A054429 -id:A054429 - cp's OEIS frontend

A007305 Numerators of Farey (or Stern-Brocot) tree fractions.

0, 1, 1, 1, 2, 1, 2, 3, 3, 1, 2, 3, 3, 4, 5, 5, 4, 1, 2, 3, 3, 4, 5, 5, 4, 5, 7, 8, 7, 7, 8, 7, 5, 1, 2, 3, 3, 4, 5, 5, 4, 5, 7, 8, 7, 7, 8, 7, 5, 6, 9, 11, 10, 11, 13, 12, 9, 9, 12, 13, 11, 10, 11, 9, 6, 1, 2, 3, 3, 4, 5, 5, 4, 5, 7, 8, 7, 7, 8, 7, 5, 6, 9, 11, 10, 11, 13, 12, 9, 9, 12, 13, 11

Offset: 0

N. J. A. Sloane

From Yosu Yurramendi, Jun 25 2014: (Start)
If the terms (n>0) are written as an array (left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...
   1,
   1,2,
   1,2,3,3,
   1,2,3,3,4,5,5,4,
   1,2,3,3,4,5,5,4,5,7,8,7,7,8,7,5,
   1,2,3,3,4,5,5,4,5,7,8,7,7,8,7,5,6,9,11,10,11,13,12,9,9,12,13,11,10,11,9,6,
then the sum of the m-th row is 3^m (m = 0,1,2,), each column k is constant, and the constants are from A007306, denominators of Farey (or Stern-Brocot) tree fractions (see formula).
If the rows are written in a right-aligned fashion:
                                                                          1,
                                                                        1,2,
                                                                   1, 2,3,3,
                                                       1, 2, 3, 3, 4, 5,5,4,
                                  1,2, 3, 3, 4, 5, 5,4,5, 7, 8, 7, 7, 8,7,5,
  1,2,3,3,4,5,5,4,5,7,8,7,7,8,7,5,6,9,11,10,11,13,12,9,9,12,13,11,10,11,9,6,
then each column is an arithmetic sequence. The differences of the arithmetic sequences also give the sequence A007306 (see formula). The first terms of columns are from A007305 itself (a(A004761(n+1)) = a(n), n>0), and the second ones from A049448 (a(A004761(n+1)+2^A070941(n)) = A049448(n), n>0). (End)
If the sequence is considered in blocks of length 2^m, m = 0,1,2,..., the blocks are the reverse of the blocks of A047679: (a(2^m+1+k) = A047679(2^(m+1)-2-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1). - Yosu Yurramendi, Jun 30 2014

			A007305/A007306 = [ 0/1; 1/1; ] 1/2; 1/3, 2/3; 1/4, 2/5, 3/5, 3/4; 1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5, ...
Another version of Stern-Brocot is A007305/A047679 = 1, 2, 1/2, 3, 1/3, 3/2, 2/3, 4, 1/4, 4/3, 3/4, 5/2, 2/5, 5/3, 3/5, 5, 1/5, 5/4, 4/5, ...

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 117.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 23.
J. C. Lagarias, Number Theory and Dynamical Systems, pp. 35-72 of S. A. Burr, ed., The Unreasonable Effectiveness of Number Theory, Proc. Sympos. Appl. Math., 46 (1992). Amer. Math. Soc.
W. J. LeVeque, Topics in Number Theory. Addison-Wesley, Reading, MA, 2 vols., 1956, Vol. 1, p. 154.
I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 141.
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..4096
A. Bogomolny, Stern-Brocot Tree
A. Bogomolny, Inspiration for Maple code
A. Brocot, Calcul des rouages par approximation, nouvelle méthode, Revue Chonométrique 3, 186-194, 1861.
G. A. Jones, The Farey graph, Séminaire Lotharingien de Combinatoire, B18e (1987), 2 pp.
Shin-ichi Katayama, Modified Farey trees and Pythagorean triples, Journal of mathematics, the University of Tokushima, 47, 2013.
G. Melançon, Lyndon factorization of sturmian words, Discr. Math., 210 (2000), 137-149.
Hugo Pfoertner, Ratio A007305(n)/A007306(n) vs n, using Plot 2.
N. J. A. Sloane, Stern-Brocot or Farey Tree
Noam Zimhoni, A forest of Eisensteinian triplets, arXiv:1904.11782 [math.NT], 2019.
Index entries for fraction trees
Index entries for sequences related to Stern's sequences

Cf. A007306, A006842, A006843, A047679, A054424, A057114, A152975.

