A086592 -id:A086592 - cp's OEIS frontend

A086593 Bisection of A086592, denominators of the left-hand half of Kepler's tree of fractions.

2, 3, 4, 5, 5, 7, 7, 8, 6, 9, 10, 11, 9, 12, 11, 13, 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, 26, 25, 29, 13, 20, 23, 25, 22, 29, 26, 31, 17, 25, 27, 30, 23, 31, 29, 34, 9, 15, 19, 20, 21, 27, 23, 28, 21, 30

Offset: 1

Antti Karttunen, Aug 28 2003

nonn

Also denominator of alternate fractions in Kepler's tree as shown in A294442. - N. J. A. Sloane, Nov 20 2017

Antti Karttunen, Table of n, a(n) for n = 1..2048 (computed from b-file of A020650 provided by _T. D. Noe_)

Cf. A000045, A020650, A020651, A071585, A071766, A086592, A294442.

(* 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[1] = 2; a[n_] := b[4 n - 4]; Array[a, 100] (* Jean-François Alcover, Jan 22 2016, after Yosu Yurramendi's formula for A020650 *)

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]
}
a <- vector()
for(m in 0:maxlevel) for(k in 0:(2^m-1)) a[2^m+k] <- d[2^(m+1)+k]
  a[1:63]
# Yosu Yurramendi, May 16 2018

a(n) = A086592(2n-1) = A020650(4n-2).
a(n+1) = A071585(n) + A071766(n), n >= 0. - Yosu Yurramendi, Jun 30 2014
From Yosu Yurramendi, Jan 04 2016: (Start)
a(2^(m+1)+k+1) - a(2^m+k+1) = A071585(k),  m >= 0, 0 <= k < 2^m.
a(2^(m+2)-k) = a(2^(m+1)-k) + a(2^m-k),   m > 0, 0 <= k < 2^m-1.
(End)
a(2^n) = A000045(n+3). - Antti Karttunen, Jan 29 2016, based on above.
a(n) = A020651(4n-1), a(n+1) = A020651(4n+1), n > 0. - Yosu Yurramendi, May 08 2018
a(2^m+k) = A071585(2^(m+1)+k),  m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, May 16 2018

