A007306 -id:A007306 - cp's OEIS frontend

A065658 The table of permutations of N, each row induced by the rotation (to the right) of the n-th node in the infinite binary "decimal" fraction tree.

7, 25, 1, 31, 22, 1, 1, 3, 2, 1, 223, 10, 247, 2, 1, 15, 94, 4, 3, 2815, 1, 127, 6, 5, 4, 3, 2, 1, 5, 7, 28, 5, 4, 115, 2, 1, 385, 20479, 127, 6, 94, 4, 3, 2, 1, 13, 175, 8, 7, 6, 5, 4, 3, 2, 1, 1792, 46, 9, 280, 7, 234881023, 5, 4, 3, 322, 1, 61, 382, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

Offset: 0

Antti Karttunen, Nov 22 2001

Consider the following infinite binary tree, where the nodes are numbered in breadth-first, left-to-right fashion from the top as in A065625 and then assigned the following rational values:
--------------------------------------(0.1)---------------------------------------
----------------(0.01)-------------------------------------(0.11)-----------------
-----(0.001)--------------(0.011)---------------(0.101)--------------(0.111)------
(0.0001)-(0.0011)----(0.0101)-(0.0111)-----(0.1001)-(0.1011)-----(0.1101)-(0.1111)
i.e., the elements (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/16, 3/16, ..., of the Quasicyclic group Z+((2a+1)/(2^b)) for prime 2) listed here in their binary "decimal" fraction form. Subjecting this tree to any similar binary tree rotation as used in A065625 induces a permutation of the rationals in range ]0,1[ (i.e., including also the ones having infinite binary expansions, corresponding to infinite paths in above tree), which we then convert to permutations of N by taking the positions of the mapped values at the ]0,1[ side of the Stern-Brocot Tree (A007305/A007306). See example at A065670.

Index entries for sequences related to Stern's sequences

The first row (rotate the top node right): A065660, 2nd row (rotate the top node's left child): A065662, 3rd row (rotate the top node's right child): A065664, 4th row: A065666, 5th row: A065668, 6th row: A065670, 7th row: A065672. For the other needed Maple procedures follow A065625, A047679, A054424 and A054429. Cf. also A065674-A065676. Inverse permutations are given in A065659.

Cf. also A065934-A065935.

[seq(RotateBinFracRightTable(j),j=0..119)]; RotateBinFracRightTable := n -> RotateBinFracNodeRight(1+(n-((trinv(n)*(trinv(n)-1))/2)),(((trinv(n)-1)*(((1/2)*trinv(n))+1))-n)+1);
RotateBinFracNodeRight := (t,n) -> frac2position_in_0_1_SB_tree(RotateBinFracNodeRight_x(t,SternBrocot0_1frac(n)));
RotateBinFracNodeRight_x := proc(t,x) local num,den; den := 2^(1+floor_log_2(t)); num := (2*(t-(den/2)))+1; if((x <= (num-1)/den) or (x >= (num+1)/den)) then RETURN(x); fi; if(x <= ((2*(num-1))+1)/(2*den)) then RETURN((2*(x - ((num-1)/den))) + ((num-1)/den)); fi; if(x < (num/den)) then RETURN(x + (1/(2*den))); fi; RETURN((num/den) + ((x-((num-1)/den))/2)); end;
SternBrocot0_1frac := proc(n) local m; m := n + 2^floor_log_2(n); SternBrocotTreeNum(m)/SternBrocotTreeDen(m); end;
frac2position_in_0_1_SB_tree := r -> RETURN(ReflectBinTreePermutation(cfrac2binexp(convert(1/r,confrac))));

A277324 Odd bisection of A260443 (the even terms): a(n) = A260443((2*n)+1).

Original entry on oeis.org

2, 6, 18, 30, 90, 270, 450, 210, 630, 6750, 20250, 9450, 15750, 47250, 22050, 2310, 6930, 330750, 3543750, 1653750, 4961250, 53156250, 24806250, 727650, 1212750, 57881250, 173643750, 18191250, 8489250, 25467750, 2668050, 30030, 90090, 40020750, 1910081250, 891371250, 9550406250, 455814843750, 212713593750

Offset: 0

Antti Karttunen, Oct 10 2016

nonn

From David A. Corneth, Oct 22 2016: (Start)
The exponents of the prime factorization of a(n) are first nondecreasing, then nonincreasing.
The exponent of 2 in the prime factorization of a(n) is 1. (End)

