A053645 -id:A053645 - cp's OEIS frontend

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.

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

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

A007931 Numbers that contain only 1's and 2's. Nonempty binary strings of length n in lexicographic order.

Original entry on oeis.org

1, 2, 11, 12, 21, 22, 111, 112, 121, 122, 211, 212, 221, 222, 1111, 1112, 1121, 1122, 1211, 1212, 1221, 1222, 2111, 2112, 2121, 2122, 2211, 2212, 2221, 2222, 11111, 11112, 11121, 11122, 11211, 11212, 11221, 11222, 12111, 12112, 12121, 12122

Offset: 1

R. Muller

Numbers written in the dyadic system [Smullyan, Stillwell]. - N. J. A. Sloane, Feb 13 2019
Logic-binary sequence: prefix it by the empty word to have all binary words on the alphabet {1,2}.
The least binary word of length k is a(2^k - 1).
See Mathematica program for logic-binary sequence using (0,1) in place of (1,2); the sequence starts with 0,1,00,01,10. - Clark Kimberling, Feb 09 2012
A007953(a(n)) = A014701(n+1); A007954(a(n)) = A048896(n). - Reinhard Zumkeller, Oct 26 2012
a(n) is n written in base 2 where zeros are not allowed but twos are. The two distinct digits used are 1, 2 instead of 0, 1. To obtain this sequence from the "canonical" base 2 sequence with zeros allowed, just replace any 0 with a 2 and then subtract one from the group of digits situated on the left: (10-->2; 100-->12; 110-->22; 1000-->112; 1010-->122). - Robin Garcia, Jan 31 2014
For numbers made of only two different digits, see also A007088 (digits 0 & 1), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256340(digits 7 & 8), A256341 (digits 8 & 9), and A032804-A032816 (in other bases). Numbers with exactly two distinct (but unspecified) digits in base 10 are listed in A031955, for other bases in A031948-A031954. - M. F. Hasler, Apr 04 2015
The variant with digits {0, 1} instead of {1, 2} is obtained by deleting all initial digits in sequence A007088 (numbers written in base 2). - M. F. Hasler, Nov 03 2020

			Positive numbers may not start with 0 in the OEIS, otherwise this sequence would have been written as: 0, 1, 00, 01, 10, 11, 000, 001, 010, 011, 100, 101, 110, 111, 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 00000, 00001, 00010, 00011, 00100, 00101, 00110, 00111, 01000, 01001, 01010, 01011, ...
From _Hieronymus Fischer_, Jun 06 2012: (Start)
a(10)   = 122.
a(100)  = 211212.
a(10^3) = 222212112.
a(10^4) = 1122211121112.
a(10^5) = 2111122121211112.
a(10^6) = 2221211112112111112.
a(10^7) = 11221112112122121111112.
a(10^8) = 12222212122221111211111112.
a(10^9) = 22122211221212211212111111112. (End)

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 2. - From N. J. A. Sloane, Jul 26 2012
K. Atanassov, On the 97th, 98th and the 99th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 3, 89-93.
R. M. Smullyan, Theory of Formal Systems, Princeton, 1961.
John Stillwell, Reverse Mathematics, Princeton, 2018. See p. 90.

Hieronymus Fischer, Table of n, a(n) for n = 1..10000 (terms up to 2^10-2 from T. D. Noe, corrected by Sean A. Irvine, April 18 2019)
K. Atanassov, On Some of Smarandache's Problems, American Research Press, 1999, 16-21.
EMIS, Mirror site for Southwest Journal of Pure and Applied Mathematics
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1, 1995.
R. R. Forslund, Positive Integer Pages
James E. Foster, A Number System without a Zero-Symbol, Mathematics Magazine, Vol. 21, No. 1. (1947), pp. 39-41.
F. Smarandache, Only Problems, Not Solutions!.
Index entries for 10-automatic sequences.

Cf. A007932 (digits 1-3), A059893, A045670, A052382 (digits 1-9), A059939, A059941, A059943, A032924, A084544, A084545, A046034 (prime digits 2,3,5,7), A089581, A084984 (no prime digits); A001742, A001743, A001744: loops; A202267 (digits 0, 1 and primes), A202268 (digits 1,4,6,8,9), A014261 (odd digits), A014263 (even digits).
Cf. A007088 (digits 0 & 1), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256340 (digits 7 & 8), A256341 (digits 8 & 9), and A032804-A032816 (in other bases).
Cf. A020450 (primes).

a007931 n = f (n + 1) where
   f x = if x < 2 then 0 else (10 * f x') + m + 1
     where (x', m) = divMod x 2
-- Reinhard Zumkeller, Oct 26 2012

[n: n in [1..100000] | Set(Intseq(n)) subset {1,2}]; // Vincenzo Librandi, Aug 19 2016

# Maple program to produce the sequence:
a:= proc(n) local m, r, d; m, r:= n, 0;
      while m>0 do d:= irem(m, 2, 'm');
        if d=0 then d:=2; m:= m-1 fi;
        r:= d, r
      od; parse(cat(r))/10
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 26 2016
# Maple program to invert this sequence: given a(n), it returns n. - N. J. A. Sloane, Jul 09 2012
invert7931:=proc(u)
local t1,t2,i;
t1:=convert(u,base,10);
[seq(t1[i]-1,i=1..nops(t1))];
[op(%),1];
t2:=convert(%,base,2,10);
add(t2[i]*10^(i-1),i=1..nops(t2))-1;
end;

f[n_] := FromDigits[Rest@IntegerDigits[n + 1, 2] + 1]; Array[f, 42] (* Robert G. Wilson v Sep 14 2006 *)
(* Next, A007931 using (0,1) instead of (1,2) *)
d[n_] := FromDigits[Rest@IntegerDigits[n + 1, 2] + 1]; Array[FromCharacterCode[ToCharacterCode[ToString[d[#]]] - 1] &, 100] (* Peter J. C. Moses, at request of Clark Kimberling, Feb 09 2012 *)
Flatten[Table[FromDigits/@Tuples[{1,2},n],{n,5}]] (* Harvey P. Dale, Sep 13 2014 *)

apply( {A007931(n)=fromdigits([d+1|d<-binary(n+1)[^1]])}, [1..44]) \\ M. F. Hasler, Nov 03 2020, replacing older code from Mar 26 2015