A007305 := proc(n) local b; b := proc(n) option remember; local msb, r;
if n < 3 then return 1 fi; msb := ilog2(n); r := n - 2^msb;
if ilog2(r) = msb - 1 then b(r) + b(3*2^(msb-1) - r - 1) else b(2^(msb - 1) + r) fi end: if n = 0 then 0 else b(n-1) fi end: # Antti Karttunen, Mar 19 2000 [Corrected and rewritten by Peter Luschny, Apr 24 2024]
seq(A007305(n), n = 0..92);

sbt[n_] := Module[{R,L,Y}, R={{1,0},{1,1}}; L={{1,1},{0,1}}; Y={{1,0},{0,1}}; w[b_] := Fold[ #1.If[ #2 == 0,L,R] &,Y,b]; u[a_] := {a[[2,1]]+a[[2,2]],a[[1,1]]+a[[1,2]]}; Map[u,Map[w,Tuples[{0,1},n]]]]
A007305(n) = Flatten[Append[{0,1},Table[Map[First,sbt[i]],{i,0,5}]]]
A047679(n) = Flatten[Table[Map[Last,sbt[i]],{i,0,5}]]
(* Peter Luschny, Apr 27 2009 *)

a <- 1
for(m in 1:6) for(k in 0:(2^(m-1)-1)) {
  a[2^m+        k] <- a[2^(m-1)+k]
  a[2^m+2^(m-1)+k] <- a[2^(m-1)+k] + a[2^m-k-1]
}
a
# Yosu Yurramendi, Jun 25 2014

a(n) = SternBrocotTreeNum(n-1) # n starting from 2 gives the sequence from 1, 1, 2, 1, 2, 3, 3, 1, 2, 3, 3, 4, 5, 5, 4, 1, ...
From Reinhard Zumkeller, Dec 22 2008: (Start)
For n > 1: a(n+2) = if A025480(n-1) != 0 and A025480(n) != 0 then a(A025480(n-1)+2) + a(A025480(n)+2) else if A025480(n)=0 then a(A025480(n-1)+2)+1 else 0 + a(A025480(n-1)+2).
a(A054429(n)+2) = A047679(n).
a(n+2) = A047679(A054429(n)).
A153036(n+1) = floor(a(n+2)/A047679(n)). (End)
From Yosu Yurramendi, Jun 25 2014: (Start)
For m = 1,2,3,..., and k = 0,1,2,...,2^(m-1)-1, with a(1)=1:
a(2^m+k) = a(2^(m-1)+k);
a(2^m+2^(m-1)+k) = a(2^(m-1)+k) + a(2^m-k-1). (End)
a(2^(m+2)-k) = A007306(2^(m+1)-k), m=0,1,2,..., k=0,1,2,...,2^m-1. - Yosu Yurramendi, Jul 04 2014
a(2^(m+1)+2^m+k) - a(2^m+k) = A007306(2^m-k+1), m=1,2,..., k=1,2,...,2^(m-1). - Yosu Yurramendi, Jul 05 2014
From Yosu Yurramendi, Jan 01 2015: (Start)
a(2^m+2^q-1) = q+1, q = 0, 1, 2,..., m = q, q+1, q+2,...
a(2^m+2^q)   = q+2, q = 0, 1, 2,..., m = q+1, q+2, q+3,... (End)
a(2^m+k) = A007306(k+1), m >= 0, 0 <= k < 2*m.  - Yosu Yurramendi, May 20 2019
a(n) = A002487(A059893(n)), n > 0. - Yosu Yurramendi, Sep 29 2021

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

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Jun 19 2001

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

Harry J. Smith, Table of n, a(n) for n = 0..1000
L. K. Arnold, S. J. Benkoski and B. J. McCabe, The discriminator (a simple application of Bertrand's postulate). Amer. Math. Monthly 92 (1985), 275-277.
Sajed Haque, Chapter 2.6.1 of Discriminators of Integer Sequences, 2017, See p. 33.
S. Haque and J. Shallit, Discriminators and k-regular sequences, arXiv:1605.00092 [cs.DM], 2016.
Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625, Dec 08, 2020
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Wikipedia, Lucas's theorem
Index to divisibility sequences

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

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

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

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

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

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

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

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

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

A010059 Another version of the Thue-Morse sequence: let A_k denote the first 2^k terms; then A_0 = 1 and for k >= 0, A_{k+1} = A_k B_k, where B_k is obtained from A_k by interchanging 0's and 1's.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Characteristic function of A001969 (evil numbers). - Ralf Stephan, Jun 20 2003
From Gary W. Adamson, Aug 24 2008: (Start)
Parity of A143579 (Odious numbers (A000069) interleaved with Evil numbers (A001969)).
Two conjectures: If n is even, the ratio of 1's to 0's = 1:1.
There are no three adjacent terms of the same parity. (End)
Conjecture (verified for the first 280000 entries): this is the characteristic function of A001969. - R. J. Mathar, Sep 05 2008
From Michel Dekking, Jan 05 2021: (Start)
Proof of these three conjectures: the first two follow directly from the third, because the sequence A010059 is the binary complement of the Thue-Morse sequence A010060.
For the third conjecture: the odious and evil numbers occur as quadruples EOOE and OEEO, simply by their definition. To obtain the mod 2 version of the interleave of the odious and evil numbers we therefore have to apply a transformation
      EOOE -> OEOE, OEEO -> OEOE to these quadruples.
But this changes the parities from the corresponding 4n, 4n+1, 4n+2, 4n+3 quadruples from 0101 to 1001 in the first case, and from 0101 to 0110 in the second case. Since the quadruples EOOE and OEEO occur in a Thue Morse pattern, then also the quadruples 1001 and 0110 occur in a Thue Morse pattern, finishing the proof.
(End)

			The evolution starting at 1 is:
.1
.1, 0
.1, 0, 0, 1,
.1, 0, 0, 1, 0, 1, 1, 0
.1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1
.1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0
...........