			A method to find terms of this sequence, explained by an example to find a(7). To find k = a(7), we find k such that A048675(k) = 2*7+1 = 15. 7 has the binary partitions: {[7, 0, 0], [5, 1, 0], [3, 2, 0], [1, 3, 0], [3, 0, 1], [1, 1, 1]}. To each of those, we prepend a 1. This gives the binary partitions of 15 starting with a 1. For example, for the first we get [1, 7, 0, 0]. We see that only [1, 5, 1, 0], [1, 3, 2, 0] and [1, 1, 1, 1] start nondecreasing, then nonincreasing, so we only check those. These numbers will be the exponents in a prime factorization. [1, 5, 1, 0] corresponds to prime(1)^1 * prime(2)^5 * prime(3)^1 * prime(4)^0 = 2430. We find that [1, 1, 1, 1] gives k = 210 for which A048675(k) = 15 so a(7) = 210. - _David A. Corneth_, Oct 22 2016

Antti Karttunen, Table of n, a(n) for n = 0..1024

Cf. A005811, A005940, A007306, A007949, A156552, A260443, A277189, A277323, A283484, A283975, A284267, A284268, A284563, A284564, A284573.

Cf. A277200 (same sequence sorted into ascending order).

a[n_] := a[n] = Which[n < 2, n + 1, EvenQ@ n, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &@ a[n/2], True, a[#] a[# + 1] &[(n - 1)/2]]; Table[a[2 n + 1], {n, 0, 38}] (* Michael De Vlieger, Apr 05 2017 *)

from sympy import factorint, prime, primepi
from operator import mul
def a003961(n):
    F=factorint(n)
    return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**F[i] for i in F])
def a260443(n): return n + 1 if n<2 else a003961(a260443(n//2)) if n%2==0 else a260443((n - 1)//2)*a260443((n + 1)//2)
def a(n): return a260443(2*n + 1)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 21 2017

a(n) = A260443((2*n)+1).
a(0) = 2; for n >= 1, a(n) = A260443(n) * A260443(n+1).
Other identities. For all n >= 0:
A007949(a(n)) = A005811(n). [See comments in A125184.]
A156552(a(n)) = A277189(n), a(n) = A005940(1+A277189(n)).
A048675(a(n)) = 2n + 1. - David A. Corneth, Oct 22 2016
A001222(a(n)) = A007306(1+n).
A056169(a(n)) = A284267(n).
A275812(a(n)) = A284268(n).
A248663(a(n)) = A283975(n).
A000188(a(n)) = A283484(n).
A247503(a(n)) = A284563(n).
A248101(a(n)) = A284564(n).
A046523(a(n)) = A284573(n).
a(n) = A277198(n) * A284008(n).
a(n) = A284576(n) * A284578(n) = A284577(n) * A000290(A284578(n)).

More linking formulas added by Antti Karttunen, Apr 16 2017

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

A008687 Number of 1's in 2's complement representation of -n.

Original entry on oeis.org

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

Offset: 0

R. H. Hardin

a(A127904(n)) = n and a(m) < n for m < A127904(n). - Reinhard Zumkeller, Feb 05 2007
a(n) = A000120(A010078(n)), n>0; a(n) = A023416(A004754(n-1)), n>1. - Reinhard Zumkeller, Dec 04 2015
Conjecture: a(n)+1 is the length of the Hirzebruch (negative) continued fraction for the Stern-Brocot tree fraction A007305(n)/A007306(n). - Andrey Zabolotskiy, Apr 17 2020
Terms a(n); n >= 2 can be generated recursively, as follows. Let S(0) = {1}, then for k >=1, let S(k) = {S(k-1)+1, S(k-1)}, where +1 means +1 on every term of S(k-1); see Example. Each step of the recursion gives the next 2^k terms of the sequence. - David James Sycamore, Jul 15 2024

			Using the above recursion for a(n); n >= 2, we have:
 S(0) = {1} so a(2) = 1;
 S(1) = {2,1} so a(3,4) = 2,1;
 S(2) = {3,2,2,1}, so a(5,6,7,8) = 3,2,2,1;
As irregular table the sequence for n >= 2 begins:
  1;
  2,1;
  3,2,2,1;
  4,3,3,2,3,2,2,1;
  5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1;
  6,5,5,4,5,4,4,3,5,4,3,3,4,3,3,2,5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1;
and so on (the k-th row contains 2^k terms; k>=0). - _David James Sycamore_, Jul 15 2024