/* inverse function */ apply( {A007931_inv(N)=fromdigits([d-1|d<-digits(N)],2)+2<M. F. Hasler, Nov 09 2020

def a(n): return int(bin(n+1)[3:].replace('1', '2').replace('0', '1'))
print([a(n) for n in range(1, 45)]) # Michael S. Branicky, May 13 2021

def A007931(n): return int(s:=bin(n+1)[3:])+(10**(len(s))-1)//9 # Chai Wah Wu, Jun 13 2025

To get a(n), write n+1 in base 2, remove initial 1, add 1 to all remaining digits: e.g., eleven (11) in base 2 is 1011; remove initial 1 and add 1 to remaining digits: a(10)=122. - Clark Kimberling, Mar 11 2003
Conversely, given a(n), to get n: subtract 1 from all digits, prefix with an initial 1, convert this binary number to base 10, subtract 1. E.g., a(6)=22 -> 11 -> 111 -> 7 -> 6. - N. J. A. Sloane, Jul 09 2012
a(n) = A053645(n+1)+A002275(A000523(n)) = a(n-2^b(n))+10^b(n) where b(n) = A059939(n) = floor(log_2(n+1)-1). - Henry Bottomley, Feb 14 2001
From Hieronymus Fischer, Jun 06 2012 and Jun 08 2012: (Start)
The formulas are designed to calculate base-10 numbers only using the digits 1 and 2.
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 2)*10^j, where m = floor(log_2(n+1)), b(j) = floor((n+1-2^m)/(2^j)).
Special values:
a(k*(2^n-1)) = k*(10^n-1)/9, k= 1,2.
a(3*2^n-2) = (11*10^n-2)/9 = 10^n+2*(10^n-1)/9.
a(2^n-2) = 2*(10^(n-1)-1)/9, n>1.
Inequalities:
a(n) <= (10^log_2(n+1)-1)/9, equality holds for n=2^k-1, k>0.
a(n) > (2/10)*(10^log_2(n+1)-1)/9.
Lower and upper limits:
lim inf a(n)/10^log_2(n) = 1/45, for n --> infinity.
lim sup a(n)/10^log_2(n) = 1/9, for n --> infinity.
G.f.: g(x) = (1/(x(1-x)))*sum_{j=0..infinity} 10^j* x^(2*2^j)*(1 + 2 x^2^j)/(1 + x^2^j).
Also: g(x) = (1/(1-x))*(h_(2,0)(x) + h_(2,1)(x) - 2*h_(2,2)(x)), where h_(2,k)(x) = sum_{j>=0} 10^j*x^(2^(j+1)-1)*x^(k*2^j)/(1-x^2^(j+1)).
Also: g(x) = (1/(1-x)) sum_{j>=0} (1 - 3(x^2^j)^2 + 2(x^2^j)^3)*x^2^j*f_j(x)/(1-x^2^j), where f_j(x) = 10^j*x^(2^j-1)/(1-(x^2^j)^2). The f_j obey the recurrence f_0(x) = 1/(1-x^2), f_(j+1)(x) = 10x*f_j(x^2). (End)

Some crossrefs added by Hieronymus Fischer, Jun 06 2012

Edited by M. F. Hasler, Mar 26 2015

A225620 Indices of partitions in the table of compositions of A228351.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 10, 12, 14, 15, 16, 20, 24, 26, 28, 30, 31, 32, 36, 40, 42, 48, 52, 56, 58, 60, 62, 63, 64, 72, 80, 84, 96, 100, 104, 106, 112, 116, 120, 122, 124, 126, 127, 128, 136, 144, 160, 164, 168, 170, 192, 200, 208, 212, 224, 228, 232, 234, 240, 244, 248, 250, 252, 254, 255

Offset: 1

Omar E. Pol, Aug 03 2013

Also triangle read by rows in which T(n,k) is the decimal representation of a binary number whose mirror represents the k-th partition of n according with the list of juxtaposed reverse-lexicographically ordered partitions of the positive integers (A026792).
In order to construct this sequence as a triangle we use the following rules:
- In the list of A026792 we replace each part of size j of the k-th partition of n by concatenation of j - 1 zeros and only one 1.
- Then replace this new set of parts by the concatenation of its parts.
- Then replace this string by its mirror version which is a binary number.
T(n,k) is the decimal value of this binary number, which represents the k-th partition of n (see example).
The partitions of n are represented by a subsequence with A000041(n) integers starting with 2^(n-1) and ending with 2^n - 1, n >= 1. The odd numbers of the sequence are in A000225.
First differs from A065609 at a(23).
Conjecture: this sequence is a sorted version of b(n) where b(2^k) = 2^k for k >= 0, b(n) = A080100(n)*(2*b(A053645(n)) + 1) otherwise. - Mikhail Kurkov, Oct 21 2023

