arXiv:1308.0403 [cs.CG], 2013.
Klaus Brockhaus, Illustration of A038712 and A080277.
David Eppstein, 1317131 and majorization by subsequences.
Fabrizio Frati, M. Patrignani, and V. Roselli, LR-Drawings of Ordered Rooted Binary Trees and Near-Linear Area Drawings of Outerplanar Graphs, arXiv preprint arXiv:1610.02841 [cs.CG], 2016.
Malgorzata J. Krawczyk, Paweł Oświęcimka, and Krzysztof Kułakowski, Ordered Avalanches on the Bethe Lattice, Entropy, Vol. 21, (2019) 968.
Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple ordinary generating functions.
Ralf Stephan, Table of generating functions.
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Index entries for sequences related to binary expansion of n.
Index entries for sequences related to Nim-sums.

A038713 translated from binary, diagonals of A003987 on either side of main diagonal.
Cf. A062383. Partial sums give A080277.
Bisection of A089312. Cf. A088837.
a(n)-1 is exponent of 2 in A089893(n).
Cf. A130093.
This is Guy Steele's sequence GS(6, 2) (see A135416).
Cf. A001620, A168312, A220466, A361019 (Dirichlet inverse).

int a(int n) { return n ^ (n-1); } // Russ Cox, May 15 2007

import Data.Bits (xor)
a038712 n = n `xor` (n - 1) :: Integer  -- Reinhard Zumkeller, Apr 23 2012

nmax:=98: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := 2^(p+1)-1 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 01 2013
# second Maple program:
a:= n-> Bits[Xor](n, n-1):
seq(a(n), n=1..98);  # Alois P. Heinz, Feb 02 2023

Table[Denominator[DivisorSigma[1, 2*n]/DivisorSigma[1, n]], {n, 1, 128}]
Table[BitXor[(n + 1), n], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 19 2011 *)

vector(66,n,bitxor(n-1,n)) \\ Joerg Arndt, Sep 01 2013; corrected by Michel Marcus, Aug 02 2018

A038712(n) = ((1<<(1+valuation(n,2)))-1); \\ Antti Karttunen, Nov 24 2024

def A038712(n): return n^(n-1) # Chai Wah Wu, Jul 05 2022

a(n) = A110654(n-1) XOR A008619(n). - Reinhard Zumkeller, Feb 05 2007
a(n) = 2^A001511(n) - 1 = 2*A006519(n) - 1 = 2^(A007814(n)+1) - 1.
Multiplicative with a(2^e) = 2^(e+1)-1, a(p^e) = 1, p > 2. - Vladeta Jovovic, Nov 06 2001; corrected by Jianing Song, Aug 03 2018
Sum_{n>0} a(n)*x^n/(1+x^n) = Sum_{n>0} x^n/(1-x^n). Inverse Moebius transform of A048298. - Vladeta Jovovic, Jan 02 2003
From Ralf Stephan, Jun 15 2003: (Start)
G.f.: Sum_{k>=0} 2^k*x^2^k/(1 - x^2^k).
a(2*n+1) = 1, a(2*n) = 2*a(n)+1. (End)
Equals A130093 * [1, 2, 3, ...]. - Gary W. Adamson, May 13 2007
Sum_{i=1..n} (-1)^A000120(n-i)*a(i) = (-1)^(A000120(n)-1)*n. - Vladimir Shevelev, Mar 17 2009
Dirichlet g.f.: zeta(s)/(1 - 2^(1-s)). - R. J. Mathar, Mar 10 2011
a(n) = A086799(2*n) - 2*n. - Reinhard Zumkeller, Aug 07 2011
a((2*n-1)*2^p) = 2^(p+1)-1, p >= 0. - Johannes W. Meijer, Feb 01 2013
a(n) = A000225(A001511(n)). - Omar E. Pol, Aug 31 2013
a(n) = A000203(n)/A000593(n). - Ivan N. Ianakiev and Omar E. Pol, Dec 14 2017
L.g.f.: -log(Product_{k>=0} (1 - x^(2^k))) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, Mar 15 2018
a(n) = 2^(1 + (A183063(n)/A001227(n))) - 1. - Omar E. Pol, Nov 06 2018
a(n) = sigma(n)/(sigma(2*n) - 2*sigma(n)) = 3*sigma(n)/(sigma(4*n) - 4*sigma(n)) = 7*sigma(n)/(sigma(8*n) - 8*sigma(n)), where sigma(n) = A000203(n). - Peter Bala, Jun 10 2022
Sum_{k=1..n} a(k) ~ n*log_2(n) + (1/2 + (gamma - 1)/log(2))*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 24 2022
a(n) = Sum_{d divides n} m(d)*phi(d), where m(n) = Sum_{d divides n} (-1)^(d+1)* mobius(d). - Peter Bala, Jan 23 2024