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Michael Gilleland, Some Self-Similar Integer Sequences
Wikipedia, Two's complement

This is Guy Steele's sequence GS(4, 3) (see A135416).
Cf. A000120, A004754, A010078, A023416, A290251.
Cf. A007305, A007306, A061313, A088696.

a008687 n = a008687_list !! n
a008687_list = 0 : 1 : c [1] where c (e:es) = e : c (es ++ [e+1,e])
-- Reinhard Zumkeller, Mar 07 2011

a[0] = 0; a[1] = 1; a[n_] := a[n] = Mod[n,2] + a[Mod[n,2] + Floor[n/2]]; Array[a, 96, 0] (* Jean-François Alcover, Aug 12 2017, after Reinhard Zumkeller *)

a(n) = if(n<2,n, n--; logint(n,2) - hammingweight(n) + 2); \\ Kevin Ryde, Apr 14 2021

a(n) = A023416(n-1) + 1.
a(n) = if n<=1 then n else (n mod 2) + a((n mod 2) + floor(n/2)). - Reinhard Zumkeller, Feb 05 2007
a(n) = if n<2 then n else a(ceiling(n/2)) + n mod 2. - Reinhard Zumkeller, Jul 25 2006
Min{m: a(m)=n} = if n>0 then A083318(n-1) else 0. - Reinhard Zumkeller, Jul 25 2006

A049455 Triangle read by rows: T(n,k) = numerator of fraction in k-th term of n-th row of variant of Farey series.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 2, 1, 3, 2, 3, 1, 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1, 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1, 6, 5, 9

Offset: 1

N. J. A. Sloane

Stern's diatomic array read by rows (version 4, the 0,1 version).
This sequence divided by A049456 gives another version of the Stern-Brocot tree.
Row n has length 2^n + 1.
Define mediant of a/b and c/d to be (a+c)/(b+d). We get A006842/A006843 if we omit terms from n-th row in which denominator exceeds n.
Largest term of n-th row = A000045(n), Fibonacci numbers. - Reinhard Zumkeller, Apr 02 2014

			0/1, 1/1; 0/1, 1/2, 1/1; 0/1, 1/3, 1/2, 2/3, 1/1; 0/1, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 1/1; 0/1, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, ... = A049455/A049456
The 0,1 version of Stern's diatomic array (cf. A002487) begins:
0,1,
0,1,1,
0,1,1,2,1,
0,1,1,2,1,3,2,3,1,
0,1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,1,
0,1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,1,5,4,7,3,8,5,7,2,7,5,3,3,7,4,5,1,
...

Martin Gardner, Colossal Book of Mathematics, Classic Puzzles, Paradoxes, and Problems, Chapter 25, Aleph-Null and Aleph-One, p. 328, W. W. Norton & Company, NY, 2001.
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.

Robert G. Wilson v, Table of n, a(n) for n = 1..10000 (first 8204 terms from Reinhard Zumkeller)
C. Giuli and R. Giuli, A primer on Stern's diatomic sequence, Fib. Quart., 17 (1979), 103-108, 246-248 and 318-320 (but beware errors).
Jennifer Lansing, Largest Values for the Stern Sequence, J. Integer Seqs., 17 (2014), #14.7.5.
M. Shrader-Frechette, Modified Farey sequences and continued fractions, Math. Mag., 54 (1981), 60-63.
N. J. A. Sloane, Stern-Brocot or Farey Tree
Index entries for sequences related to Stern's sequences

Cf. A049456. Also A007305, A007306, A006842, A006843, A070878, A070879.
Row sums are A007051.
Cf. A000051 (row lengths), A293165 (distinct terms).

import Data.List (transpose)
import Data.Ratio ((%), numerator, denominator)
a049455 n k = a049455_tabf !! (n-1) !! (k-1)
a049455_row n = a049455_tabf !! (n-1)
a049455_tabf = map (map numerator) $ iterate
   (\row -> concat $ transpose [row, zipWith (+/+) row $ tail row]) [0, 1]
   where u +/+ v = (numerator u + numerator v) %
                   (denominator u + denominator v)
-- Reinhard Zumkeller, Apr 02 2014

f[l_List] := Block[{k = Length@l, j = l}, While[k > 1, j = Insert[j, j[[k]] + j[[k - 1]], k]; k--]; j]; NestList[f, {0, 1}, 6] // Flatten (* Robert G. Wilson v, Nov 10 2019 *)

mediant(x, y) = (numerator(x)+numerator(y))/(denominator(x)+denominator(y));
newrow(rowa) = {my(rowb = []); for (i=1, #rowa-1, rowb = concat(rowb, rowa[i]); rowb = concat(rowb, mediant(rowa[i], rowa[i+1]));); concat(rowb, rowa[#rowa]);}
rows(nn) = {my(rowa); for (n=1, nn, if (n==1, rowa = [0, 1], rowa = newrow(rowa)); print(apply(x->numerator(x), rowa)););} \\ Michel Marcus, Apr 03 2019

