A020650 -id:A020650 - cp's OEIS frontend

A284459 Permutation of the positive integers: this permutation transforms the enumeration system of positive irreducible fractions A002487/A002487' (Calkin-Wilf) into the enumeration system A245327/A245328, and A162911/A162912 (Drib) into A020651/A020650 (Yu-Ting inverted).

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

Offset: 1

Yosu Yurramendi, Mar 27 2017

nonn

The inverse permutation is A284460.

Yosu Yurramendi, Table of n, a(n) for n = 1..16383
Index entries for sequences that are permutations of the natural numbers

Cf. A002487, A020650, A020651, A162911, A162912, A245327, A245328, A284460.

maxrow <- 12 # by choice
a <- 1
b01 <- 1
for(m in 0:maxrow){
  b01 <- c(b01, c(1-b01[2^m:(2^(m+1)-1)], b01[2^m:(2^(m+1)-1)]) )
  for(k in 0:(2^m-1)){
    a[2^(m+1) +       k] <- a[2^m + k] + 2^(m + b01[2^(m+1) +       k])
    a[2^(m+1) + 2^m + k] <- a[2^m + k] + 2^(m + b01[2^(m+1) + 2^m + k])
}}
a
# Yosu Yurramendi, Mar 27 2017

maxblock <- 7 # by choice
a <- 1:3
for(n in 4:2^maxblock){
ones <- which(as.integer(intToBits(n)) == 1)
nbit <- as.integer(intToBits(n))[1:tail(ones, n = 1)]
anbit <- nbit
for(i in 2:(length(anbit) - 1))
   anbit[i] <- 1 - bitwXor(anbit[i], anbit[i-1])
a <- c(a, sum(anbit*2^(0:(length(anbit) - 1))))
}
a
# Yosu Yurramendi, Apr 25 2021

a(n) = A258996(A231551(n)) = A231551(A092569(n)), n > 0 . - Yosu Yurramendi, Apr 10 2017

A284460 Permutation of the positive integers: this permutation transforms the enumeration system of positive irreducible fractions A245327/A245328 into the enumeration system A002487/A002487' (Calkin-Wilf), and A020651/A020650 (Yu-Ting inverted) into A162911/A162912(Drib).

Original entry on oeis.org

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

Offset: 1

Yosu Yurramendi, Mar 28 2017

nonn

The inverse permutation is A284459.

Yosu Yurramendi, Table of n, a(n) for n = 1..16383
Index entries for sequences that are permutations of the natural numbers

Cf. A002487, A020650, A020651, A162911, A162912, A245327, A245328, A284459.

maxrow <- 4 # by choice
a <- 1
b01 <- 1
for(m in 0:maxrow){
  b01 <- c(b01,rep(1,2^(m+1))); b01[(2^(m+1)+2^m-2^(m-1)):(2^(m+1)+2^m+2^(m-1)-1)] <- 0
  for(k in 0:(2^m-1)){
    a[2^(m+1) +       k] <- a[2^m + k] + 2^(m + b01[2^(m+1) +       k])
    a[2^(m+1) + 2^m + k] <- a[2^m + k] + 2^(m + b01[2^(m+1) + 2^m + k])
}}
a
# Yosu Yurramendi, Mar 28 2017

a(n) = A231550(A258996(n)) = A092569(A231550(n)), n > 0 . - Yosu Yurramendi, Apr 10 2017

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

Original entry on oeis.org

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

Offset: 1

Antti Karttunen

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

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

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

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

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

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

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

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

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

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

A083318 a(0) = 1; for n>0, a(n) = 2^n + 1.

Original entry on oeis.org

1, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025, 2049, 4097, 8193, 16385, 32769, 65537, 131073, 262145, 524289, 1048577, 2097153, 4194305, 8388609, 16777217, 33554433, 67108865, 134217729, 268435457, 536870913, 1073741825, 2147483649

Offset: 0

Paul Barry, Apr 25 2003

Inverse binomial transform of A005056.
Also, A000533 interpreted as binary numbers, written in base 10. Numbers whose representation in base 2 is has n+1 digits and the digit "1" is the initial and final digit and if n>1 then the internal digits are "0" (see example). - Omar E. Pol, Feb 24 2008
a(n) equals the number of ternary sequences of length n such that no two consecutive terms differ by 1. - David Nacin, May 31 2017

			From _Omar E. Pol_, Feb 24 2008: (Start)
------------------------------
n .... a(n) .. a(n) in base 2
------------------------------
0 ..... 1 ..... 1
1 ..... 3 ..... 11
2 ..... 5 ..... 101
3 ..... 9 ..... 1001
4 .... 17 ..... 10001
5 .... 33 ..... 100001
6 .... 65 ..... 1000001
7 ... 129 ..... 10000001
8 ... 257 ..... 100000001
9 ... 513 ..... 1000000001
(End)
G.f. = 1 + 3*x + 5*x^2 + 9*x^3 + 17*x^4 + 33*x^5 + 65*x^6 + 129*x^7 + ... - _Michael Somos_, Jun 04 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Ramon Carbó-Dorca, Boolean Hypercubes: The Origin of a Tagged Recursive Logic and the Limits of Artificial Intelligence, Universitat de Girona (Spain, 2020).
Quynh Nguyen, Jean Pedersen, and Hien T. Vu, New Integer Sequences Arising From 3-Period Folding Numbers, Vol. 19 (2016), Article 16.3.1. Cites this sequence.
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Except for the leading term, the same as A000051.