			T(6,8) = 58 because 58 in base 2 is 111010 whose mirror is 010111 which is the concatenation of 01, 01, 1, 1, whose number of digits are 2, 2, 1, 1, which are also the 8th partition of 6.
Illustration of initial terms:
The sequence represents a table of partitions (see below):
--------------------------------------------------------
.            Binary                        Partitions
n  k  T(n,k) number  Mirror   Diagram       (A026792)
.                                          1 2 3 4 5 6
--------------------------------------------------------
.                             _
1  1     1       1    1        |           1,
.                             _ _
1  1     2      10    01      _  |           2,
2  2     3      11    11       | |         1,1,
.                             _ _ _
3  1     4     100    001     _ _  |           3,
3  2     6     110    011     _  | |         2,1,
3  3     7     111    111      | | |       1,1,1,
.                             _ _ _ _
4  1     8    1000    0001    _ _    |           4,
4  2    10    1010    0101    _ _|_  |         2,2,
4  3    12    1100    0011    _ _  | |         3,1,
4  4    14    1110    0111    _  | | |       2,1,1,
4  5    15    1111    1111     | | | |     1,1,1,1,
.                             _ _ _ _ _
5  1    16   10000    00001   _ _ _    |           5,
5  2    20   10100    00101   _ _ _|_  |         3,2,
5  3    24   11000    00011   _ _    | |         4,1,
5  4    26   11010    01011   _ _|_  | |       2,2,1,
5  5    28   11100    00111   _ _  | | |       3,1,1,
5  6    30   11110    01111   _  | | | |     2,1,1,1,
5  7    31   11111    11111    | | | | |   1,1,1,1,1,
.                             _ _ _ _ _ _
6  1    32  100000    000001  _ _ _      |           6
6  2    36  100100    001001  _ _ _|_    |         3,3,
6  3    40  101000    000101  _ _    |   |         4,2,
6  4    42  101010    010101  _ _|_ _|_  |       2,2,2,
6  5    48  110000    000011  _ _ _    | |         5,1,
6  6    52  110100    001011  _ _ _|_  | |       3,2,1,
6  7    56  111000    000111  _ _    | | |       4,1,1,
6  8    58  111010    010111  _ _|_  | | |     2,2,1,1,
6  9    60  111100    001111  _ _  | | | |     3,1,1,1,
6  10   62  111110    011111  _  | | | | |   2,1,1,1,1,
6  11   63  111111    111111   | | | | | | 1,1,1,1,1,1,
.
Triangle begins:
  1;
  2,   3;
  4,   6,  7;
  8,  10, 12, 14, 15;
  16, 20, 24, 26, 28, 30, 31;
  32, 36, 40, 42, 48, 52, 56, 58, 60, 62, 63;
  ...
From _Gus Wiseman_, Apr 01 2020: (Start)
Using the encoding of A066099, this sequence ranks all finite nonempty multisets, as follows.
   1: {1}
   2: {2}
   3: {1,1}
   4: {3}
   6: {1,2}
   7: {1,1,1}
   8: {4}
  10: {2,2}
  12: {1,3}
  14: {1,1,2}
  15: {1,1,1,1}
  16: {5}
  20: {2,3}
  24: {1,4}
  26: {1,2,2}
  28: {1,1,3}
  30: {1,1,1,2}
  31: {1,1,1,1,1}
(End)

Column 1 is A000079. Row n has length A000041(n). Right border gives A000225.
Cf. A026792, A065609, A135010, A141285, A186114, A194446, A194546, A206437, A207779, A211978, A225600, A225610, A228351.
The case covering an initial interval is A333379 or A333380.
All of the following pertain to compositions in the order of A066099.
- The weakly increasing version is this sequence.
- The weakly decreasing version is A114994.
- The strictly increasing version is A333255.
- The strictly decreasing version is A333256.
- The unequal version is A233564.
- The equal version is A272919.
- The case covering an initial interval is A333217.
- Initial intervals are ranked by A164894.
- Reversed initial intervals are ranked by A246534.
Cf. A000120, A029931, A048793, A070939, A124766, A233249, A333219, A333220.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Select[Range[0,100],LessEqual@@stc[#]&] (* Gus Wiseman, Apr 01 2020 *)

Conjecture: a(A000070(m) - k) = 2^m - A228354(k) for m > 0, 0 < k <= A000041(m). - Mikhail Kurkov, Oct 20 2023

A004760 List of numbers whose binary expansion does not begin 10.

Original entry on oeis.org

0, 1, 3, 6, 7, 12, 13, 14, 15, 24, 25, 26, 27, 28, 29, 30, 31, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122

Offset: 1

N. J. A. Sloane

For n >= 2 sequence {a(n+2)} is the minimal recursive such that A007814(a(n+2))=A007814(n). - Vladimir Shevelev, Apr 27 2009
A053645(a(n)) = n-1 for n > 0. - Reinhard Zumkeller, May 20 2009
a(n+1) is also the number of nodes in a complete binary tree with n nodes in the bottommost level. - Jacob Jona Fahlenkamp, Feb 01 2023

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Vladimir Shevelev, Several results on sequences which are similar to the positive integers, arXiv:0904.2101 [math.NT], 2009. [_Vladimir Shevelev_, Apr 15 2009]

Cf. A004761, A007814, A010060, A159559, A159560, A159615, A159619, A159629, A159698, A004759.

0,1,seq(seq(3*2^d+x,x=0..2^d-1),d=0..6); # Robert Israel, Aug 03 2016

Select[Range@ 125, If[Length@ # < 2, #, Take[#, 2]] &@ IntegerDigits[#, 2] != {1, 0} &] (* Michael De Vlieger, Aug 02 2016 *)

is(n)=n<2 || binary(n)[2] \\ Charles R Greathouse IV, Sep 23 2012

print1("0, 1");for(i=0,5,for(n=3<Charles R Greathouse IV, Sep 23 2012

a(n) = if(n<=2,n-1, (n-=2) + 2<Kevin Ryde, Jul 22 2022

def A004760(n): return m+(1<0 else n-1 # Chai Wah Wu, Jul 26 2023

maxrow <- 8 # by choice
b01 <- 1
for(m in 0:maxrow){
  b01 <- c(b01,rep(1,2^(m+1))); b01[2^(m+1):(2^(m+1)+2^m-1)] <- 0
}
a <- which(b01 == 1)
# Yosu Yurramendi, Mar 30 2017

For n > 0, a(n) = 3n - 2 - A006257(n-1). - Ralf Stephan, Sep 16 2003
a(0) = 0, a(1) = 1, for n > 0: a(2n) = 2*a(n) + 1, a(2n+1) = 2*a(n+1). - Philippe Deléham, Feb 29 2004
For n >= 3, A007814(a(n)) = A007814(n-2). - Vladimir Shevelev, Apr 15 2009
a(n+2) = min{m>a(n+1): A007814(m)=A007814(n)}; A010060(a(n+2)) = 1-A010060(n). - Vladimir Shevelev, Apr 27 2009
a(1)=0, a(2)=1, a(2^m+k+2) = 2^(m+1) + 2^m+k, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Jul 30 2016
G.f.: x/(1-x)^2 + (x/(1-x))*Sum_{k>=0} 2^k*x^(2^k). - Robert Israel, Aug 03 2016
a(2^m+k) = A004761(2^m+k) + 2^m, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Aug 08 2016
For n > 0, a(n+1) = n + 2^ceiling(log_2(n)) - 1. - Jacob Jona Fahlenkamp, Feb 01 2023

Offset changed to 1, b-file corrected. - N. J. A. Sloane, Aug 07 2016

A076478 The binary Champernowne sequence: concatenate binary vectors of lengths 1, 2, 3, ... in numerical order.