Row 1 is 0/1, 1/1. Obtain row n from row n-1 by inserting mediants between each pair of terms.

More terms from Larry Reeves (larryr(AT)acm.org), Apr 12 2000

A054424 Permutation of natural numbers: maps the canonical list of fractions (A020652/A020653) to whole Stern-Brocot (Farey) tree (top = 1/1 and both sides < 1 and > 1, but excluding the "fractions" 0/1 and 1/0).

Original entry on oeis.org

1, 2, 3, 4, 7, 8, 5, 6, 15, 16, 31, 32, 9, 11, 12, 14, 63, 64, 10, 13, 127, 128, 17, 23, 24, 30, 255, 256, 19, 28, 511, 512, 33, 18, 20, 47, 48, 27, 29, 62, 1023, 1024, 22, 25, 2047, 2048, 65, 35, 39, 21, 95, 96, 26, 56, 60, 126, 4095, 4096, 34, 40, 55, 61, 8191, 8192

Offset: 1

Antti Karttunen

nonn

			Whole Stern-Brocot tree: 1/1 1/2 2/1 1/3 2/3 3/2 3/1 1/4 2/5 3/5 3/4 4/3 5/3 5/2 4/1 1/5 2/7
Canonical fractions: 1/1 1/2 2/1 1/3 3/1 1/4 2/3 3/2 4/1 1/5 5/1 1/6 2/5 3/4 4/3 5/2 6/1

Cf. A047679, A007305, A007306, A054427, A057114. In table form: A054425. Inverse permutation: A054426.

cfrac2binexp := proc(c) local i,e,n; n := 0; for i from 1 to nops(c) do e := c[i]; if(i = nops(c)) then e := e-1; fi; n := ((2^e)*n) + ((i mod 2)*((2^e)-1)); od; RETURN(n); end;
frac2position_in_whole_SB_tree := proc(r) local k,msb; if(1 = r) then RETURN(1); else if(r > 1) then k := cfrac2binexp(convert(r,confrac)); else k := ReflectBinTreePermutation(cfrac2binexp(convert(1/r,confrac))); fi; msb := floor_log_2(k); if(r > 1) then RETURN(k + (2^(msb+1))); else RETURN(k + (2^(msb+1)) - (2^msb)); fi; fi; end;
canonical_fractions_to_whole_SternBrocot_permutation := proc(u) local a,n,i; a := []; for n from 2 to u do for i from 1 to n-1 do if (1 = igcd(n,i)) then a := [op(a),frac2position_in_whole_SB_tree(i/(n-i))]; fi; od; od; RETURN(a); end; # ReflectBinTreePermutation and floor_log_2 given in A054429

canonical_fractions_to_whole_SternBrocot_permutation(30);

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

A065625 Table of permutations of N, each row being a generator of the "rotation group" of infinite planar binary tree. Inverse generators are given in A065626.

Original entry on oeis.org

3, 1, 1, 7, 5, 1, 2, 3, 2, 1, 6, 2, 7, 2, 1, 14, 11, 4, 3, 2, 1, 15, 6, 5, 9, 3, 2, 1, 4, 7, 3, 5, 4, 3, 2, 1, 5, 4, 15, 6, 11, 4, 3, 2, 1, 12, 10, 8, 7, 6, 5, 4, 3, 2, 1, 13, 22, 9, 4, 7, 13, 5, 4, 3, 2, 1, 28, 23, 10, 19, 8, 7, 6, 5, 4, 3, 2, 1, 29, 12, 11, 10, 9, 8, 15, 6, 5, 4, 3, 2, 1, 30, 13, 6, 11, 5, 9, 8, 7, 6, 5, 4, 3, 2, 1, 31, 14, 14, 12, 23, 10, 9, 17, 7, 6, 5, 4, 3, 2