Cf. A000533, A083319, A132749.

Concatenation([1], List([1..40], n-> 2^n +1)); # G. C. Greubel, Nov 20 2019

[2^n+1-0^n : n in [0..40]]; // Vincenzo Librandi, Sep 01 2011

seq(`if`(n=0, 1, 2^n + 1), n=0..40); # G. C. Greubel, Nov 20 2019

Join[{1},2^Range[40]+1] (* Harvey P. Dale, May 17 2013 *)

{a(n) = if( n<1, n==0, 2^n + 1)}; /* Michael Somos, Jun 04 2016 */

[1]+[2^n +1 for n in (1..40)] # G. C. Greubel, Nov 20 2019

a(n) = 2^n + 1^n - 0^n.
G.f.: (1-2*x^2)/((1-x)*(1-2x)).
E.g.f.: exp(2*x) + exp(x) - exp(0).
a(n) = Sum_{k=0..n} 0^(k*(n-k))*2^(n-k). - Paul Barry, Feb 09 2005
a(n) = Min{m: A008687(m) = n+1}. - Reinhard Zumkeller, Jul 25 2006
Row sums of triangle A132749; = binomial transform of [1, 2, 0, 2, 0, 2, 0, 2, ...]. - Gary W. Adamson, Aug 28 2007
A020650(a(n)) = 1. - Yosu Yurramendi, Jun 01 2016

Edited by N. J. A. Sloane, Sep 28 2007

A065190 Self-inverse permutation of the positive integers: 1 is fixed, followed by an infinite number of adjacent transpositions (n n+1).

Original entry on oeis.org

1, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14, 17, 16, 19, 18, 21, 20, 23, 22, 25, 24, 27, 26, 29, 28, 31, 30, 33, 32, 35, 34, 37, 36, 39, 38, 41, 40, 43, 42, 45, 44, 47, 46, 49, 48, 51, 50, 53, 52, 55, 54, 57, 56, 59, 58, 61, 60, 63, 62, 65, 64, 67, 66, 69, 68, 71, 70, 73

Offset: 1

Antti Karttunen, Oct 19 2001

Also, a lexicographically minimal sequence of distinct positive integers such that a(n) is coprime to n. - Ivan Neretin, Apr 18 2015
The larger term of the pair (a(n), a(n+1)) is always odd. Had we started the sequence with a(1) = 0, it would be the lexicographically first sequence with this property if always extented with the smallest integer not yet present. - Eric Angelini, Feb 17 2017
From Yosu Yurramendi, Mar 21 2017: (Start)
This sequence is self-inverse. Except for the fixed point 1, it consists completely of 2-cycles: (2n, 2n+1), n > 0.
A020651(a(n)) = A020650(n), A020650(a(n)) = A020651(n), n > 0.
A245327(a(n)) = A245328(n), A245328(a(n)) = A245327(n), n > 0.
A063946(a(n)) = a(A063946(n)), n > 0.
A054429(a(n)) = a(A054429(n)) = A092569(n), n > 0.
A258996(a(n)) = a(A258996(n)), n > 0.
A258746(a(n)) = a(A258746(n)), n > 0. (End)
From Enrique Navarrete, Nov 13 2017: (Start)
With a(0)=0, and the rest of the sequence appended, a(n) is the smallest positive number not yet in the sequence such that the arithmetic mean of the first n+1 terms a(0), a(1), ..., a(n) is not an integer; i.e., the sequence is 0, 1, 3, 2, 5, 4, 7, 6, 9, 8, ...
Example: for n=5, (0 + 1 + 3 + 2 + 5)/5 is not an integer.
Fixed points are odd numbers >= 3 and also a(n) = n-2 for even n >= 4. (End)

Harry J. Smith, Table of n, a(n) for n = 1..1000
F. M. Dekking, Permutations of N generated by left-right filling algorithms, arXiv:2001.08915 [math.CO], 2020.
Index entries for sequences that are permutations of the natural numbers
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A004442, A065190 o A014681 = A065168, A014681 o A065190 = A065164.

[1] cat [n+(-1)^n: n in [2..80]]; // Vincenzo Librandi, Apr 18 2015

[seq(f(j),j=1..120)]; f := (n) -> `if`((n < 2), n,n+((-1)^n));

f[n_] := Rest@ Flatten@ Transpose[{Range[1, n + 1, 2], {1}~Join~Range[2, n, 2]}]; f@ 72 (* Michael De Vlieger, Apr 18 2015 *)
Rest@ CoefficientList[Series[x (x^3 - 2 x^2 + 2 x + 1)/((x - 1)^2*(x + 1)), {x, 0, 72}], x] (* Michael De Vlieger, Feb 17 2017 *)
Join[{1},LinearRecurrence[{1,1,-1},{3,2,5},80]] (* Harvey P. Dale, Feb 24 2021 *)