W. H. Gottschalk and G. A. Hedlund, Topological Dynamics. American Mathematical Society, Colloquium Publications, Vol. 36, Providence, RI, 1955, p. 105.
M. Lothaire, Combinatorics on Words. Addison-Wesley, Reading, MA, 1983, p. 23.
A. Salomaa, Jewels of Formal Language Theory. Computer Science Press, Rockville, MD, 1981, p. 6.

Robert Israel, Table of n, a(n) for n = 0..10000
J.-P. Allouche and Jeffrey Shallit, The Ubiquitous Prouhet-Thue-Morse Sequence, in C. Ding. T. Helleseth and H. Niederreiter, eds., Sequences and Their Applications: Proceedings of SETA '98, Springer-Verlag, 1999, pp. 1-16.
Scott Balchin and Dan Rust, Computations for Symbolic Substitutions, Journal of Integer Sequences, Vol. 20 (2017), Article 17.4.1.
Françoise Dejean, Sur un Théorème de Thue, J. Combinatorial Theory, vol. 13 A, iss. 1 (1972) 90-99.
F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
Michael Gilleland, Some Self-Similar Integer Sequences
G. A. Hedlund, Remarks on the work of Axel Thue on sequences, Nordisk Mat. Tid., 15 (1967), 148-150.
Tanya Khovanova, There are no coincidences, arXiv preprint 1410.2193 [math.CO], 2014.
H. M. Morse, Recurrent geodesics on a surface of negative curvature, Trans. Amer. Math. Soc., 22 (1921), 84-100.
Stephen Wolfram, A New Kind Of Science | Online.
Index entries for sequences related to binary expansion of n
Index entries for characteristic functions

Cf. A001285 (1, 2 version), A010060 (0, 1 version), A106400 (+1, -1 version), A059448 (with reversed subsections).
Cf. also A000069, A026147, A159481.
Cf. A000120, A001969, A143579.

a010059 = (1 -) . a010060  -- Reinhard Zumkeller, Feb 04 2013

A010059 := n->1-A010060(n);
map(t->49-t,convert(StringTools[ThueMorse](1000),bytes)); # Robert Israel, Feb 02 2016
# second Maple program:
a := n -> ifelse(type(add(convert(n, base, 2)), even), 1, 0):
seq(a(n), n = 0 .. 104); # Peter Luschny, Mar 11 2024

Mod[ CoefficientList[Series[(1 + Sqrt[(1 - 3x)/(1 + x)])/(2(1 + x)), {x, 0, 111}], x], 2] (* Stephan Wolfram *)
CoefficientList[ Series[1/(1 - x) + Product[1 - x^2^k, {k, 0, 10}], {x, 0, 111}]/2, x] (* Robert G. Wilson v, Jul 16 2004 *)
Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {1}, 7] (* Robert G. Wilson v Sep 26 2006 *)
od = Select[ Range[0, 129], OddQ@ DigitCount[ #, 2, 1] &]; ev = Select[ Range[0, 129], EvenQ@ DigitCount[ #, 2, 1] &]; Mod[ Flatten@ Transpose[{od, ev}], 2] (* Robert G. Wilson v, Apr 14 2009 *)
Nest[ Join[ #, Mod[2# + 1, 3]] &, {1}, 7] (* Robert G. Wilson v, Jul 27 2014 *)
{{1}}~Join~SubstitutionSystem[{0 -> {0, 1}, 1 -> {1, 0}}, {0}, 6] // Flatten (* Michael De Vlieger, Aug 15 2016, Version 10.2 *)
1 - ThueMorse[Range[0, 100]] (* Paolo Xausa, Oct 25 2024 *)

a(n)=!(hammingweight(n)%2); \\ Charles R Greathouse IV, Mar 29 2013

def A010059(n): return n.bit_count()&1^1 # Chai Wah Wu, Mar 01 2023

maxrow <- 8 # by choice
b01 <- 0
for(m in 0:maxrow) for(k in 0:(2^m-1)){
b01[2^(m+1)+    k] <-   b01[2^m+k]
b01[2^(m+1)+2^m+k] <- 1-b01[2^m+k]
}
(b01 <- c(1,b01))
# Yosu Yurramendi, Apr 10 2017