Offset: 0

Antti Karttunen, Nov 08 2001

Consider the following infinite binary tree, where the nodes are numbered in breadth-first, left-to-right fashion from the top as:
.............................1............................
.............2...............................3............
.....4...............5...............6...............7....
.8.......9.......10.....11.......12.....13.......14.....15
etc., i.e. the node Y is a descendant of the node X, iff its binary expansion (the most significant bits) begin with the binary expansion of X.
In this table the n-th row is a permutation induced by the rotation of the node n right and in the table A065626 the corresponding row gives the inverse of that permutation, induced by rotation of the node n left. Particular realizations of this tree are the Christoffel tree and the Stern-Brocot tree (A007305/A007306), thus each such rotation, or composition of such rotations (e.g. A065249) induces a particular bijective function on rationals and such functions form the "group A" of the order-preserving permutations of the rational numbers as defined by Cameron.

The first row (rotate the top node right): A057114, 2nd row (rotate the top node's left child): A065627, 3rd row (rotate the top node's right child): A065629, 4th row: A065631, 5th row: A065633, 6th row: A065635, 7th row: A065637, 8th row: A065639. Maple procedure floor_log_2 given in A054429, for trinv, follow A065167.

Variant of the same idea: A065658.

[seq(RotateRightTable(j),j=0..119)];
RotateRightTable := n -> RotateNodeRight(1+(n-((trinv(n)*(trinv(n)-1))/2)),(((trinv(n)-1)*(((1/2)*trinv(n))+1))-n)+1);
# Rewrites t-prefixed x's in the following way: t -> t1, t1... -> t11..., t0 -> t, t01... -> t10..., t00... -> t0... and leaves other x's intact.
RotateNodeRight := proc(t,x) local u,y; u := floor_log_2(t)+1; y := floor_log_2(x)+1; if(y < u) then RETURN(x); fi; if(floor(x/(2^(y-u))) <> t) then RETURN(x); fi; if(x = t) then RETURN((2*x)+1); fi; if(1 = (floor(x/(2^(y-u-1))) mod 2)) then RETURN(x + (t * 2^(y-u)) + 2^(y-u)); fi; if(y = (u+1)) then RETURN(x/2); fi; if(1 = (floor(x/(2^(y-u-2))) mod 2)) then RETURN(x + 2^(y-u-2)); fi; RETURN(x - (t * 2^(y-u-1))); end;

A287896 a(n) = A002487(n)*A001511(n).

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 3, 4, 4, 6, 5, 6, 5, 6, 4, 5, 5, 8, 7, 9, 8, 10, 7, 8, 7, 10, 8, 9, 7, 8, 5, 6, 6, 10, 9, 12, 11, 14, 10, 12, 11, 16, 13, 15, 12, 14, 9, 10, 9, 14, 12, 15, 13, 16, 11, 12, 10, 14, 11, 12, 9, 10, 6, 7, 7, 12, 11, 15, 14, 18, 13, 16, 15, 22, 18, 21, 17, 20, 13, 15, 14, 22, 19, 24, 21, 26, 18, 20, 17

Offset: 1

I. V. Serov, Jun 02 2017

nonn

Proposed name: N-fusc.
Each number n>0 appears in this sequence exactly n times.
From Yosu Yurramendi, Apr 08 2019: (Start)
The terms (n>0) may be written as a left-justified array with rows of length 2^m:
  1,
  2, 2,
  3, 3,  4,  3,
  4, 4,  6,  5,  6,  5,  6,  4,
  5, 5,  8,  7,  9,  8, 10,  7,  8,  7, 10,  8,  9,  7,  8,  5,
  6, 6, 10,  9, 12, 11, 14, 10, 12, 11, 16, 13, 15, 12, 14,  9, 10,  9, ...
  ...
as well as right-justified fashion:
                                                                      1,
                                                                   2, 2,
                                                            3, 3,  4, 3,
                                            4,  4,  6,  5,  6, 5,  6, 4,
             5, 5,  8,  7,  9,  8, 10,  7,  8,  7, 10,  8,  9, 7,  8, 5,
... 14,  9, 10, 9, 14, 12, 15, 13, 16, 11, 12, 10, 14, 11, 12, 9, 10, 6,
From these two dispositions interesting properties can be induced (see FORMULA section)
(End)

I. V. Serov, Table of n, a(n) for n = 1..8192

Cf. A001511, A002487, A007306, A288002.