{ for (n=1, 1000, if (n>1, a=n + (-1)^n, a=1); write("b065190.txt", n, " ", a) ) } \\ Harry J. Smith, Oct 13 2009

x='x+O('x^100); Vec(x*(x^3-2*x^2+2*x+1)/((x-1)^2*(x+1))) \\ Altug Alkan, Feb 04 2016

def a(n): return 1 if n<2 else n + (-1)**n # Indranil Ghosh, Mar 22 2017

maxrow <- 8 # by choice
a <- c(1,3,2) # If it were c(1,2,3), it would be A000027
  for(m in 1:maxrow) for(k in 0:(2^m-1)){
a[2^(m+1)+    k] = a[2^m+k] + 2^m
a[2^(m+1)+2^m+k] = a[2^m+k] + 2^(m+1)
}
a
# Yosu Yurramendi, Apr 10 2017

a(1) = 1, a(n) = n+(-1)^n.
From Colin Barker, Feb 18 2013: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) for n>4.
G.f.: x*(x^3 - 2*x^2 + 2*x + 1) / ((x-1)^2*(x+1)). (End)
a(n)^a(n) == 1 (mod n). - Thomas Ordowski, Jan 04 2016
E.g.f.: x*(1+exp(x)) - 1 + exp(-x). - Robert Israel, Feb 04 2016
a(n) = A014681(n-1) + 1. - Michel Marcus, Dec 10 2016
a(1) = 1, for n > 0 a(2*n) = 2*a(a(n)) + 1, a(2*n + 1) = 2*a(a(n)). - Yosu Yurramendi, Dec 12 2020

A020651 Denominators in recursive bijection from positive integers to positive rationals (the bijection is f(1) = 1, f(2n) = f(n)+1, f(2n+1) = 1/(f(n)+1)).

Original entry on oeis.org

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

Offset: 1

David W. Wilson

If we insert an initial 1, this is the sequence of numerators in left-hand half of Kepler's tree of fractions. Form a tree of fractions by beginning with 1/1 and then giving every node i/j two descendants labeled i/(i+j) and j/(i+j). See A086592 for denominators. See A294442 for the Kepler tree itself.
Level n of the tree consists of 2^n nodes: 1/2; 1/3, 2/3; 1/4, 3/4, 2/5, 3/5; 1 /5, 4/5, 3/7, 4/7, 2/7, 5/7, 3/8, 5/8; ... Fibonacci numbers occur at the right edge this tree, i.e., a(A000225(n)) = A000045(n+1). The fractions are given in their reduced form, thus gcd(A020650(n), A020651(n)) = 1 and gcd(A020651(n), A086592(n)) = 1 for all n. - Antti Karttunen, May 26 2004
A generalization which includes the "rabbit tree" (A226080) and "all rationals tree" (A226130) follows. Suppose that a,b,c,d,e,f,g,h are complex numbers. Let S be the set of numbers defined by these rules: (1) 1 is in S; (2) if x is in S and cx+d is not 0, then U(x) = (ax+b)/(cx+d) is in S; (3) if x is in S and gx+h is not 0, then D(x) = (ex+f)/(gx+h) is in S. If an infinite path in the resulting tree has convergent nodes, then there is some node after which the path is "updown zigzag" ((UoD)o(UoD)o ...) or "downup zigzag" (DoU)o(DoU)o ...). If ag+ch is not 0, then the updown zigzag limit is invariant of x and equals [ae + cf - bg - dh + sqrt(X)]/(2(ag + ch)), where X = (ae + cf - bg - dh)^2 + 4(be + df + ag + ch). If ce + dg is not 0, then the downup zigzag limit is invariant of x and equals [ae + bg - cf - dh + sqrt(Y)]/(2(ce + dg)), where Y = (ae + bg - cf - dh)^2 + 4(af + bh)(ce + dg)) = X. Thus, for the tree A020651, the updown zigzag limit is -1 + sqrt(2) and the downup zigzag limit, sqrt(2). - Clark Kimberling, Nov 10 2013
From Yosu Yurramendi, Jul 13 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,
   1,3,2,3,
   1,4,3,4,2,5,3,5,
   1,5,4,5,3,7,4,7,2, 7,5, 7,3, 8,5, 8,
   1,6,5,6,4,9,5,9,3,10,7,10,4,11,7,11,2,9,7,9,5,12,7,12,3,11,8,11,5,13,8,13,
then the sum of the m-th row is 3^m (m = 0,1,2,), and each column is an arithmetic sequence. The differences of the arithmetic sequences, except the first on the left, give the sequence A093873 (Numerators in Kepler's tree of harmonic fractions) (a(2^(m+1)-1-k) - a(2^m-1-k) = A093873(k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
If the rows are written in a right-aligned fashion:
                                                                          1,
                                                                       1, 2,
                                                                  1, 3,2, 3,
                                                        1, 4,3, 4,2, 5,3, 5,
                                      1,5,4,5,3, 7,4, 7,2, 7,5, 7,3, 8,5, 8,
  1,6,5,6,4,9,5,9,3,10,7,10,4,11,7,11,2,9,7,9,5,12,7,12,3,11,8,11,5,13,8,13,
then each column k is a Fibonacci sequence. (End)
For m >= 0, a(2^m) = 1 and a(3*2^m) = 2. For n >= 0, a(A070875(n)) = 3 (for m >= 0, a(5*2^m) = 3 and a(7*2^m) = 3). - Yosu Yurramendi, Jun 02 2016

			1, 2, 1/2, 3, 1/3, 3/2, 2/3, 4, 1/4, 4/3, ...