G.f.: (1/2) * (1/(1-x) + Product_{k>=0} (1 - x^2^k)). - Ralf Stephan, Jun 20 2003
a(n) = A143579(n) mod 2. - Gary W. Adamson, Aug 24 2008
a(n) + A010060(n) = 1 for all n.
a(n) = A159481(n+1) - A159481(n). - Reinhard Zumkeller, Apr 16 2009
a(n) + A026147(n-1) = 2n for n >= 1. - Clark Kimberling, Oct 06 2014
a(n) = A000069(n+1) (mod 2). - John M. Campbell, Jun 30 2016
a(n) = A059448(A054429(n)) = (A106400(n)+1)/2 = (1+A008836(A005940(1+n)))/2. - Antti Karttunen, May 30 2017
If A(n)=(a(0),a(2),...,a(2^n-1)), then A(n+1)=(A(n),1-A(n)). - Arie Bos, Jul 27 2022
a(n) = (1 + (-1)^A000120(n))/2. - Lorenzo Sauras Altuzarra, Mar 10 2024

A135141 a(1)=1, a(p_n)=2a(n), a(c_n)=2a(n)+1, where p_n = n-th prime, c_n = n-th composite number.

Original entry on oeis.org

1, 2, 4, 3, 8, 5, 6, 9, 7, 17, 16, 11, 10, 13, 19, 15, 12, 35, 18, 33, 23, 21, 14, 27, 39, 31, 25, 71, 34, 37, 32, 67, 47, 43, 29, 55, 22, 79, 63, 51, 20, 143, 26, 69, 75, 65, 38, 135, 95, 87, 59, 111, 30, 45, 159, 127, 103, 41, 24, 287, 70, 53, 139, 151, 131, 77, 36, 271, 191

Offset: 1

Katarzyna Matylla, Feb 13 2008

A permutation of the positive integers, related to A078442.
a(p) is even when p is prime and is divisible by 2^(prime order of p).
From Robert G. Wilson v, Feb 16 2008: (Start)
What is the length of the cycle containing 10? Is it infinite? The cycle begins 10, 17, 12, 11, 16, 15, 19, 18, 35, 29, 34, 43, 26, 31, 32, 67, 36, 55, 159, 1055, 441, 563, 100, 447, 7935, 274726911, 1013992070762272391167, ... Implementation in Mmca: NestList[a(AT)# &, 10, 26] Furthermore, it appears that any non-single-digit number has an infinite cycle.
Records: 1, 2, 4, 8, 9, 17, 19, 35, 39, 71, 79, 143, 159, 287, 319, 575, 639, 1151, 1279, 2303, 2559, 4607, 5119, 9215, 10239, 18431, 20479, 36863, 40959, 73727, 81919, 147455, 163839, 294911, 327679, 589823, 655359, ..., . (End)

			a(20) = 33 = 2*16 + 1 because 20 is 11th composite and a(11)=16. Or, a(20)=33=100001(bin). In other words it is a composite number, its index is a prime number, whose index is a prime....

Robert G. Wilson v, Table of n, a(n) for n = 1..10000
Antti Karttunen, Entanglement Permutations, 2016-2017
Index entries for sequences computed from indices in prime factorization
Index entries for sequences that are permutations of the natural numbers

Cf. A000720, A007814, A010051, A026238, A078442.
Cf. A246346, A246347 (record positions and values).
Cf. A227413 (inverse).
Cf. A071574, A245701, A245702, A245703, A245704, A246377, A236854, A237427 for related and similar permutations.

import Data.List (genericIndex)
a135141 n = genericIndex a135141_list (n-1)
a135141_list = 1 : map f [2..] where
   f x | iprime == 0 = 2 * (a135141 $ a066246 x) + 1
       | otherwise   = 2 * (a135141 iprime)
       where iprime = a049084 x
-- Reinhard Zumkeller, Jan 29 2014

a[1] = 1; a[n_] := If[PrimeQ@n, 2*a[PrimePi[n]], 2*a[n - 1 - PrimePi@n] + 1]; Array[a, 69] (* Robert G. Wilson v, Feb 16 2008 *)

/* Let pc = prime count (which prime it is), cc = composite count: */
pc[1]:0;
cc[1]:0;
pc[2]:1;
cc[4]:1;
pc[n]:=if primep(n) then 1+pc[prev_prime(n)] else 0;
cc[n]:=if primep(n) then 0 else if primep(n-1) then 1+cc[n-2] else 1+cc[n-1];
a[1]:1;
a[n]:=if primep(n) then 2*a[pc[n]] else 1+2*a[cc[n]];

A135141(n) = if(1==n, 1, if(isprime(n), 2*A135141(primepi(n)), 1+(2*A135141(n-primepi(n)-1)))); \\ Antti Karttunen, Dec 09 2019

from sympy import isprime, primepi
def a(n): return 1 if n==1 else 2*a(primepi(n)) if isprime(n) else 2*a(n - 1 - primepi(n)) + 1 # Indranil Ghosh, Jun 11 2017, after Mathematica code

a(n) = 2*A135141((A049084(n))*chip + A066246(n)*(1-chip)) + 1 - chip, where chip = A010051(n). - Reinhard Zumkeller, Jan 29 2014
From Antti Karttunen, Dec 09 2019: (Start)
A007814(a(n)) = A078442(n).
A070939(a(n)) = A246348(n).
A080791(a(n)) = A246370(n).
A054429(a(n)) = A246377(n).
A245702(a(n)) = A245703(n).
a(A245704(n)) = A245701(n). (End)

A047679 Denominators in full Stern-Brocot tree.

