A086593 -id:A086593 - cp's OEIS frontend

A071766 Denominator of the continued fraction expansion whose terms are the first-order differences of exponents in the binary representation of 4n, with the exponents of 2 being listed in descending order.

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

Offset: 0

Paul D. Hanna, Jun 04 2002

If the terms (n>0) are written as an array:
 1,
 1, 2,
 1, 2, 3, 3,
 1, 2, 3, 3, 4, 5, 4, 5,
 1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8,
 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,
then the sum of the m-th row is 3^m (m = 0,1,2,3,...), each column is constant and the terms are from A071585 (a(2^m+k) = A071585(k), k = 0,1,2,...).
If the rows are written in a right-aligned fashion:
                                                                                 1,
                                                                             1,  2,
                                                                     1,  2,  3,  3,
                                                       1, 2,  3,  3, 4,  5,  4,  5,
                          1, 2,  3,  3, 4,  5,  4,  5, 5, 7,  7,  8, 5,  7,  7,  8,
..., 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^(m+2)+k) = a(2^(m+1)+k) + a(2^m+k), m = 0,1,2,..., k = 0,1,2,...,2^m-1 with a_k(1) = A071766(k) and a_k(2) = A086593(k) being the first two terms of each column sequence). - Yosu Yurramendi, Jun 23 2014

			a(37) = 5 as it is the denominator 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].
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, ...

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

Cf. A071585, A086593.

{1}~Join~Table[Denominator@ FromContinuedFraction@ Append[Abs@ Differences@ #, Last@ #] &@ Log2[NumberExpand[4 n, 2] /. 0 -> Nothing], {n, 120}] (* Version 11, or *)
{1}~Join~Table[Denominator@ 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 <- 6 # arbitrary
a <- 1
for(m in 0:blocklevel) for(k in 0:(2^(m-1)-1)){
  a[2^(m+1)+k]             <- a[2^m+k]
  a[2^(m+1)+2^(m-1)+k]     <- a[2^m+2^(m-1)+k]
  a[2^(m+1)+2^m+k]         <- a[2^(m+1)+k]     +  a[2^(m+1)+2^(m-1)+k]
  a[2^(m+1)+2^m+2^(m-1)+k] <- a[2^(m+1)+2^m+k]
}
a
# Yosu Yurramendi, Jul 11 2014

a(n) = A071585(m), where m = n - floor(log_2(n));
a(0) = 1, a(2^k) = 1, a(2^k + 1) = 2.
a(2^k - 1) = Fibonacci(k+1) = A000045(k+1).
a(2^m+k) = A071585(k), m=0,1,2,..., k=0,1,2,...,2^m-1. - Yosu Yurramendi, Jun 23 2014
a(2^m-k) = F_k(m), k=0,1,2,..., m > log_2(k). F_k(m) is a Fibonacci sequence, where F_k(1) = a(2^(m_0(k))-1-k), F_k(2) = a(2^(m_0(k)+1)-1-k), m_0(k) = ceiling(log_2(k+1))+1 = A070941(k). - Yosu Yurramendi, Jun 23 2014
a(n) = A002487(A059893(A233279(n))) = A002487(1+A059893(A006068(n))), n > 0. - Yosu Yurramendi, Sep 29 2021

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

A229742 a(n) = A071585(n) - A071766(n).

Original entry on oeis.org

0, 1, 2, 1, 3, 3, 1, 2, 4, 5, 4, 5, 1, 2, 3, 3, 5, 7, 7, 8, 5, 7, 7, 8, 1, 2, 3, 3, 4, 5, 4, 5, 6, 9, 10, 11, 9, 12, 11, 13, 6, 9, 10, 11, 9, 12, 11, 13, 1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8, 7, 11, 13, 14, 13, 17, 15, 18, 11, 16, 17, 19, 14