T. D. Noe, Table of n, a(n) for n = 1..10000
D. N. Andreev, On a Wonderful Numbering of Positive Rational Numbers, Matematicheskoe Prosveshchenie, Series 3, volume 1, 1997, pages 126-134 (in Russian). a(n) = denominator of r(n).
Godofredo Iommi and Mario Ponce, Odometers in non-compact spaces, arXiv:2404.03768 [math.DS], 2024. See p. 19.
Johannes Kepler, Excerpt from the Chapter II of the Book III of the Harmony of the World: On the seven harmonic divisions of the string (illustrates the A020651/A086592-tree).
Pelegrí Viader, Jaume Paradís and Lluís Bibiloni, A New Light on Minkowski's ?(x) Function, J. Number Theory, 73 (2) (1998), 212-227. See p. 215.
Shen Yu-Ting, A Natural Enumeration of Non-Negative Rational Numbers -- An Informal Discussion, American Mathematical Monthly, volume 87, number 1, January 1980, pages 25-29. a(n) = denominator of gamma_n.
Index entries for fraction trees
Index entries for sequences related to music

See A294442 and A093873/A093875 for two different versions of the Kepler tree.

Cf. A020650, A086592.

import Data.List (transpose); import Data.Ratio (denominator)
a020651_list = map denominator ks where
   ks = 1 : concat (transpose [map (+ 1) ks, map (recip . (+ 1)) ks])
-- Reinhard Zumkeller, Feb 22 2014

A020651 := n -> `if`((n < 2),n,`if`(type(n,even), A020651(n/2), A020650(n-1)));

f[1] = 1; f[n_?EvenQ] := f[n] = f[n/2]+1; f[n_?OddQ] := f[n] = 1/(f[(n-1)/2]+1); a[n_] := Denominator[f[n]]; Table[a[n], {n, 1, 94}] (* Jean-François Alcover, Nov 22 2011 *)

N <- 25 # arbitrary
a <- c(1,1,2)
for(n in 1:N){
  a[4*n]   <- a[2*n]
  a[4*n+1] <- a[2*n] + a[2*n+1]
  a[4*n+2] <-          a[2*n+1]
  a[4*n+3] <- a[2*n] + a[2*n+1]
}
a
# Yosu Yurramendi, Jul 13 2014

a(1) = 1, a(2n) = a(n), a(2n+1) = A020650(2n). - Antti Karttunen, May 26 2004
a(2n) = A020650(2n+1). - Yosu Yurramendi, Jul 17 2014
a(2^m + k) = A093873(2^(m+1) + k) = A093873(2^(m+1) + 2^m + k), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, May 18 2016
a(2^m + 2^r + k) = A093873(2^r + k)*(m-(r-1)) + A093873(k), m >= 0, r <= m-1, 0 <= k < 2^r. For k=0 A093873(0) = 0 is needed. - Yosu Yurramendi, Jul 30 2016
a((2n+1)*2^m) = A086592(n), m >= 0, n > 0. For n = 0 A086592(0) = 1 is needed. - Yosu Yurramendi, Feb 14 2017
a(4n+2) = a(4n+1) - a(4n) = a(2n+1) = a(4n+1) - a(n), n > 0. - Yosu Yurramendi, May 08 2018
a(1) = 1, a(n+1) = 2*floor(1/a(n))+1-1/a(n). - Jan Malý, Jul 30 2019
a(n) = A002487(A231551(n)), n > 0. - Yosu Yurramendi, Jul 15 2021

Entry revised by N. J. A. Sloane, May 24 2004

A071585 Numerator of the continued fraction expansion whose terms are the first-order differences of exponents in the binary representation of 4*n, with the exponents of 2 being listed in descending order.

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8, 6, 9, 10, 11, 9, 12, 11, 13, 6, 9, 10, 11, 9, 12, 11, 13, 7, 11, 13, 14, 13, 17, 15, 18, 11, 16, 17, 19, 14, 19, 18, 21, 7, 11, 13, 14, 13, 17, 15, 18, 11, 16, 17, 19, 14, 19, 18, 21, 8, 13, 16, 17, 17, 22, 19, 23, 16, 23, 24, 27, 19