Original entry on oeis.org

1, 2, 1, 3, 3, 2, 1, 4, 5, 5, 4, 3, 3, 2, 1, 5, 7, 8, 7, 7, 8, 7, 5, 4, 5, 5, 4, 3, 3, 2, 1, 6, 9, 11, 10, 11, 13, 12, 9, 9, 12, 13, 11, 10, 11, 9, 6, 5, 7, 8, 7, 7, 8, 7, 5, 4, 5, 5, 4, 3, 3, 2, 1, 7, 11, 14, 13, 15, 18, 17, 13, 14, 19, 21, 18, 17, 19, 16, 11, 11, 16, 19, 17, 18

Offset: 0

Wouter Meeussen

Numerators are A007305.
Write n in binary; list run lengths; add 1 to last run length; make into continued fraction. Sequence gives denominator of fraction obtained.
From Reinhard Zumkeller, Dec 22 2008: (Start)
  For n > 1: a(n) = if A025480(n-1) != 0 and A025480(n) != 0 then = a(A025480(n-1)) + a(A025480(n)) else if A025480(n)=0 then a(A025480(n-1))+0 else 1+a(A025480(n-1));
  a(n) = A007305(A054429(n)+2) and a(A054429(n)) = A007305(n+2);
  A153036(n+1) = floor(A007305(n+2)/a(n)). (End)
From Yosu Yurramendi, Jun 25 2014 and Jun 30 2014: (Start)
If the terms are written as an array a(m, k) = a(2^(m-1)-1+k) with m >= 1 and k = 0, 1, ..., 2^(m-1)-1:
 1,
 2,1,
 3,3, 2, 1,
 4,5, 5, 4, 3, 3, 2,1,
 5,7, 8, 7, 7, 8, 7,5,4, 5, 5, 4, 3, 3,2,1,
 6,9,11,10,11,13,12,9,9,12,13,11,10,11,9,6,5,7,8,7,7,8,7,5,4,5,5,4,3,3,2,1,
then the sum of the m-th row is 3^(m-1), and each column is an arithmetic sequence. The differences of these arithmetic sequences give the sequence A007306(k+1). The first terms of columns are 1 for k = 0 and a(k-1) for k >= 1.
In a row reversed version A(m, k) = a(m, m-(k+1)):
 1
 1,2
 1,2,3,3,
 1,2,3,3,4,5,5,4
 1,2,3,3,4,5,5,4,5,7,8,7,7,8,7,5
 1,2,3,3,4,5,5,4,5,7,8,7,7,8,7,5,6,9,11,10,11,13,12,12,9,9,12,13,11,10,11,9,6
each column k >= 0 is constant, namely A007306(k+1).
This row reversed version coincides with the array for A007305 (see the Jun 25 2014 comment there). (End)
Looking at the plot, the sequence clearly shows a fractal structure. (The repeating pattern oddly resembles the [first completed] facade of the Sagrada Familia!) - Daniel Forgues, Nov 15 2019

			E.g., 57->111001->[ 3,2,1 ]->[ 3,2,2 ]->3 + 1/(2 + 1/(2) ) = 17/2. For n=1,2, ... we get 2, 3/2, 3, 4/3, 5/3, 5/2, 4, 5/4, 7/5, 8/5, ...
1; 2,1; 3,3,2,1; 4,5,5,4,3,3,2,1; ....
Another version of Stern-Brocot is A007305/A047679 = 1, 2, 1/2, 3, 1/3, 3/2, 2/3, 4, 1/4, 4/3, 3/4, 5/2, 2/5, 5/3, 3/5, 5, 1/5, 5/4, 4/5, ...

Alois P. Heinz, Table of n, a(n) for n = 0..10000
N. J. A. Sloane, Stern-Brocot or Farey Tree
Index entries for fraction trees
Index entries for sequences related to Stern's sequences

Cf. A007305, A007306, A054424, A152976.