Definition corrected by N. J. A. Sloane, Sep 07 2015 at the suggestion of Marc LeBrun

Name corrected by Wolfdieter Lang, Aug 30 2016

A020330 Numbers whose base-2 representation is the juxtaposition of two identical strings.

Original entry on oeis.org

3, 10, 15, 36, 45, 54, 63, 136, 153, 170, 187, 204, 221, 238, 255, 528, 561, 594, 627, 660, 693, 726, 759, 792, 825, 858, 891, 924, 957, 990, 1023, 2080, 2145, 2210, 2275, 2340, 2405, 2470, 2535, 2600, 2665, 2730, 2795, 2860, 2925, 2990, 3055, 3120, 3185, 3250

Offset: 1

David W. Wilson, Melia Aldridge (ma38(AT)spruce.evansville.edu)

All differences are in union of A000051 and A001576. - Vladimir Shevelev, Dec 07 2013

			36 is a term because 36 = 100100_2, which is 100 followed by 100.

Reinhard Zumkeller, Table of n, a(n) for n = 1..8191
Daniel M. Kane, Carlo Sanna, and Jeffrey Shallit, Waring's Theorem for Binary Powers, Combinatorica, Vol. 39, No. 6 (2019), pp. 1335-1350, arXiv preprint, arXiv:1801.04483 [math.NT], 2018.
Parthasarathy Madhusudan, Dirk Nowotka, Aayush Rajasekaran, and Jeffrey Shallit, Lagrange's Theorem for Binary Squares, arXiv:1710.04247 [math.NT], 2017-2018.
Manfred Madritsch and Stephan Wagner, A central limit theorem for integer partitions, Monatshefte für Mathematik, Vol. 161, No. 1 (2010), pp. 85-114, alternative link.
Aayush Rajasekaran, Using Automata Theory to Solve Problems in Additive Number Theory, MS thesis, University of Waterloo, 2018.

Subsequence of A121016.
Cf. A000051, A001576, A007088, A030308, A062383, A070939, A330157.
Column k=0 of A246830, column k=1 of A246834.

a020330 n = foldr (\d v -> 2 * v + d) 0 (bs ++ bs) where
   bs = a030308_row n
-- Reinhard Zumkeller, Feb 19 2013

[n+2*n*2^Floor(Log(2, n)): n in [1..50]]; // Vincenzo Librandi, Apr 05 2018

a:= n-> (l-> Bits[Join]([l[],l[]]))(Bits[Split](n)):
seq(a(n), n=1..50);  # Alois P. Heinz, Aug 24 2024

Table[n + 2 n 2^Floor[Log[2, n]], {n, 50}] (* T. D. Noe, Dec 10 2013 *)
FromDigits[#, 2] & /@ (# <> # & /@ IntegerString[Range@100, 2]) (* Hans Rudolf Widmer, Aug 24 2024 *)

a(n)=n+n<<#binary(n) \\ Charles R Greathouse IV, Mar 29 2013

is(n)=my(L=#binary(n)\2); n>>L==bitand(n,2^L-1) \\ Charles R Greathouse IV, Mar 29 2013

def a(n): return int(bin(n)[2:]*2, 2)
print([a(n) for n in range(1, 51)]) # Michael S. Branicky, Mar 10 2021

def A020330(n): return (n<Chai Wah Wu, Feb 28 2023

a(n) = n + 2*n*2^floor(log_2(n)). - Ralf Stephan, Dec 07 2004
Sum_{n>=1} 1/a(n) = A330157. - Amiram Eldar, Oct 22 2020
a(n) = n * (2^A070939(n) + 1). - Jianing Song, Apr 10 2021

A035928 Numbers n such that BCR(n) = n, where BCR = binary-complement-and-reverse = take one's complement then reverse bit order.