Offset: 0

Paul D. Hanna, Jun 01 2002

Thus a(n)/a(m) = d_1 + 1/(d_2 + 1/(d_3 + ... + 1/d_k)) where m = n - 2^floor(log_2(n)) + 1 and where d_j = b_j - b_(j+1) are the differences of the binary exponents b_j > b_(j+1) defined by: 4*n = 2^b_1 + 2^b_2 + 2^b_3 + ... 2^b_k.
All the rationals are uniquely represented by this sequence - compare Stern's diatomic sequence A002487.
This sequence lists the rationals >= 1 in order by the sum of the terms of their continued fraction expansions. For example, the numerators generated from partitions of 5 that do not end with 1 are listed together as 5, 7, 7, 8, 5, 7, 7, 8, since: 5/1 = [5]; 7/2 = [3;2]; 7/3 = [2;3]; 8/3 = [2;1,2]; 5/4 = [1;4]; 7/5 = [1;2,2]; 7/4 = [1;1,3]; 8/5 = [1;1,1,2].
From Yosu Yurramendi, Jun 23 2014: (Start)
If the terms (n>0) are written as an array:
  1,
  2,
  3, 3,
  4, 5, 4, 5,
  5, 7, 7, 8, 5, 7, 7, 8,
  6, 9,10,11, 9,12,11,13, 6, 9,10,11, 9,12,11,13,
  7,11,13,14,13,17,15,18,11,16,17,19,14,19,18,21,7,11,13,14,13,17,15,18,11, ...
then the sum of the k-th row is 2*3^(k-2) for k>1, each column is an arithmetic progression. The differences of the arithmetic sequences give the sequence A071585 itself: a(2^(p+1)+k) - a(2^p+k) = a(k). A002487 and A007306 also have these properties. The first terms of columns, excluding a(0), give A086593.
If the rows (n>0) are written on right:
                                                         1;
                                                         2;
                                                     3,  3;
                                             4,  5,  4,  5;
                               5, 7,  7,  8, 5,  7,  7,  8;
  6, 9, 10, 11, 9, 12, 11, 13, 6, 9, 10, 11, 9, 12, 11, 13;
then each column is a Fibonacci sequence: a(2^(p+2)+k) = a(2^(p+1)+k) + a(2^p+k). The first terms of columns, excluding a(0), give A086593. (End)
n>1 occurs in this sequence phi(n) = A000010(n) times, as it occurs in A007306 (Franklin T. Adams-Watters's comment), that is the sequence obtained by adding numerator and denominator in the Calkin-Wilf enumeration system of positive rationals. A229742(n)/A071766(n) is also an enumeration system of all positive rationals (HCS system), and in each level m >= 0 (ranks between 2^m and 2^(m+1)-1) rationals are the same in both systems. Thus a(n) (A229742(n)+A071766(n)) has the same terms in each level as A007306. The same property occurs in all numerator+denominator sequences of enumeration systems of positive rationals, as, for example, A007306 (A007305+A047679), A086592 (A020650+A020651), and A268087 (A162909+A162910). - Yosu Yurramendi, Apr 06 2016
a(n) = A086592(A059893(n)), a(A059893(n)) = A086592(n), n > 0. - Yosu Yurramendi, May 30 2017

			a(37)=17 as it is the numerator of 17/5 = 3 + 1/(2 + 1/2), which is a continued fraction that can be derived from the binary expansion of 4*37 = 2^7 + 2^4 + 2^2; the binary exponents are {7, 4, 2}, thus the differences of these exponents are {3, 2, 2}; giving the continued fraction expansion of 17/5=[3,2,2].
Illustration of Sum_{m=0..2^(k-1)-1} a(2^k + m) = 3^k:
k=2: 3^2 = a(2^2) + a(2^2 + 1) = 4 + 5;
k=3: 3^3 = a(2^3) + a(2^3 + 1) + a(2^3 + 2) + a(2^3 + 3) = 5 + 7 + 7 + 8;
k=4: 3^4 = a(2^4) + a(2^4+1) + a(2^4+2) + a(2^4+3) + a(2^4+4) + a(2^4+5) + a(2^4+6) + a(2^4+7) = 6 + 9 + 10 + 11 + 9 + 12 + 11 + 13.
1, 2, 3, 3/2, 4, 5/2, 4/3, 5/3, 5, 7/2, 7/3, 8/3, 5/4, 7/5, 7/4, 8/5, 6, ...
From _Yosu Yurramendi_, Jun 27 2014: (Start)
a(0) =             = 1;
a(1) = a(0) + a(0) = 2;
a(2) = a(0) + a(1) = 3;
a(3) = a(1) + a(0) = 3;
a(4) = a(0) + a(2) = 4;
a(5) = a(1) + a(3) = 5;
a(6) = a(2) + a(0) = 4;
a(7) = a(3) + a(1) = 5;
a(8) = a(0) + a(4) = 5;
a(9) = a(1) + a(5) = 7;
a(10) = a(2) + a(6) = 7;
a(11) = a(3) + a(7) = 8;
a(12) = a(4) + a(0) = 5;
a(13) = a(5) + a(1) = 7;
a(14) = a(6) + a(2) = 7;
a(15) = a(7) + a(3) = 8. (End)

Paul D. Hanna, Table of n, a(n) for n = 0..10000
Index entries for fraction trees

Cf. A071766.