Original entry on oeis.org

0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0

Offset: 0

N. J. A. Sloane, Nov 10 2002

Can also be seen as triangle where row n contains all binary vectors of length n+1. - Reinhard Zumkeller, Aug 18 2015
From Clark Kimberling, Jul 18 2021: (Start)
In the following list, W represents the sequence of words w(n) represented by A076478. The list includes five partitions and two self-inverse permutations of the positive integers.
length of w(n): A000523
positions in W of words w(n) such that # 0's = # 1's: A258410;
positions in W of words w(n) such that # 0's < # 1's: A346299;
positions in W of words w(n) such that # 0's > # 1's: A346300;
positions in W of words w(n) that end with 0: A005498;
positions in W of words w(n) that end with 1: A005843;
positions in W of words w(n) such that first digit = last digit: A346301;
positions in W of words w(n) such that first digit != last digit: A346302;
positions in W of words w(n) such that 1st digit = 0 and last digit 0: A171757;
positions in W of words w(n) such that 1st digit = 0 and last digit 1: A346303;
positions in W of words w(n) such that 1st digit = 1 and last digit 0: A346304;
positions in W of words w(n) such that 1st digit = 1 and last digit 1: A346305;
position in W of n-th positive integer (base 2):  A206332;
positions in W of binary complement of w(n):  A346306;
sum of digits in w(n): A048881;
number of runs in w(n): A346307;
positions in W of palindromes: A346308;
positions in W of words such that #0's - #1's is odd: A346309;
positions in W of words such that #0's - #1's is even: A346310;
positions in W of the reversal of the n-th word in W: A081241. (End)

			0,
1,
0,0,
0,1,
1,0,
1,1,
0,0,0,
0,0,1,
0,1,0,
0,1,1,
1,0,0,
1,0,1,
...

Bodil Branner, Dynamics, Chap. IV.14 of The Princeton Companion to Mathematics, ed. T. Gowers, p. 499.
K. Dajani and C. Kraaikamp, Ergodic Theory of Numbers, Math. Assoc. America, 2002, p. 72.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Michael Barnsley and Andrew Vince, Self-similar polygonal tiling, The American Mathematical Monthly 124.10 (2017): 905-921. See page 917.
Igor Pak, Complexity problems in enumerative combinatorics, arXiv:1803.06636 [math.CO], 2018.

Cf. A007931, A030308, A053645.

Cf. A341256, A341258, A341334, A342753, A342910.

import Data.List (unfoldr)
a076478 n = a076478_list !! n
a076478_list = concat $ tail $ map (tail . reverse . unfoldr
   (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2 )) [1..]
-- Reinhard Zumkeller, Feb 08 2012

a076478_row n = a076478_tabf !! n :: [[Int]]
a076478_tabf = tail $ iterate (\bs -> map (0 :) bs ++ map (1 :) bs) [[]]
a076478_list' = concat $ concat a076478_tabf
-- Reinhard Zumkeller, Aug 18 2015

d[n_] := Rest@IntegerDigits[n + 1, 2] + 1; -1 + Flatten[Array[d, 50]] (* Clark Kimberling, Feb 07 2012 *)
z = 1000;
t1 = Table[Tuples[{0, 1}, n], {n, 1, 10}];
"All binary words, lexicographic order:"
tt = Flatten[t1, 1]; (* all binary words, lexicographic order *)
"All binary words, flattened:"
Flatten[tt];
w[n_] := tt[[n]];
"List tt of all binary words:"
tt = Table[w[n], {n, 1, z}]; (*  all the binary words *)
u1 = Flatten[tt]; (* words, concatenated, A076478, binary Champernowne sequence *)
u2 = Map[Length, tt];
"Positions of 0^n:"
Flatten[Position[Map[Union, tt], {0}]]
"Positions of 1^n:"
Flatten[Position[Map[Union, tt], {1}]]
"Positions of words in which #0's = #1's:"  (* A258410 *)
"This and the next two sequences partition N."
u3 = Select[Range[Length[tt]], Count[tt[[#]], 0] == Count[tt[[#]], 1] &]
"Positions of words in which #0's < #1's:"  (* A346299 *)
u4 = Select[Range[Length[tt]], Count[tt[[#]], 0] < Count[tt[[#]], 1] &]
"Positions of words in which #0's > #1's:"  (* A346300 *)
u5 = Select[Range[Length[tt]], Count[tt[[#]], 0] > Count[tt[[#]], 1] &]
"Positions of words ending with 0:" (* A005498 *)
u6 = Select[Range[Length[tt]], Last[tt[[#]]] == 0 &]
"Positions of words ending with 1:" (* A005843 *)
u7 = Select[Range[Length[tt]], Last[tt[[#]]] == 1 &]
"Positions of words starting and ending with same digit:" (* A346301 *)
u8 = Select[Range[Length[tt]], First[tt[[#]]] == Last[tt[[#]]] &]
"Positions of words starting and ending with opposite digits:" (* A346302  *)
u9 = Select[Range[Length[tt]], First[tt[[#]]] != Last[tt[[#]]] &]
"Positions of words starting with 0 and ending with 0:" (* A346303 *)
"This and the next three sequences partition N."
u10 = Select[Range[Length[tt]], First[tt[[#]]] == 0 && Last[tt[[#]]] == 0 &]
"Positions of words starting with 0 and ending with 1:" (* A171757 *)
u11 = Select[Range[Length[tt]], First[tt[[#]]] == 0 && Last[tt[[#]]] == 1 &]
"Positions of words starting with 1 and ending with 0:" (* A346304 *)
u12 = Select[Range[Length[tt]], First[tt[[#]]] == 1 && Last[tt[[#]]] == 0 &]
"Positions of words starting with 1 and ending with 1:" (* A346305 *)
u13 = Select[Range[Length[tt]], First[tt[[#]]] == 1 && Last[tt[[#]]] == 1 &]
"Position of n-th positive integer (base 2) in tt:"
d[n_] := If[First[w[n]] == 1, FromDigits[w[n], 2]];
u14 = Flatten[Table[Position[Table[d[n], {n, 1, 200}], n], {n, 1, 200}]] (* A206332 *)
"Position of binary complement of w(n):"
u15 = comp = Flatten[Table[Position[tt, 1 - w[n]], {n, 1, 50}]] (* A346306 *)
"Sum of digits of w(n):"
u16 = Table[Total[w[n]], {n, 1, 100}] (* A048881 *)
"Number of runs in w(n):"
u17 = Map[Length, Table[Map[Length, Split[w[n]]], {n, 1, 100}]] (* A346307 *)
"Palindromes:"
Select[tt, # == Reverse[#] &]
"Positions of palindromes:"
u18 = Select[Range[Length[tt]], tt[[#]] == Reverse[tt[[#]]] &] (* A346308 *)
"Positions of words in which #0's - #1's is odd:"
u19 = Select[Range[Length[tt]], OddQ[Count[w[#], 0] - Count[w[#], 1]] &] (* A346309 *)
"Positions of words in which #0's - #1's is even:"
u20 = Select[Range[Length[tt]], EvenQ[Count[w[#], 0] - Count[w[#], 1]] &] (* A346310 *)
"Position of the reversal of the n-th word:"  (* A081241 *)
u21 = Flatten[Table[Position[tt, Reverse[w[n]]], {n, 1, 150}]]
(* Clark Kimberling, Jul 18 2011 *)