CFruns[ n_Integer ] := Fold[ #2+1/#1&, Infinity, Reverse[ MapAt[ #+1&, Length/@Split[ IntegerDigits[ n, 2 ] ], {-1} ] ] ]
(* second program: *)
a[n_] := Module[{LL = Length /@ Split[IntegerDigits[n, 2]]}, LL[[-1]] += 1; FromContinuedFraction[LL] // Denominator]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Feb 25 2016 *)

{a(n) = local(v, w); v = binary(n++); w = [1]; for( n=2, #v, if( v[n] != v[n-1], w = concat(w, 1), w[#w]++)); w[#w]++; contfracpnqn(w)[2, 1]} /* Michael Somos, Jul 22 2011 */

a <- 1
for(m in 1:6) for(k in 0:(2^(m-1)-1)) {
  a[2^m+        k] = a[2^(m-1)+k] + a[2^m-k-1]
  a[2^m+2^(m-1)+k] = a[2^(m-1)+k]
}
a
# Yosu Yurramendi, Dec 31 2014

a(n) = SternBrocotTreeDen(n) # n starting from 1.
From Yosu Yurramendi, Jul 02 2014: (Start)
For m >0 and 0 <= k < 2^(m-1), with a(0)=1, a(1)=2:
a(2^m+k-1) = a(2^(m-1)+k-1) + a((2^m-1)-k-1);
a(2^m+2^(m-1)+k-1) = a(2^(m-1)+k-1). (End)
a(2^m-2^q  ) = q+1, q >= 0, m > q
a(2^m-2^q-1) = q+2, q >= 0, m > q+1. - Yosu Yurramendi, Jan 01 2015
a(2^(m+1)-1-k) = A007306(k+1), m >= 0, 0 <= k <= 2^m. - Yosu Yurramendi, May 20 2019
a(n) = A002487(1+A059893(n)), n > 0. - Yosu Yurramendi, Sep 29 2021

Edited by Wolfdieter Lang, Mar 31 2015

A056539 Self-inverse permutation: reverse the bits in binary expansion of n and also complement them (0->1, 1->0) if the run count (A005811) is even.

Original entry on oeis.org

0, 1, 2, 3, 6, 5, 4, 7, 14, 9, 10, 13, 12, 11, 8, 15, 30, 17, 22, 25, 26, 21, 18, 29, 28, 19, 20, 27, 24, 23, 16, 31, 62, 33, 46, 49, 54, 41, 38, 57, 58, 37, 42, 53, 50, 45, 34, 61, 60, 35, 44, 51, 52, 43, 36, 59, 56, 39, 40, 55, 48, 47, 32, 63, 126, 65, 94, 97, 110, 81, 78

Offset: 0

Antti Karttunen, Jun 20 2000

			n:                     0, 1,  2,  3,   4,   5,   6,   7,   8,   9,  10,  11,  12,  13,  14,  15
binary expansion:      0, 1, 10, 11, 100, 101, 110, 111,1000,1001,1010,1011,1100,1101,1110,1111
reversed/complemented: 0, 1, 10, 11, 110, 101, 100, 111,1110,1001,1010,1101,1100,1011,1000,1111

Cf. A054429.

When restricted to A014486 induces another permutation, A057164. A105726 is a "deep" variant.

[seq(runcounts2binexp(reverse(binexp2runcounts(j))),j=0..511)];
runcounts2binexp := proc(c) local i,e,n; n := 0; for i from 1 to nops(c) do e := c[i]; n := ((2^e)*n) + ((i mod 2)*((2^e)-1)); od; RETURN(n); end;
binexp2runcounts := proc(nn) local n,a,p,c; n := nn; a := []; p := (`mod`(n,2)); c := 0; while(n > 0) do c := c+1; n := floor(n/2); if((`mod`(n,2)) <> p) then a := [c,op(a)]; c := 0; p := (`mod`(p+1,2)); fi; od; RETURN(a); end;
# reverse given in A056538

A056539[n_] := If[n == 0, 0, FromDigits[Reverse[If[Last[#] == 1, #, 1-#]], 2] & [IntegerDigits[n, 2]]];
Array[A056539, 100, 0] (* Paolo Xausa, Nov 28 2024 *)

def a005811(n): return bin(n^(n>>1))[2:].count("1")
def a(n):
    if n==0: return 0
    x=bin(n)[2:][::-1]
    if a005811(n)%2==1: return int(x, 2)
    z=''.join('1' if i == '0' else '0' for i in x)
    return int(z, 2) # Indranil Ghosh, Apr 29 2017

a(2n) = A036044(2n), a(2n+1) = A030101(2n+1). - Antti Karttunen, Feb 14 2003

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)

A135416 a(n) = A036987(n)*(n+1)/2.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane, based on a message from Guy Steele and Don Knuth, Mar 01 2008

nonn

Guy Steele defines a family of 36 integer sequences, denoted here by GS(i,j) for 1 <= i, j <= 6, as follows. a[1]=1; a[2n] = i-th term of {0,1,a[n],a[n]+1,2a[n],2a[n]+1}; a[2n+1] = j-th term of {0,1,a[n],a[n]+1,2a[n],2a[n]+1}. The present sequence is GS(1,5).
The full list of 36 sequences:
GS(1,1) = A000007
GS(1,2) = A000035
GS(1,3) = A036987
GS(1,4) = A007814
GS(1,5) = A135416 (the present sequence)
GS(1,6) = A135481
GS(2,1) = A135528
GS(2,2) = A000012
GS(2,3) = A000012
GS(2,4) = A091090
GS(2,5) = A135517
GS(2,6) = A135521
GS(3,1) = A036987
GS(3,2) = A000012
GS(3,3) = A000012
GS(3,4) = A000120
GS(3,5) = A048896
GS(3,6) = A038573
GS(4,1) = A135523
GS(4,2) = A001511
GS(4,3) = A008687
GS(4,4) = A070939
GS(4,5) = A135529
GS(4,6) = A135533
GS(5,1) = A048298
GS(5,2) = A006519
GS(5,3) = A080100
GS(5,4) = A087808
GS(5,5) = A053644
GS(5,6) = A000027
GS(6,1) = A135534
GS(6,2) = A038712
GS(6,3) = A135540
GS(6,4) = A135542
GS(6,5) = A054429
GS(6,6) = A003817
(with a(0)=1): Moebius transform of A038712.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences related to binary expansion of n