ncf[n_]:=Module[{br=Reverse[Flatten[Position[Reverse[IntegerDigits[4 n,2]],1]-1]]}, Numerator[FromContinuedFraction[Flatten[Join[{Abs[ Differences[ br]],Last[br]}]]]]]; Join[{1},Array[ncf,80]] (* Harvey P. Dale, Jul 01 2012 *)
{1}~Join~Table[Numerator@ FromContinuedFraction@ Append[Abs@ Differences@ #, Last@ #] &@ Log2[NumberExpand[4 n, 2] /. 0 -> Nothing], {n, 120}] (* Version 11, or *)
{1}~Join~Table[Numerator@ FromContinuedFraction@ Append[Abs@ Differences@ #, Last@ #] &@ Log2@ DeleteCases[# Reverse[2^Range[0, Length@ # - 1]] &@ IntegerDigits[4 n, 2], k_ /; k == 0], {n, 120}] (* Michael De Vlieger, Aug 15 2016 *)

blocklevel <- 7  # arbitrary
a <- c(1,2)
for(k in 1:blocklevel)
a <- c(a, a + c(a[((length(a)/2)+1):length(a)],a[1:(length(a)/2)]))
a
# Yosu Yurramendi, Jun 26 2014

blocklevel <- 7  # arbitrary
a <- c(1,2)
for(p in 0:blocklevel)
  for(k in 1:2^(p+1)){
    if (k <=  2^p) a[k + 2^(p+1)] = a[k] + a[k + 2^p]
    else           a[k + 2^(p+1)] = a[k] + a[k - 2^p]
}
a
# Yosu Yurramendi, Jun 27 2014

a(2^k + 2^j + m) = (k-j)*a(2^j + m) + a(m) when 2^k > 2^j > m >= 0.
a(0) = 1, a(2^k) = k + 2,
a(2^k + 1) = 2*k + 1 (k>0),
a(2^k + 2) = 3*k - 2 (k>1),
a(2^k + 3) = 3*k - 1 (k>1),
a(2^k + 4) = 4*k - 7 (k>2).
a(2^k - 1) = Fibonacci(k+2) = A000045(k+2).
Sum_{m=0..2^(k-1)-1} a(2^k + m) = 3^k (k>0).
From Yosu Yurramendi, Jun 27 2014: (Start)
Write n = k + 2^(m+1), k = 0,1,2,...,2^(m+1)-1, m = 0,1,2,...
if 0 <= k < 2^m, a(k + 2^(m+1)) = a(k) + a(k + 2^m).
if 2^m <= k < 2^(m+1), a(k + 2^(m+1)) = a(k) + a(k - 2^m).
with a(0)=1, a(1)=2. (End)
a(n) = A059893(A086592(n)), n>0. - Yosu Yurramendi, Apr 09 2016
a(n) = A093873(n) + A093875(n), n > 0. - Yosu Yurramendi, Jul 22 2016
a(n) = A093873(2n) + A093873(2n+1), n > 0; a(n) = A093875(2n) = A093875(2n+1), n > 0. - Yosu Yurramendi, Jul 25 2016
a(n) = sqrt(A071766(2^(m+1)+n)*A229742(2^(m+1)+n) - A071766(2^m+n)*A229742(2^m+n)), for n > 0, where m = floor(log_2(n)+1). - Yosu Yurramendi, Jun 10 2019
a(n) = A007306(A059893(A233279(n))), n > 0. - Yosu Yurramendi, Aug 07 2021
a(n) = A007306(A059894(A006068(n))), n > 0. - Yosu Yurramendi, Sep 29 2021
Conjecture: a(n) = a(floor(n/2)) + Sum_{k=1..A000120(n)} a(b(n, k))*(-1)^(k-1) for n > 0 with a(0) = 1 where b(n, k) = A025480(b(n, k-1) - 1) for n > 0, k > 0 with b(n, 0) = n. - Mikhail Kurkov, Feb 20 2023

A086592 Denominators in left-hand half of Kepler's tree of fractions.

Original entry on oeis.org

2, 3, 3, 4, 4, 5, 5, 5, 5, 7, 7, 7, 7, 8, 8, 6, 6, 9, 9, 10, 10, 11, 11, 9, 9, 12, 12, 11, 11, 13, 13, 7, 7, 11, 11, 13, 13, 14, 14, 13, 13, 17, 17, 15, 15, 18, 18, 11, 11, 16, 16, 17, 17, 19, 19, 14, 14, 19, 19, 18, 18, 21, 21, 8, 8, 13, 13, 16, 16, 17, 17, 17, 17, 22, 22, 19, 19, 23

Offset: 1

Antti Karttunen, Aug 28 2003

Form a tree of fractions by beginning with 1/1 and then giving every node i/j two descendants labeled i/(i+j) and j/(i+j).
Level n of the left-hand half of the tree consists of 2^(n-1) nodes: 1/2; 1/3, 2/3; 1/4, 3/4, 2/5, 3/5; 1/5, 4/5, 3/7, 4/7, 2/7, 5/7, 3/8, 5/8; ... .
The right-hand half is identical to the left-hand half. - Michel Dekking, Oct 05 2017
n>1 occurs in this sequence phi(n) = A000010(n) times, as it occurs in A007306 (Franklin T. Adams-Watters' comment), that is the sequence obtained by adding numerator and denominator in the Calkin-Wilf enumeration system of positive rationals. A020650(n)/A020651(n) is also an enumeration system of all positive rationals (Yu-Ting system), and in each level m >= 0 (ranks between 2^m and 2^(m+1)-1) rationals are the same in both systems. Thus a(n) has the same terms in each level as A007306. The same property occurs in all numerator+denominator sequences of enumeration systems of positive rationals, as, for example, A007306 (A007305+A047679), A071585 (A229742+A071766), and A268087 (A162909+A162910). - Yosu Yurramendi, Apr 06 2016

Johannes Kepler, Mysterium cosmographicum, Tuebingen, 1596, 1621, Caput XII.
Johannes Kepler, Harmonice Mundi, Linz, 1619, Liber III, Caput II.
Johannes Kepler, The Harmony of the World [1619], trans. E. J. Aiton, A. M. Duncan and J. V. Field, American Philosophical Society, Philadelphia, 1997, p. 163.