{m=5; for(d=1,m, for(k=0,2^d-1,v=binary(k); while(matsize(v)[2]

listn(n)= my(a=List(), i=0, s=0); while(s<=n, listput(~a, binary(i++)[^1]); s+=#a[#a]); concat(a)[1..n+1]; \\ Ruud H.G. van Tol, Mar 17 2025

from itertools import count, product
def agen():
    for digits in count(1):
        for b in product([0, 1], repeat=digits):
            yield from b
g = agen()
print([next(g) for n in range(105)]) # Michael S. Branicky, Jul 18 2021

To get the m-th binary vector, write m+1 in base 2 and remove the initial 1. - Clark Kimberling, Feb 07 2010

Extended by Klaus Brockhaus, Nov 11 2002

A153141 Permutation of nonnegative integers: A059893-conjugate of A153151.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Dec 20 2008

This permutation is induced by a wreath recursion a = s(a,b), b = (b,b) (i.e., binary transducer, where s means that the bits at that state are toggled: 0 <-> 1) given on page 103 of the Bondarenko, Grigorchuk, et al. paper, starting from the active (swapping) state a and rewriting bits from the second most significant bit to the least significant end, continuing complementing as long as the first 1-bit is reached, which is the last bit to be complemented.
The automorphism group of infinite binary tree (isomorphic to an infinitely iterated wreath product of cyclic groups of two elements) embeds naturally into the group of "size-preserving Catalan bijections". Scheme-function psi gives an isomorphism that maps this kind of permutation to the corresponding Catalan automorphism/bijection (that acts on S-expressions). The following identities hold: *A069770 = psi(A063946) (just swap the left and right subtrees of the root), *A057163 = psi(A054429) (reflect the whole tree), *A069767 = psi(A153141), *A069768 = psi(A153142), *A122353 = psi(A006068), *A122354 = psi(A003188), *A122301 = psi(A154435), *A122302 = psi(A154436) and from *A154449 = psi(A154439) up to *A154458 = psi(A154448). See also comments at A153246 and A153830.
a(1) to a(2^n) is the sequence of row sequency numbers in a Hadamard-Walsh matrix of order 2^n, when constructed to give "dyadic" or Payley sequency ordering. - Ross Drewe, Mar 15 2014
In the Stern-Brocot enumeration system for positive rationals (A007305/A047679), this permutation converts the denominator into the numerator: A007305(n) = A047679(a(n)). - Yosu Yurramendi, Aug 01 2020

			18 = 10010 in binary and after complementing the second, third and fourth most significant bits at positions 3, 2 and 1, we get 1110, at which point we stop (because bit-1 was originally 1) and fix the rest, so we get 11100 (28 in binary), thus a(18)=28. This is the inverse of "binary adding machine". See pages 8, 9 and 103 in the Bondarenko, Grigorchuk, et al. paper.
19 = 10011 in binary. By complementing bits in (zero-based) positions 3, 2 and 1 we get 11101 in binary, which is 29 in decimal, thus a(19)=29.

A. Karttunen, Table of n, a(n) for n = 0..2047
Ievgen Bondarenko, Rostislav Grigorchuk, Rostyslav Kravchenko, Yevgen Muntyan, Volodymyr Nekrashevych, Dmytro Savchuk, and Zoran Sunic, Classification of groups generated by 3-state automata over a 2-letter alphabet, arXiv:0803.3555 [math.GR], 2008, pp. 8--9 & 103.
S. Wolfram and R. Lamy, Discussion on the NKS Forum
Index entries for sequences that are permutations of the natural numbers

Inverse: A153142. a(n) = A059893(A153151(A059893(n))) = A059894(A153152(A059894(n))) = A154440(A154445(n)) = A154442(A154443(n)). Corresponds to A069767 in the group of Catalan bijections. Cf. also A154435-A154436, A154439-A154448, A072376.
Differs from A006068 for the first time at n=14, where a(14)=10 while A006068(14)=11.
A240908-A240910 these give "natural" instead of "dyadic" sequency ordering values for Hadamard-Walsh matrices, orders 8,16,32. - Ross Drewe, Mar 15 2014

def ok(n): return n&(n - 1)==0
def a153151(n): return n if n<2 else 2*n - 1 if ok(n) else n - 1
def A(n): return (int(bin(n)[2:][::-1], 2) - 1)/2
def msb(n): return n if n<3 else msb(n/2)*2
def a059893(n): return A(n) + msb(n)
def a(n): return 0 if n==0 else a059893(a153151(a059893(n))) # Indranil Ghosh, Jun 09 2017

maxlevel <- 5 # by choice
a <- 1
for(m in 1:maxlevel){
a[2^m    ] <- 2^(m+1) - 1
a[2^m + 1] <- 2^(m+1) - 2
for (k in 1:(2^m-1)){
   a[2^(m+1) + 2*k    ] <- 2*a[2^m + k]
   a[2^(m+1) + 2*k + 1] <- 2*a[2^m + k] + 1}
}
a <- c(0,a)
# Yosu Yurramendi, Aug 01 2020