A002487 Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2n) = a(n), a(2n+1) = a(n) + a(n+1).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Also called fusc(n) [Dijkstra].
a(n)/a(n+1) runs through all the reduced nonnegative rationals exactly once [Stern; Calkin and Wilf].
If the terms are written as an array:
  column 0 1 2 3 4 5 6 7 8 9 ...
  row 0: 0
  row 1: 1
  row 2: 1,2
  row 3: 1,3,2,3
  row 4: 1,4,3,5,2,5,3,4
  row 5: 1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5
  row 6: 1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,...
  ...
then (ignoring row 0) the sum of the k-th row is 3^(k-1), each column is an arithmetic progression and the steps are nothing but the original sequence. - Takashi Tokita (butaneko(AT)fa2.so-net.ne.jp), Mar 08 2003
From N. J. A. Sloane, Oct 15 2017: (Start)
The above observation can be made more precise. Let A(n,k), n >= 0, 0 <= k <= 2^(n-1)-1 for k > 0, denote the entry in row n and column k of the left-justified array above.
The equations for columns 0,1,2,3,4,... are successively (ignoring row 0):
     1 (n >= 1),
     n (n >= 2),
   n-1 (n >= 3),
  2n-3 (n >= 3),
   n-2 (n >= 4),
  3n-7 (n >= 4),
  ...
and in general column k > 0 is given by
A(n,k) = a(k)*n - A156140(k) for n >= ceiling(log_2(k+1))+1, and 0 otherwise.
(End)
a(n) is the number of odd Stirling numbers S_2(n+1, 2r+1) [Carlitz].
Moshe Newman proved that the fraction a(n+1)/a(n+2) can be generated from the previous fraction a(n)/a(n+1) = x by 1/(2*floor(x) + 1 - x). The successor function f(x) = 1/(floor(x) + 1 - frac(x)) can also be used.
a(n+1) = number of alternating bit sets in n [Finch].
If f(x) = 1/(1 + floor(x) - frac(x)) then f(a(n-1)/a(n)) = a(n)/a(n+1) for n >= 1. If T(x) = -1/x and f(x) = y, then f(T(y)) = T(x) for x > 0. - Michael Somos, Sep 03 2006
a(n+1) is the number of ways of writing n as a sum of powers of 2, each power being used at most twice (the number of hyperbinary representations of n) [Carlitz; Lind].
a(n+1) is the number of partitions of the n-th integer expressible as the sum of distinct even-subscripted Fibonacci numbers (= A054204(n)), into sums of distinct Fibonacci numbers [Bicknell-Johnson, theorem 2.1].
a(n+1) is the number of odd binomial(n-k, k), 0 <= 2*k <= n. [Carlitz], corrected by Alessandro De Luca, Jun 11 2014
a(2^k) = 1. a(3*2^k) = a(2^(k+1) + 2^k) = 2. Sequences of terms between a(2^k) = 1 and a(2^(k+1)) = 1 are palindromes of length 2^k-1 with a(2^k + 2^(k-1)) = 2 in the middle. a(2^(k-1) + 1) = a(2^k - 1) = k+1 for k > 1. - Alexander Adamchuk, Oct 10 2006
The coefficients of the inverse of the g.f. of this sequence form A073469 and are related to binary partitions A000123. - Philippe Flajolet, Sep 06 2008
It appears that the terms of this sequence are the number of odd entries in the diagonals of Pascal's triangle at 45 degrees slope. - Javier Torres (adaycalledzero(AT)hotmail.com), Aug 06 2009
Let M be an infinite lower triangular matrix with (1, 1, 1, 0, 0, 0, ...) in every column shifted down twice:
   1;
   1, 0;
   1, 1, 0;
   0, 1, 0, 0;
   0, 1, 1, 0, 0;
   0, 0, 1, 0, 0, 0;
   0, 0, 1, 1, 0, 0, 0;
   ...
   Then this sequence A002487 (without initial 0) is the first column of lim_{n->oo} M^n. (Cf. A026741.) - Gary W. Adamson, Dec 11 2009 [Edited by M. F. Hasler, Feb 12 2017]
Member of the infinite family of sequences of the form a(n) = a(2*n); a(2*n+1) = r*a(n) + a(n+1), r = 1 for A002487 = row 1 in the array of A178239. - Gary W. Adamson, May 23 2010
Equals row 1 in an infinite array shown in A178568, sequences of the form
a(2*n) = r*a(n), a(2*n+1) = a(n) + a(n+1); r = 1. - Gary W. Adamson, May 29 2010
Row sums of A125184, the Stern polynomials. Equivalently, B(n,1), the n-th Stern polynomial evaluated at x = 1. - T. D. Noe, Feb 28 2011
The Kn1y and Kn2y triangle sums, see A180662 for their definitions, of A047999 lead to the sequence given above, e.g., Kn11(n) = A002487(n+1) - A000004(n), Kn12(n) = A002487(n+3) - A000012(n), Kn13(n) = A002487(n+5) - A000034(n+1) and Kn14(n) = A002487(n+7) - A157810(n+1). For the general case of the knight triangle sums see the Stern-Sierpiński triangle A191372. This triangle not only leads to Stern's diatomic series but also to snippets of this sequence and, quite surprisingly, their reverse. - Johannes W. Meijer, Jun 05 2011
Maximum of terms between a(2^k) = 1 and a(2^(k+1)) = 1 is the Fibonacci number F(k+2). - Leonid Bedratyuk, Jul 04 2012
Probably the number of different entries per antidiagonal of A223541. That would mean there are exactly a(n+1) numbers that can be expressed as a nim-product 2^x*2^y with x + y = n. - Tilman Piesk, Mar 27 2013
Let f(m,n) be the frequency of the integer n in the interval [a(2^(m-1)), a(2^m-1)]. Let phi(n) be Euler's totient function (A000010). Conjecture: for all integers m,n n<=m f(m,n) = phi(n). - Yosu Yurramendi, Sep 08 2014
Back in May 1995, it was proved that A000360 is the modulo 3 mapping, (+1,-1,+0)/2, of this sequence A002487 (without initial 0). - M. Jeremie Lafitte (Levitas), Apr 24 2017
Define a sequence chf(n) of Christoffel words over an alphabet {-,+}: chf(1) = '-'; chf(2*n+0) = negate(chf(n)); chf(2*n+1) = negate(concatenate(chf(n),chf(n+1))). Then the length of the chf(n) word is fusc(n) = a(n); the number of '-'-signs in the chf(n) word is c-fusc(n) = A287729(n); the number of '+'-signs in the chf(n) word is s-fusc(n) = A287730(n). See examples below. - I. V. Serov, Jun 01 2017
The sequence can be extended so that a(n) = a(-n), a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1) for all n in Z. - Michael Somos, Jun 25 2019
Named after the German mathematician Moritz Abraham Stern (1807-1894), and sometimes also after the French clockmaker and amateur mathematician Achille Brocot (1817-1878). - Amiram Eldar, Jun 06 2021
It appears that a(n) is equal to the multiplicative inverse of A007305(n+1) mod A007306(n+1).  For example, a(12) is 2, the multiplicative inverse of A007305(13) mod A007306(13), where A007305(13) is 4 and A007306(13) is 7. - Gary W. Adamson, Dec 18 2023