Johannes Kepler, Harmonices mundi libri V ... (A Latin original scanned in Internet Archive. The fraction-tree is illustrated on the page 27 of the third book (Liber III), which is on the page 117 of the PDF-document.)
Johannes Kepler, Excerpt from the Chapter II of the Book III of the Harmony of the World: On the seven harmonic divisions of the string (Illustrates the A020651/A086592-tree).
OEIS Wiki, Historical sequences
Pelegrí Viader, Jaume Paradís and Lluís Bibiloni, A New Light on Minkowski's ?(x) Function, J. Number Theory, 73 (2) (1998), 212-227. See p. 215.
Index entries for fraction trees
Index entries for sequences related to music

Bisection of A020650.
See A093873/A093875 for the full tree.
A020651 gives the numerators. Bisection: A086593. Cf. A002487, A004169.

(* b = A020650 *) b[1] = 1; b[2] = 2; b[3] = 1; b[n_] := b[n] = Switch[ Mod[n, 4], 0, b[n/2 + 1] + b[n/2], 1, b[(n - 1)/2 + 1], 2, b[(n - 2)/2 + 1] + b[(n - 2)/2], 3, b[(n - 3)/2]]; a[n_] := b[2n]; Array[a, 100] (* Jean-François Alcover, Jan 22 2016 *)

maxlevel <- 15
d <- c(1,2)
for(m in 0:maxlevel)
for(k in 1:2^m) {
   d[2^(m+1)    +k] <- d[k] + d[2^m+k]
   d[2^(m+1)+2^m+k] <- d[2^(m+1)+k]
}
b <- vector()
for(m in 0:maxlevel) for(k in 0:(2^m-1)) b[2^m+k] <- d[2^(m+1)+k]
a <- vector()
for(n in 1:2^maxlevel) {a[2*n-1] <- b[n]; a[2*n] <- b[n+1]}
a[1:128]
# Yosu Yurramendi, May 16 2018

a(n) = A020650(n) + A020651(n) = A020650(2n).
a(n) = A071585(A059893(n)), a(A059893(n)) = A071585(n), n > 0. - Yosu Yurramendi, May 30 2017
a(2*n-1) = A086593(n); a(2*n) = A086593(n+1), n > 0. - Yosu Yurramendi, May 16 2018
a(n) = A007306(A231551(n)), n > 0. - Yosu Yurramendi, Aug 07 2021

Entry revised by N. J. A. Sloane, May 24 2004

A092569 Permutation of integers a(a(n)) = n. In binary representation of n, transformation of inner bits, 1 <-> 0, gives binary representation of a(n).

Original entry on oeis.org

0, 1, 2, 3, 6, 7, 4, 5, 14, 15, 12, 13, 10, 11, 8, 9, 30, 31, 28, 29, 26, 27, 24, 25, 22, 23, 20, 21, 18, 19, 16, 17, 62, 63, 60, 61, 58, 59, 56, 57, 54, 55, 52, 53, 50, 51, 48, 49, 46, 47, 44, 45, 42, 43, 40, 41, 38, 39, 36, 37, 34, 35, 32, 33, 126, 127, 124, 125, 122, 123, 120

Offset: 0

Zak Seidov, Feb 28 2004

Primes which stay primes under transformation "opposite inner bits", A092570.
This permutation transforms the enumeration system of positive irreducible fractions A020651/A020650 into the enumeration system A245327/A245326, and vice versa. - Yosu Yurramendi, Jun 16 2015
A117120(a(n)) = a(A117120(n)), n > 0.
A258996(a(n)) = a(A258996(n)), n > 0.
A258746(a(n)) = a(A258746(n)), n > 0.
A054429(a(n)) = a(A054429(n)), n > 0.
a(n) = A054429(A065190(n)) = A065190(A054429(n)), n > 0. - Yosu Yurramendi, Mar 23 2017

			a(9)=15 because 9_10 = 1001_2, transformation of inner bits gives 1001_2 -> 1111_2 = 15_10.

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Cf. A092570.

bb={0, 1, 2, 3};Do[id=IntegerDigits[n, 2];Do[id[[i]]=1-id[[i]], {i, 2, Length[id]-1}];bb=Append[bb, FromDigits[id, 2]], {n, 4, 1000}];fla=Flatten[bb]
(* Second program: *)
Table[If[n < 2, n, Function[b, FromDigits[#, 2] &@ Join[{First@ b}, Most[Rest@ b] /. { 0 -> 1, 1 -> 0}, {Last@ b}]]@ IntegerDigits[n, 2]], {n, 0, 70}] (* Michael De Vlieger, Apr 03 2017 *)