Conjecture: a(n) = f(a(f(a(A053645(n)))) + A053644(n)) for n > 0 where f(n) = A054429(n) for n > 0 with f(0) = 0. - Mikhail Kurkov, Oct 02 2023
From Mikhail Kurkov, Dec 22 2023: (Start)
a(n) < 2^k iff n < 2^k for k >= 0.
Conjectured formulas:
a(2^m + k) = f(2^m + f(k)) for m >= 0, 0 <= k < 2^m with a(0) = 0.
a(n) = f(A153142(f(n))) for n > 0 with a(0) = 0. (End)

A038572 a(n) = n rotated one binary place to the right.

Original entry on oeis.org

0, 1, 1, 3, 2, 6, 3, 7, 4, 12, 5, 13, 6, 14, 7, 15, 8, 24, 9, 25, 10, 26, 11, 27, 12, 28, 13, 29, 14, 30, 15, 31, 16, 48, 17, 49, 18, 50, 19, 51, 20, 52, 21, 53, 22, 54, 23, 55, 24, 56, 25, 57, 26, 58, 27, 59, 28, 60, 29, 61, 30, 62, 31, 63, 32, 96, 33, 97, 34, 98, 35, 99, 36, 100

Offset: 0

Marc LeBrun

Iterating a(n), a(a(n)), ... eventually leads to 2^A000120(n) - 1. - Franklin T. Adams-Watters, Apr 09 2010

			For n = 35, 35_10 = 100011_2, which after rotating one binary place to the right becomes 110001. Now, 110001_2 = 49_10. So, a(35) = 49. - _Indranil Ghosh_, Jan 21 2017

Indranil Ghosh, Table of n, a(n) for n = 0..20000 (first 1024 terms from T. D. Noe)

Cf. A006257, A088146.

a038572 0 = 0
a038572 n = a053645 n * m + n' where (n', m) = divMod n 2
-- Reinhard Zumkeller, Dec 03 2012

A038572 := 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

Table[ FromDigits[ RotateRight[ IntegerDigits[n, 2]], 2], {n, 0, 80}] (* Robert G. Wilson v *)

a(n)=if(n<2,return(n)); my(d=binary(n)); fromdigits(concat(d[#d], d[1..#d-1]),2) \\ Charles R Greathouse IV, Sep 02 2015

def A038572(n):
    x = bin(n)[2:]
    return int(x[-1]+x[:-1],2) # Indranil Ghosh, Jan 21 2017

def A038572(n): return (n>>1)+(1<Chai Wah Wu, Jan 22 2023

a(n) = A053645(n) * A000035(n) + A004526(n) = most significant bit(n) * least significant bit(n) + floor(n/2).

a(0)=0, a(1)=1, a(2n) = n, a(2n+1) = 2a(n) + 2a(n+1) - n. - Ralf Stephan, Oct 24 2003

A329369 Number of permutations of {1,2,...,m} with excedance set constructed by taking m-i (0 < i < m) if b(i-1) = 1 where b(k)b(k-1)...b(1)b(0) (0 <= k < m-1) is the binary expansion of n.

Original entry on oeis.org

1, 1, 3, 1, 7, 3, 7, 1, 15, 7, 17, 3, 31, 7, 15, 1, 31, 15, 37, 7, 69, 17, 37, 3, 115, 31, 69, 7, 115, 15, 31, 1, 63, 31, 77, 15, 145, 37, 81, 7, 245, 69, 155, 17, 261, 37, 77, 3, 391, 115, 261, 31, 445, 69, 145, 7, 675, 115, 245, 15, 391, 31, 63, 1, 127, 63

Offset: 0

Mikhail Kurkov, Nov 12 2019

Another version of A152884.
The excedance set of a permutation p of {1,2,...,m} is the set of indices i such that p(i) > i; it is a subset of {1,2,...,m-1}.
Great work on this subject was done by R. Ehrenborg and E. Steingrimsson, so most of the formulas given below are just their results translated into the language of the sequences which are related to the binary expansion of n.
Conjecture 1: equivalently, number of open tours by a biased rook on a specific f(n) X 1 board, which ends on a white cell, where f(n) = A070941(n) = floor(log_2(2n)) + 1 and cells are colored white or black according to the binary representation of 2n. A cell is colored white if the binary digit is 0 and a cell is colored black if the binary digit is 1. A biased rook on a white cell moves only to the left and otherwise moves only to the right. - Mikhail Kurkov, May 18 2021
Conjecture 2: this sequence is an inverse modulo 2 binomial transform of A284005. - Mikhail Kurkov, Dec 15 2021

			a(1) = 1 because the 1st excedance set is {m-1} and the permutations of {1,2,...,m} with such excedance set are 21, 132, 1243, 12354 and so on, i.e., for a given m we always have 1 permutation.
a(2) = 3 because the 2nd excedance set is {m-2} and the permutations of {1,2,...,m} with such excedance set are 213, 312, 321, 1324, 1423, 1432, 12435, 12534, 12543 and so on, i.e., for a given m we always have 3 permutations.
a(3) = 1 because the 3rd excedance set is {m-2, m-1} and the permutations of {1,2,...,m} with such excedance set are 231, 1342, 12453 and so on, i.e., for a given m we always have 1 permutation.

Mikhail Kurkov, Table of n, a(n) for n = 0..8191
Max Alekseyev, Recursion for the sum with Stirling numbers of both kinds, answer to question on MathOverflow (2024).
R. Ehrenborg and E. Steingrimsson, The excedance set of a permutation, Advances in Appl. Math., 24, 284-299, 2000.
Peter J. Taylor, Correctness of the algorithm for the A329369, A347205 and related sequences, answer to question on MathOverflow (2024).