Original entry on oeis.org

2, 10, 12, 38, 42, 52, 56, 142, 150, 170, 178, 204, 212, 232, 240, 542, 558, 598, 614, 666, 682, 722, 738, 796, 812, 852, 868, 920, 936, 976, 992, 2110, 2142, 2222, 2254, 2358, 2390, 2470, 2502, 2618, 2650, 2730, 2762, 2866, 2898, 2978, 3010, 3132, 3164, 3244

Offset: 1

Mike Keith

Numbers n such that A036044(n) = n.
Also: numbers such that n+BR(n) is in A000225={2^k-1} (with BR = binary reversed). - M. F. Hasler, Dec 17 2007
Also called "antipalindromes". - Jeffrey Shallit, Feb 04 2022

			38 is such a number because 38=100110; complement to get 011001, then reverse bit order to get 100110.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
James Haoyu Bai, Joseph Meleshko, Samin Riasat, and Jeffrey Shallit, Quotients of Palindromic and Antipalindromic Numbers, arXiv:2202.13694 [math.NT], 2022.
Aayush Rajasekaran, Jeffrey Shallit, and Tim Smith, Sums of Palindromes: an Approach via Nested-Word Automata, preprint arXiv:1706.10206 [cs.FL], June 30 2017.

Cf. A061855.
Cf. A000225, A036044, A062383.
Intersection of A195064 and A195066; cf. A195063, A195065.

a035928 n = a035928_list !! (n-1)
a035928_list = filter (\x -> a036044 x == x) [0,2..]
-- Reinhard Zumkeller, Sep 16 2011

[seq(ReflectBinSeq(j,(floor_log_2(j)+1)),j=1..256)];
ReflectBinSeq := (x,n) -> (((2^n)*x)+binrevcompl(x));
binrevcompl := proc(nn) local n,z; n := nn; z := 0; while(n <> 0) do z := 2*z + ((n+1) mod 2); n := floor(n/2); od; RETURN(z); 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; # Computes essentially the same as floor(log[2](n))
# alternative Maple program:
q:= n-> (l-> is(n=add((1-l[-i])*2^(i-1), i=1..nops(l))))(Bits[Split](n)):
select(q, [$1..3333])[];  # Alois P. Heinz, Feb 10 2021

bcrQ[n_]:=Module[{idn2=IntegerDigits[n,2]},Reverse[idn2/.{1->0,0->1}] == idn2]; Select[Range[3200],bcrQ] (* Harvey P. Dale, May 24 2012 *)

for(n=1,1000,l=length(binary(n)); b=binary(n); if(sum(i=1,l,abs(component(b,i)-component(b,l+1-i)))==l,print1(n,",")))

for(i=1,999,if(Set(vecextract(t=binary(i),"-1..1")+t)==[1],print1(i","))) \\ M. F. Hasler, Dec 17 2007

a(n) = my (b=binary(n)); (n+1)*2^#b-fromdigits(Vecrev(b),2)-1 \\ Rémy Sigrist, Mar 15 2021

def comp(s): z, o = ord('0'), ord('1'); return s.translate({z:o, o:z})
def BCR(n): return int(comp(bin(n)[2:])[::-1], 2)
def aupto(limit): return [m for m in range(limit+1) if BCR(m) == m]
print(aupto(3244)) # Michael S. Branicky, Feb 10 2021

from itertools import count, islice
def A035928_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:n==int(format(~n&(1<<(m:=n.bit_length()))-1,'0'+str(m)+'b')[::-1],2),count(max(startvalue,1)))
A035928_list = list(islice(A035928_gen(),30)) # Chai Wah Wu, Jun 30 2022

If offset were 0, a(2n+1) - a(2n) = 2^floor(log_2(n)+1).

a(n) = n * A062383(n) + A036044(n). - Rémy Sigrist, Jun 11 2022

More terms from Erich Friedman

A003817 a(0) = 0, a(n) = a(n-1) OR n.