Offset: 0

N. J. A. Sloane, Oct 05 2013, at the suggestion of Kevin Ryde

From Yosu Yurramendi, Jun 30 2014: (Start)
If the terms (n>0) are written as an array (left-aligned fashion):
  1,
  2,1,
  3,3, 1, 2,
  4,5, 4, 5,1, 2, 3, 3,
  5,7, 7, 8,5, 7, 7, 8,1,2, 3, 3,4, 5, 4, 5,
  6,9,10,11,9,12,11,13,6,9,10,11,9,12,11,13,1,2,3,3,4,5,4,5,5,7,7,8,5,7,7,8,
then the sum of the k-th row is 3^(k-1) and each column is an arithmetic sequence. The differences of the arithmetic sequences gives the sequence A071585 (a(2^(p+1)+k) - a(2^p+k) = A071585(k), p = 0,1,2,..., k = 0,1,2,...,2^p-1).
The first terms of each column give A071766. The second terms of each column give A086593. So, A086593(n) = A071585(n) + A071766(n).
If the rows (n>0) are written in a right-aligned fashion:
                                                                           1,
                                                                         2,1,
                                                                     3,3,1,2,
                                                             4,5,4,5,1,2,3,3,
                                             5,7,7,8,5,7,7,8,1,2,3,3,4,5,4,5,
   6,9,10,11,9,12,11,13,6,9,10,11,9,12,11,13,1,2,3,3,4,5,4,5,5,7,7,8,5,7,7,8,
then each column is a Fibonacci sequence (a(2^(p+2)+k) = a(2^(p+1)+k) + a(2^p+k) p = 0,1,2,..., k = 0,1,2,...,2^p-1, with a_k(1) = A071585(k) and a_k(2) = A071766(k) being the first two terms of each column sequence). (End)

			A229742/A071766 = 0, 1, 2, 1/2, 3, 3/2, 1/3, 2/3, 4, 5/2, 4/3, 5/3, 1/4, 2/5, 3/4, 3/5, 5, 7/2, 7/3, 8/3, 5/4, 7/5, 7/4, 8/5, ... (this is the HCS form of the Stern-Brocot tree).

Index entries for fraction trees

Cf. A071585, A071766, A086593, A238837.

blocklevel <- 6 # arbitrary
a <- 1
for(m in 0:blocklevel) for(k in 0:(2^(m-1)-1)){
  a[2^(m+1)+k]             <- a[2^m+k] + a[2^m+2^(m-1)+k]
  a[2^(m+1)+2^(m-1)+k]     <- a[2^(m+1)+k]
  a[2^(m+1)+2^m+k]         <- a[2^m+2^(m-1)+k]
  a[2^(m+1)+2^m+2^(m-1)+k] <- a[2^m+k]
}
a
# Yosu Yurramendi, Jul 11 2014

From Yosu Yurramendi, May 26 2019: (Start)
a(2^(m+1)+2^m+k) = A071585(    k)
a(2^(m+1)    +k) = A071585(2^m+k), m >= 0, 0 <= k < 2^m. (End)
a(n) = A002487(A059893(A006068(n))) = A002487(1+A059893(A233279(n))), n > 0. - Yosu Yurramendi, Sep 29 2021

A294442 Kepler's tree of fractions, read across rows (the fraction i/j is represented as the pair i,j).

Original entry on oeis.org

1, 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: 0

N. J. A. Sloane, Nov 20 2017

The first row contains the single fraction 1/1,
the second row contains the single fraction 1/2,
and thereafter below each fraction i/j we write two fractions i/(i+j), j/(i+j).
If we just look at the numerators we recover the same sequence, and if we just look at the denominators we get A086593 with the terms (after the first) repeated.
Sequence A020651 is almost the same as this, except that it lacks one of the initial 1's, and the definition focuses on single numbers rather than pairs of numbers or fractions. For that reason it seems to be best to have a separate entry (this sequence) for the actual tree.

			The tree begins as follows:
..............1/1
...............|
..............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

Michael De Vlieger, Table of n, a(n) for n = 0..16383
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.
Richard J. Mathar, The Kepler binary tree of reduced fractions, 2017.
Index entries for fraction trees.
Index entries for sequences related to music.

Cf. A020651, A086593.
A different version of the Kepler tree is described in A093873.
See A294446 for the tree of Farey fractions.

# S[n] is the list of fractions, written as pairs [i,j], in row n of Kepler's triangle
S[0]:=[[1,1]]; S[1]:=[[1,2]];
for n from 2 to 10 do
S[n]:=[];
for k from 1 to nops(S[n-1]) do
t1:=S[n-1][k];
a:=[t1[1],t1[1]+t1[2]];
b:=[t1[2],t1[1]+t1[2]];
S[n]:=[op(S[n]),a,b];
od:
lprint(S[n]);
od:

Map[{Numerator@ #, Denominator@ #} &, #] &@ Flatten@ Nest[Append[#, Flatten@ Map[{#1/(#1 + #2), #2/(#1 + #2)} & @@ {Numerator@ #, Denominator@ #} &, Last@ #]] &, {{1/1}, {1/2}}, 5] // Flatten (* Michael De Vlieger, Apr 18 2018 *)

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

cp's OEIS Frontend

A071766 Denominator of the continued fraction expansion whose terms are the first-order differences of exponents in the binary representation of 4n, with the exponents of 2 being listed in descending order.

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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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A229742 a(n) = A071585(n) - A071766(n).

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A294442 Kepler's tree of fractions, read across rows (the fraction i/j is represented as the pair i,j).

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

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