Cf. A000120, A000523, A001511, A007814, A008292, A023416, A036987, A053645, A062253, A062254, A062255, A063250, A080791, A091344, A091347, A091348, A110501, A136126, A152884 (lexicographic order), A173018, A329718, A344902, A344947, A357990, A358612, A373183, A379817.
Similar recurrences: A006014, A124758, A217924, A243499, A284005, A290615, A341392, A347204, A347205.
Cf. A360288.

g:= proc(n) option remember;  2^padic[ordp](n, 2) end:
a:= proc(n) option remember; `if`(n=0, 1, (h-> a(h)+
     `if`(n::odd, 0, (t-> a(h-t)+a(n-t))(g(h))))(iquo(n, 2)))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Jan 30 2023

a[n_] := a[n] = Which[n == 0, 1, OddQ[n], a[(n-1)/2], True, a[n/2] + a[n/2 - 2^IntegerExponent[n/2, 2]] + a[n - 2^IntegerExponent[n/2, 2]]];
a /@ Range[0, 65] (* Jean-François Alcover, Feb 13 2020 *)

upto(n) = my(A, v1); v1 = vector(n+1, i, 0); v1[1] = 1; for(i=1, n, v1[i+1] = v1[i\2+1] + if(i%2, 0, A = 1 << valuation(i/2, 2); v1[i/2-A+1] + v1[i-A+1])); v1 \\ Mikhail Kurkov, Jun 06 2024