Equals A048298(n+1)/2. Cf. A036987, A182660.

GS:=proc(i,j,M) local a,n; a:=array(1..2*M+1); a[1]:=1;
for n from 1 to M do
a[2*n] :=[0,1,a[n],a[n]+1,2*a[n],2*a[n]+1][i];
a[2*n+1]:=[0,1,a[n],a[n]+1,2*a[n],2*a[n]+1][j];
od: a:=convert(a,list); RETURN(a); end;
GS(1,5,200):

i = 1; j = 5; Clear[a]; a[1] = 1; a[n_?EvenQ] := a[n] = {0, 1, a[n/2], a[n/2]+1, 2*a[n/2], 2*a[n/2]+1}[[i]]; a[n_?OddQ] := a[n] = {0, 1, a[(n-1)/2], a[(n-1)/2]+1, 2*a[(n-1)/2], 2*a[(n-1)/2]+1}[[j]]; Array[a, 105] (* Jean-François Alcover, Sep 12 2013 *)

A048298(n) = if(!n,0,if(!bitand(n,n-1),n,0));
A135416(n) = (A048298(n+1)/2); \\ Antti Karttunen, Jul 22 2018

def A135416(n): return int(not(n&(n+1)))*(n+1)>>1 # Chai Wah Wu, Jul 06 2022

G.f.: sum{k>=1, 2^(k-1)*x^(2^k-1) }.

Recurrence: a(2n+1) = 2a(n), a(2n) = 0, starting a(1) = 1.

Formulae and comments by Ralf Stephan, Jun 20 2014

A233271 a(0)=0; thereafter a(n+1) = a(n) + 1 + number of 0's in binary representation of a(n), counted with A080791.

Original entry on oeis.org

0, 1, 2, 4, 7, 8, 12, 15, 16, 21, 24, 28, 31, 32, 38, 42, 46, 49, 53, 56, 60, 63, 64, 71, 75, 79, 82, 87, 90, 94, 97, 102, 106, 110, 113, 117, 120, 124, 127, 128, 136, 143, 147, 152, 158, 162, 168, 174, 178, 183, 186, 190, 193, 199, 203, 207, 210, 215, 218, 222

Offset: 0

Antti Karttunen, Dec 12 2013

These are iterates of A233272: a(0)=0, and for n>0, a(n) = A233272(a(n-1)). The difference from A216431 stems from the fact that it uses A023416 to count the 0-bits in the binary expansion of n, while this sequence uses A080791, which results a slightly different start for the iteration, and a much better alignment with sequences related to "infinite trunk of binary beanstalk", A179016.

Apart from term a(2)=2, it seems that each term a(n) >= A179016(n). Please see their ratio plotted with Plot2, and also their differences: A233270.

Antti Karttunen, Table of n, a(n) for n = 0..8727
A. Karttunen Ratio A233271/A179016 plotted with OEIS Plot2-script

Differs from A216431 only in that here 1 has been inserted into position a(1), between 0 and 2.

Cf. A080791, A010062, A233272, A179016, A233270, A218616, A054429, A213710, A218600.

a[0] = 0; a[n_] := a[n] = If[n == 1, 1, # + 1 + Last@ DigitCount[#, 2] &@ a[n - 1]]; Table[a@ n, {n, 0, 59}] (* or *)
Insert[NestList[# + 1 + DigitCount[#, 2, 0] &, 0, nn], 1, 2] (* Michael De Vlieger, Mar 07 2016, the latter after Harvey P. Dale at A216431 *)

a(0)=0, and for n>0, a(n) = A233272(a(n-1)).
a(0)=0, and for n>0, a(n) = a(n-1) + 1 + A080791(a(n-1)).
a(n) = A054429(A218616(n)) = A054429(A179016(A218602(n))) [This sequence can be mapped to the infinite trunk of "binary beanstalk" with involutions A054429 & A218602].
For all n, a(A213710(n)) = 2^n = A000079(n).
For n>=3, a(A218600(n)) = A000225(n).