Original entry on oeis.org

0, 1, 3, 3, 7, 7, 7, 7, 15, 15, 15, 15, 15, 15, 15, 15, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63

Offset: 0

Marc LeBrun

Also, 0+1+2+...+n in lunar arithmetic in base 2 written in base 10. - N. J. A. Sloane, Oct 02 2010

For n>0: replace all 0's with 1's in binary representation of n. - Reinhard Zumkeller, Jul 14 2003

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000 [From _Reinhard Zumkeller_, Nov 14 2009]
D. Applegate, M. LeBrun and N. J. A. Sloane, Dismal Arithmetic, arXiv:1107.1130 [math.NT], 2011. [Note: we have now changed the name from "dismal arithmetic" to "lunar arithmetic" - the old name was too depressing]
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Reinhard Zumkeller, Logical Convolutions
Index entries for sequences related to dismal (or lunar) arithmetic

This is Guy Steele's sequence GS(6, 6) (see A135416).
Cf. A000004, A142149, A086099, A142150, A142151, A001477, A062383.
Cf. A167832, A167878. - Reinhard Zumkeller, Nov 14 2009
Cf. A179526; subsequence of A007448. - Reinhard Zumkeller, Jul 18 2010
Cf. A265705.

import Data.Bits ((.|.))
a003817 n = if n == 0 then 0 else 2 * a053644 n - 1
a003817_list = scanl (.|.) 0 [1..] :: [Integer]
-- Reinhard Zumkeller, Dec 08 2012, Jan 15 2012

A003817 := n -> n + Bits:-Nand(n, n):
seq(A003817(n), n=0..61); # Peter Luschny, Sep 23 2019

a[0] = 0; a[n_] := a[n] = BitOr[ a[n-1], n]; Table[a[n], {n, 0, 61}] (* Jean-François Alcover, Dec 19 2011 *)
nxt[{n_,a_}]:={n+1,BitOr[a,n+1]}; Transpose[NestList[nxt,{0,0},70]] [[2]] (* Harvey P. Dale, May 06 2016 *)
2^BitLength[Range[0,100]]-1 (* Paolo Xausa, Feb 08 2024 *)

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

def a(n): return 0 if n==0 else 1 + 2*a(int(n/2)) # Indranil Ghosh, Apr 28 2017

def A003817(n): return (1<Chai Wah Wu, Jul 17 2024

a(n) = a(n-1) + n*(1-floor(a(n-1)/n)). If 2^(k-1) <= n < 2^k, a(n) = 2^k - 1. - Benoit Cloitre, Aug 25 2002
a(n) = 1 + 2*a(floor(n/2)) for n > 0. - Benoit Cloitre, Apr 04 2003
G.f.: (1/(1-x)) * Sum_{k>=0} 2^k*x^2^k. - Ralf Stephan, Apr 18 2003
a(n) = 2*A053644(n)-1 = A092323(n) + A053644(n). - Reinhard Zumkeller, Feb 15 2004; corrected by Anthony Browne, Jun 26 2016
a(n) = OR{k OR (n-k): 0<=k<=n}. - Reinhard Zumkeller, Jul 15 2008
For n>0: a(n+1) = A035327(n) + n = A035327(n) XOR n. - Reinhard Zumkeller, Nov 14 2009
A092323(n+1) = floor(a(n)/2). - Reinhard Zumkeller, Jul 18 2010
a(n) = A265705(n,0) = A265705(n,n). - Reinhard Zumkeller, Dec 15 2015
a(n) = A062383(n) - 1.
G.f. A(x) satisfies: A(x) = 2*A(x^2)*(1 + x) + x/(1 - x). - Ilya Gutkovskiy, Aug 31 2019
a(n) >= A175039(n) - Austin Shapiro, Dec 29 2022

cp's OEIS Frontend

A339570 Denote the van der Corput sequence of fractions 1/2, 1/4, 3/4, 1/8, 5/8, 3/8, 7/8, 1/16, ... (A030101/A062383) by v(n), n >= 1. Then a(n) = denominator of v(A014486(n)).

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A153587 a(n) = n mod (A062383(n) - n).

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C

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A029837 Binary order of n: log_2(n) rounded up to next integer.

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A030101 a(n) is the number produced when n is converted to binary digits, the binary digits are reversed and then converted back into a decimal number.

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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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Coq

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