a(2n+1) = a(n) for n >= 0.
a(2n) = a(n) + a(n - 2^f(n)) + a(2n - 2^f(n)) for n > 0 with a(0) = 1 where f(n) = A007814(n) (equivalent to proposition 2.1 at the page 286, see R. Ehrenborg and E. Steingrimsson link).
a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n) = a(2^m*n) + a(2^(m-1)*(2n+1)) + a(2^(m-1)*(4n+1)) for m > 0, n >= 0 (equivalent to proposition 2.5 at the page 287, see R. Ehrenborg and E. Steingrimsson link).
a(2n) = a(2*g(n)) + a(2n - 2^h(n)) + a(2*g(n) + 2^h(n)) for n > 0 with a(0) = 1 where g(n) = A053645(n), h(n) = A063250(n) (equivalent to proposition 2.1 at the page 286, see R. Ehrenborg and E. Steingrimsson link).
a(2n) = 2*a(n + g(n)) + a(2*g(n)) for n > 0, floor(n/3) < 2^(floor(log_2(n))-1) (in other words, for 2^m + k where 0 <= k < 2^(m-1), m > 0) with a(0) = 1 (just a special case of the previous formula, because for 2^m + k where 0 <= k < 2^(m-1), m > 0 we have 2^h(n) = n - g(n)).
a(2n) = a(f(n,-1)) + a(f(n,0)) + a(f(n,1)) for n > 0 with a(0) = 1 where f(n,k) = 2*(f(floor(n/2),k) + n mod 2) + k*A036987(n) for n > 1 with f(1,k) = abs(k) (equivalent to a(2n) = a(2*g(n)) + a(2n - 2^h(n)) + a(2*g(n) + 2^h(n))).
a(n) = Sum_{j=0..2^wt(n) - 1} (-1)^(wt(n) - wt(j)) Product_{k=0..wt(n) - 1} (1 + wt(floor(j/2^k)))^T(n,k) for n > 0 with a(0) = 1 where wt(n) = A000120(n), T(n,k) = T(floor(n/2), k - n mod 2) for k > 0 with T(n,0) = A001511(n) (equivalent to theorem 6.3 at page 296, see R. Ehrenborg and E. Steingrimsson link). Here T(n, k) - 1 for k > 0 is the length of the run of zeros between k-th pair of ones from the right side in the binary expansion of n. Conjecture 1: this formula is equivalent to inverse modulo 2 binomial transform of A284005.
Sum_{k=0..2^n-1} a(k) = (n+1)! for n >= 0.
a((4^n-1)/3) = A110501(n+1) for n >= 0.
a(2^2*(2^n-1)) = A091344(n+1),
a(2^3*(2^n-1)) = A091347(n+1),
a(2^4*(2^n-1)) = A091348(n+1).
More generally, a(2^m*(2^n-1)) = a(2^n*(2^m-1)) = S(n+1,m) for n >= 0, m >= 0 where S(n,m) = Sum_{k=1..n} k!*k^m*Stirling2(n,k)*(-1)^(n-k) (equivalent to proposition 6.5 at the page 297, see R. Ehrenborg and E. Steingrimsson link).
Conjecture 2: a(n) = (1 + A023416(n))*a(g(n)) + Sum_{k=0..floor(log_2(n))-1} (1-R(n,k))*a(g(n) + 2^k*(1 - R(n,k))) for n > 1 with a(0) = 1, a(1) = 1, where g(n) = A053645(n) and where R(n,k) = floor(n/2^k) mod 2 (at this moment this is the only formula here, which is not related to R. Ehrenborg's and E. Steingrimsson's work and arises from another definition given above, exactly conjectured definition with a biased rook). Here R(n,k) is the (k+1)-th bit from the right side in the binary expansion of n. - Mikhail Kurkov, Jun 23 2021
From Mikhail Kurkov, Jan 23 2023: (Start)
The formulas below are not related to R. Ehrenborg's and E. Steingrimsson's work.
Conjecture 3: a(n) = A357990(n, 1) for n >= 0.
Conjecture 4: a(2^m*(2k+1)) = Sum_{i=1..wt(k) + 2} i!*i^m*A358612(k, i)*(-1)^(wt(k) - i) for m >= 0, k >= 0 where wt(n) = A000120(n).
Conjecture 5: a(2^m*(2^n - 2^p - 1)) = Sum_{i=1..n} i!*i^m*(-1)^(n - i)*((i - p + 1)*Stirling2(n, i) - Stirling2(n - p, i - p) + Sum_{j=0..p-2} (p - j - 1)*Stirling2(n - p, i - j)/j! Sum_{k=0..j} (i - k)^p*binomial(j, k)*(-1)^k) for n > 2, m >= 0, 0 < p < n - 1. Here we consider that Stirling2(n, k) = 0 for n >= 0, k < 0. (End)
Conjecture 6: a(2^m*n + q) = Sum_{i=A001511(n+1)..A000120(n)+1} A373183(n, i)*a(2^m*(2^(i-1)-1) + q) for n >= 0, m >= 0, q >= 0. Note that this formula is recursive for n != 2^k - 1. Also, it is not related to R. Ehrenborg's and E. Steingrimsson's work. - Mikhail Kurkov, Jun 05 2024
From Mikhail Kurkov, Jul 10 2024: (Start)
a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=1..m+1} a(2^i*k)*(-1)^(m-i+1)*Sum_{j=i..m+1} j^n*Stirling1(j, i)*Stirling2(m+1, j) for m >= 0, n >= 0, k >= 0 with a(0) = 1.
Proof: start with a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n) given above and rewrite it as a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=0..m} binomial(m+1, i) a(2^i*(2^(n-1)*(2k+1) - 1)).
Then conjecture that a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=1..m+1} a(2^i*k)*f(n, m, i). From that it is obvious that f(0, m, i) = [i = (m+1)].
After that use a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=0..m} binomial(m+1, i) Sum_{j=1..i+1} a(2^j*k)*f(n-1, i, j) = Sum_{i=1..m+1} a(2^i*k) Sum_{j=i-1..m} binomial(m+1, j)*f(n-1, j, i). From that it is obvious that f(n, m, i) = Sum_{j=i-1..m} binomial(m+1, j)*f(n-1, j, i).
Finally, all we need is to show that basic conditions and recurrence for f(n, m, i) gives f(n, m, i) = (-1)^(m-i+1)*Sum_{j=i..m+1} j^n*Stirling1(j, i)*Stirling2(m+1, j) (see Max Alekseyev link).
a(2^m*(2k+1)) = a(2^(m-1)*k) + (m+1)*a(2^m*k) + Sum_{i=1..m-1} a(2^m*k + 2^i) for m > 0, k >= 0.
Proof: start with a(2^(m+1)*(2k+1)) = a(2^m*k) + (m+2)*a(2^(m+1)*k) + Sum_{i=1..m} a(2^(m+1)*k + 2^i).
Then use a(2^m*(4k+1)) = a(2^m*k) + (m+1)*a(2^(m+1)*k) + Sum_{i=1..m-1} a(2^(m+1)*k + 2^i).
From that we get a(2^(m+1)*(2k+1)) - a(2^m*k) - (m+2)*a(2^(m+1)*k) - a(2^(m+1)*k + 2^m) = a(2^m*(4k+1)) - a(2^m*k) - (m+1)*a(2^(m+1)*k).
Finally, a(2^(m+1)*(2k+1)) = a(2^(m+1)*k) + a(2^m*(2*k+1)) + a(2^m*(4k+1)) which agrees with the a(2^m*(2n+1)) = a(2^m*n) + a(2^(m-1)*(2n+1)) + a(2^(m-1)*(4n+1)) given above.
This formula can be considered as an alternative to a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n). There are algorithms for both these formulas that allow you to calculate them without recursion. However, even though it is necessary to calculate binomial coefficients in the mentioned formula, the triple-looped algorithm for it still works faster (see Peter J. Taylor link).
It looks like you can also change v2 in the mentioned algorithm to vector with elements a(2^m*(2^(i+A007814(n+1)-1)-1) + q) to get a(2^m*n + q) instead of a(n). This may have common causes with formula that uses A373183 given above. (End)
From Mikhail Kurkov, Jan 27 2025: (Start)
The formulas below are not related to R. Ehrenborg's and E. Steingrimsson's work.
Conjecture 7: A008292(n+1,k+1) = Sum_{i=0..2^n-1} [A000120(i) = k]*a(i) for n >= 0, k >= 0.
Conjecture 8: a(2^m*(2^n*(2k+1)-1)) = Sum_{i=0..m} Sum_{j=0..m-i} Sum_{q=0..i} binomial(m-i,j)*(m-j+1)^n*a(2^(q+1)*k)*L(m,i,q)*(-1)^j for m >= 0, n > 0, k >= 0 where L(n,k,m) = W(n-m,k-m,m+1) for n > 0, 0 <= k < n, 0 <= m <= k and where W(n,k,m) = (k+m)*W(n-1,k,m) + (n-k)*W(n-1,k-1,m) + [m > 1]*W(n,k,m-1) for 0 <= k < n, m > 0 with W(0,0,m) = 1, W(n,k,m) = 0 for n < 0 or k < 0.
In particular, W(n, k, 1) = A173018(n, k), W(n, k, 2) = A062253(n, k), W(n, k, 3) = A062254(n, k) and W(n, k, 4) = A062255(n, k).
Conjecture 9: a(n) = b(n,wt(n)) for n >= 0 where b(2n+1,k) = b(n,k) + (wt(n)-k+2)*b(n,k-1), b(2n,k) = (wt(n)-k+1)*b(2n+1,k) for n > 0, k > 0 with b(n,0) = A341392(n) for n >= 0, b(0,k) = 0 for k > 0 and where wt(n) = A000120(n) (see A379817).
More generally, a(2^m*(2k+1)) = ((m+1)!)^2*b(k,wt(k)-m) - Sum_{j=1..m} Stirling1(m+2,j+1)*a(2^(j-1)*(2k+1)) for m >= 0, k >= 0. Here we also consider that b(n,k) = 0 for k < 0. (End)
Conjecture 10: if we change b(n,0) = A341392(n) given above to b(n,0) = A341392(n)*x^n, then nonzero terms of the resulting polynomials for b(n,wt(n)) form c(n,k) such that a(n) = Sum_{k=0..A080791(n)} c(n,k) for n >= 0 where c(n,k) = (Product_{i=0..k-1} (1 + 1/A000120(floor(n/2^(A000523(n)-i))))) * Sum_{j=max{0,k-A080791(n)+A080791(A053645(n))}..A080791(A053645(n))} c(A053645(n),j) for n > 0, k >= 0 with c(0,0) = 1, c(0,k) = 0 for k > 0. - Mikhail Kurkov, Jun 19 2025

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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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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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A007931 Numbers that contain only 1's and 2's. Nonempty binary strings of length n in lexicographic order.

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