A253565 Permutation of natural numbers: a(0) = 1, a(1) = 2; after which, a(2n) = A253550(a(n)), a(2n+1) = A253560(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 9, 6, 8, 7, 25, 15, 27, 10, 18, 12, 16, 11, 49, 35, 125, 21, 75, 45, 81, 14, 50, 30, 54, 20, 36, 24, 32, 13, 121, 77, 343, 55, 245, 175, 625, 33, 147, 105, 375, 63, 225, 135, 243, 22, 98, 70, 250, 42, 150, 90, 162, 28, 100, 60, 108, 40, 72, 48, 64, 17, 169, 143, 1331, 91, 847, 539, 2401, 65, 605, 385, 1715, 275, 1225, 875, 3125, 39

Offset: 0

Antti Karttunen, Jan 03 2015

This sequence can be represented as a binary tree. Each child to the left is obtained by applying A253550 to the parent, and each child to the right is obtained by applying A253560 to the parent:
                                    1
                                    |
                 ...................2...................
                3                                       4
      5......../ \........9                   6......../ \........8
     / \                 / \                 / \                 / \
    /   \               /   \               /   \               /   \
   /     \             /     \             /     \             /     \
  7       25         15       27         10       18         12       16
11 49   35  125    21  75   45  81     14  50   30  54     20  36   24  32
etc.
Sequence A253563 is the mirror image of the same tree. Also in binary trees A005940 and A163511 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees. Of these four trees, this is the one where the left child is always smaller than the right child.
Note that the indexing of sequence starts from 0, although its range starts from one.
The term a(n) is the Heinz number of the adjusted partial sums of the n-th composition in standard order, where (1) the k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again, (2) the Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and (3) we define the adjusted partial sums of a composition to be obtained by subtracting one from all parts, taking partial sums, and adding one back to all parts. See formula for a simplification. A triangular form is A242628. The inverse is A253566. The non-adjusted version is A358170. - Gus Wiseman, Dec 17 2022

			From _Gus Wiseman_, Dec 23 2022: (Start)
This represents the following bijection between compositions and partitions. The n-th composition in standard order together with the reversed prime indices of a(n) are:
   0:        () -> ()
   1:       (1) -> (1)
   2:       (2) -> (2)
   3:     (1,1) -> (1,1)
   4:       (3) -> (3)
   5:     (2,1) -> (2,2)
   6:     (1,2) -> (2,1)
   7:   (1,1,1) -> (1,1,1)
   8:       (4) -> (4)
   9:     (3,1) -> (3,3)
  10:     (2,2) -> (3,2)
  11:   (2,1,1) -> (2,2,2)
  12:     (1,3) -> (3,1)
  13:   (1,2,1) -> (2,2,1)
  14:   (1,1,2) -> (2,1,1)
  15: (1,1,1,1) -> (1,1,1,1)
(End)

Antti Karttunen, Table of n, a(n) for n = 0..8191
Index entries for sequences that are permutations of the natural numbers

Inverse: A253566.
Cf. A252737 (row sums), A252738 (row products).
Cf. A122111, A163511, A253550, A253560, A252464, A253563, A054429.
Applying A001222 gives A000120.
A reverse version is A005940.
These are the Heinz numbers of the rows of A242628.
Sum of prime indices of a(n) is A359043, reverse A161511.
A048793 gives partial sums of reversed standard comps, Heinz number A019565.
A066099 lists standard compositions.
A112798 list prime indices, sum A056239.
A358134 gives partial sums of standard compositions, Heinz number A358170.
Cf. A029837, A059893, A070939, A241916, A358135, A358137, A358195, A359042.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Times@@Prime/@#&/@Table[Accumulate[stc[n]-1]+1,{n,0,60}] (* Gus Wiseman, Dec 17 2022 *)

a(0) = 1, a(1) = 2; after which, a(2n) = A253550(a(n)), a(2n+1) = A253560(a(n)).
As a composition of related permutations:
a(n) = A122111(A163511(n)).
a(n) = A253563(A054429(n)).
Other identities and observations. For all n >= 0:
a(2n+1) - a(2n) > 0. [See the comment above.]
If n = 2^(x_1)+...+2^(x_k) then a(n) = Product_{i=1..k} prime(x_k-x_{i-1}-k+i) where x_0 = 0. - Gus Wiseman, Dec 23 2022

cp's OEIS Frontend

A007305 Numerators of Farey (or Stern-Brocot) tree fractions.

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

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A010059 Another version of the Thue-Morse sequence: let A_k denote the first 2^k terms; then A_0 = 1 and for k >= 0, A_{k+1} = A_k B_k, where B_k is obtained from A_k by interchanging 0's and 1's.

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A135141 a(1)=1, a(p_n)=2*a(n), a(c_n)=2*a(n)+1, where p_n = n-th prime, c_n = n-th composite number.

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A047679 Denominators in full Stern-Brocot tree.

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A056539 Self-inverse permutation: reverse the bits in binary expansion of n and also complement them (0->1, 1->0) if the run count (A005811) is even.

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A135141 a(1)=1, a(p_n)=2a(n), a(c_n)=2a(n)+1, where p_n = n-th prime, c_n = n-th composite number.