Table[Block[{a = 1, b = 0, m = n}, While[m > 0, If[OddQ@ m, b = a + b, a = a + b]; m = Floor[m/2]]; b] IntegerExponent[2 n, 2], {n, 89}] (* Michael De Vlieger, Jun 14 2017, after Jean-François Alcover at A002487 *)

from functools import reduce
def A287896(n): return (n&-n).bit_length()*sum(reduce(lambda x,y:(x[0],x[0]+x[1]) if int(y) else (x[0]+x[1],x[1]),bin(n)[-1:2:-1],(1,0))) # Chai Wah Wu, Jul 14 2022

a(1) = 1; for n>1: a(n) = (A002487(n-1) + A002487(n) + A002487(n+1))/2.
a(n) = A007306(n) - A288002(n).
From Yosu Yurramendi, Apr 08 2019: (Start)
For m >= 0, 0 <= k < 2^m, a(2^(m+1)+k) - a(2^m+k) = a(k). a(0) = 1 is needed.
For m >= 0, 0 <= k < 2^m, a(2^(m+1)-1-k) - a(2^(m)-1-k) = a(k).
(End)

A061313 Minimal number of steps to get from 1 to n by (a) subtracting 1 or (b) multiplying by 2.

Original entry on oeis.org

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

Offset: 1

Henry Bottomley, Jun 06 2001

Also number of steps to get from n to 1 by process of adding 1 if odd, or dividing by 2 if even.
It is straightforward to prove that the number n appears F(n) times in this sequence, where F(n) is the n-th Fibonacci number (A000045). - Gary Gordon, May 31 2019
Conjecture: a(n)+2 is the sum of the terms of the Hirzebruch (negative) continued fraction for the Stern-Brocot tree fraction A007305(n)/A007306(n). - Andrey Zabolotskiy, Apr 17 2020

			a(2) = 1 since 2 = 1*2, a(3) = 3 since 3 = 1*2*2-1, a(11) = 6 since 11 = (1*2*2-1)*2*2-1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Index to sequences related to the complexity of n

Cf. A056792, A006577, A029837, A000045.

Cf. A007305, A007306, A008687.

a061313 n = fst $ until ((== 1) . snd) (\(u, v) -> (u + 1, f v)) (0, n)
   where f n = if r == 0 then n' else n + 1  where (n', r) = divMod n 2
-- Reinhard Zumkeller, Sep 05 2015

f[n_] := Block[{c = 0, m = n}, While[m != 1, If[ EvenQ[m], While[ EvenQ[m], m = m/2; c++ ], m++; c++ ]]; Return[c]]; Table[f[n], {n, 1, 100}]

a(n)=if(n<2,0,s=n; c=1; while((s+s%2)/(2-s%2)>1,s=(s+s%2)/(2-s%2); c++); c)

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

a(n) = if(n--,2*(logint(n,2)+1)) - hammingweight(n); \\ Kevin Ryde, Oct 21 2021

a(2n) = a(n)+1; a(2n+1) = a(n+1)+2; a(1) = 0.
Is Sum_{k=1..n} a(k) asymptotic to C*n*log(n) where 3 > C > 2? - Benoit Cloitre, Aug 31 2002
G.f.: x/(1-x) * Sum_{k>=0} (x^2^k + x^2^(k+1)/(1+x^2^k)). - Ralf Stephan, Jun 14 2003
a(n) = A080791(n-1) + A029837(n). - Ralf Stephan, Jun 14 2003
a(n) = 2*A023416(n-1) + A000120(n-1) = A023416(A062880(n)) = A023416(A000695(n)) + 1. - Ralf Stephan, Jul 16 2003
a(n) = A119477(n) - 1. - Philippe Deléham, Nov 03 2008

cp's OEIS Frontend

A065658 The table of permutations of N, each row induced by the rotation (to the right) of the n-th node in the infinite binary "decimal" fraction tree.

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A277324 Odd bisection of A260443 (the even terms): a(n) = A260443((2*n)+1).

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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.

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A008687 Number of 1's in 2's complement representation of -n.

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A049455 Triangle read by rows: T(n,k) = numerator of fraction in k-th term of n-th row of variant of Farey series.

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A054424 Permutation of natural numbers: maps the canonical list of fractions (A020652/A020653) to whole Stern-Brocot (Farey) tree (top = 1/1 and both sides < 1 and > 1, but excluding the "fractions" 0/1 and 1/0).

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

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