T(n)={pow2=2;v=binary(n);L=#v-1;forstep(k=L,2,-1,if(v[k],n-=pow2,n+=pow2);pow2*=2);return(n)};
for(n=0,70,print1(T(n),", ")) \\ Washington Bomfim, Jan 18 2011

maxrow <- 8 # by choice
a <- 1:3 # If it were c(1, 3, 2), it would be A054429
for(m in 1:maxrow) for(k in 0:(2^m-1)){
a[2^(m+1)+    k] = a[2^m+k] + 2^(m+1)
a[2^(m+1)+2^m+k] = a[2^m+k] + 2^m
}
a
# Yosu Yurramendi, Apr 10 2017

a(0) = 0, a(1) = 1, a(2) = 2, a(3) = 3, a(2^(m+1) +k) = a(2^m+k) + 2^(m+1),

a(2^(m+1)+2^m+k) = a(2^m+k) + 2^m, m >= 1, 0 <= k < 2^m. - Yosu Yurramendi, Apr 02 2017

A245327 Numerators in recursive bijection from positive integers to positive rationals, where the bijection is f(1) = 1, f(2n) = 1/(f(n)+1), f(2n+1) = f(n)+1.

Original entry on oeis.org

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

Offset: 1

Yosu Yurramendi, Jul 18 2014

a(n)/A245328(n) enumerates all the reduced nonnegative rational numbers exactly once.
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,
   2, 3,1, 3,
   3, 5,2, 5,3, 4,1, 4,
   5, 8,3, 8,5, 7,2, 7,4, 7,3, 7,4,5,1,5,
   8,13,5,13,8,11,3,11,7,12,5,12,7,9,2,9,7,11,4,11,7,10,3,10,5,9,4,9,5,6,1,6,
then the sum of the m-th row is 3^m (m = 0,1,2,), and each column k is a Fibonacci sequence.
If the rows are written in a right-aligned fashion:
                                                                        1,
                                                                      1,2,
                                                                  2,3,1,3,
                                                          3,5,2,5,3,4,1,4,
                                      5, 8,3, 8,5, 7,2, 7,4,7,3,7,4,5,1,5,
8,13,5,13,8,11,3,11,7,12,5,12,7,9,2,9,7,11,4,11,7,10,3,10,5,9,4,9,5,6,1,6,
then each column is an arithmetic sequence.
If the sequence is considered by blocks of length 2^m, m = 0,1,2,..., the blocks of this sequence are permutations of terms of blocks from A002487 (Stern's diatomic series or Stern-Brocot sequence), and, more precisely, the reverses of blocks of A020650 ( a(2^m+k) = A020650(2^(m+1)-1-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
Moreover, each block is the bit-reversed permutation of the corresponding block of A245325.

Michael De Vlieger, Table of n, a(n) for n = 1..16383, rows 1-14, flattened.
Index entries for fraction trees

Cf. A002487, A020651, A245325, A245328, A273493.

f[n_] := Which[n == 1, 1, EvenQ@ n, 1/(f[n/2] + 1), True, f[(n - 1)/2] + 1]; Table[Numerator@ f@ k, {n, 7}, {k, 2^(n - 1), 2^n - 1}] // Flatten (* Michael De Vlieger, Mar 02 2017 *)

a(n) = my(A=0); forstep(i=logint(n, 2), 0, -1, if(bittest(n, i), A++, A=1/(A+1))); numerator(A) \\ Mikhail Kurkov, Mar 12 2023

N  <- 25 # arbitrary
a <- c(1,1,2)
for(n in 1:N){
  a[4*n]   <-          a[2*n+1]
  a[4*n+1] <- a[2*n] + a[2*n+1]
  a[4*n+2] <- a[2*n]
  a[4*n+3] <- a[2*n] + a[2*n+1]
}
a

a(2n) = A245328(2n+1) , a(2n+1) = A245328(2n) , n=0,1,2,3,...
a((2*n+1)*2^m - 2) = A273493(n), n > 0, m > 0. For n = 0, m > 0, A273493(0) = 1 is needed. For n = 1, m = 0, A273493(0) = 1 is needed. For n > 1, m = 0, numerator((2*n-1) = num+den(n-1). - Yosu Yurramendi, Mar 02 2017
a(n) = A002487(A284459(n)). - Yosu Yurramendi, Aug 23 2021

cp's OEIS Frontend

A284459 Permutation of the positive integers: this permutation transforms the enumeration system of positive irreducible fractions A002487/A002487' (Calkin-Wilf) into the enumeration system A245327/A245328, and A162911/A162912 (Drib) into A020651/A020650 (Yu-Ting inverted).

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A284460 Permutation of the positive integers: this permutation transforms the enumeration system of positive irreducible fractions A245327/A245328 into the enumeration system A002487/A002487' (Calkin-Wilf), and A020651/A020650 (Yu-Ting inverted) into A162911/A162912(Drib).

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

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A083318 a(0) = 1; for n>0, a(n) = 2^n + 1.

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A065190 Self-inverse permutation of the positive integers: 1 is fixed, followed by an infinite number of adjacent transpositions (n n+1).

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Magma

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PARI

PARI

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A020651 Denominators in recursive bijection from positive integers to positive rationals (the bijection is f(1) = 1, f(2n) = f(n)+1, f(2n+1) = 1/(f(n)+1)).

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Haskell