			Stern's diatomic array begins:
  1,1,
  1,2,1,
  1,3,2,3,1,
  1,4,3,5,2,5,3,4,1,
  1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5,1,
  1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,7,11,4,9,...
  ...
a(91) = 19, because 91_10 = 1011011_2; b_6=b_4=b_3=b_1=b_0=1, b_5=b_2=0;  L=5; m_1=0, m_2=1, m_3=3, m_4=4, m_5=6; c_1=2, c_2=3, c_3=2, c_4=3; f(1)=1, f(2)=2, f(3)=5, f(4)=8, f(5)=19. - _Yosu Yurramendi_, Jul 13 2016
From _I. V. Serov_, Jun 01 2017: (Start)
a(n) is the length of the Christoffel word chf(n):
n  chf(n) A070939(n)   a(n)
1   '-'       1          1
2   '+'       2          1
3   '+-'      2          2
4   '-'       3          1
5   '--+'     3          3
6   '-+'      3          2
... (End)
G.f. = x + x^2 + 2*x^3 + x^4 + 3*x^5 + 2*x^6 + 3*x^7 + x^8 + ... - _Michael Somos_, Jun 25 2019

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Index entries for "core" sequences
Index entries for fraction trees
Index entries for sequences related to enumerating the rationals
Index entries for sequences related to Stern's sequences
Index entries for sequences related to trees

Cf. A000120, A000123, A000360, A001045, A002083, A011655, A020950, A026741, A037227, A046815, A070871, A070872, A071883, A073459, A084091, A101624, A126606, A168081, A174980, A174981, A178239, A178568, A212288, A213369, A260443, A277020, A277325, A287729, A287730, A293160.
Record values are in A212289.
If the 1's are replaced by pairs of 1's we obtain A049456.
Inverse: A020946.
Cf. a(A001045(n)) = A000045(n). a(A062092(n)) = A000032(n+1).
Cf. A064881-A064886 (Stern-Brocot subtrees).
A column of A072170.
Cf. A049455 for the 0,1 version of Stern's diatomic array.
Cf. A000119, A262097 for analogous sequences in other bases and A277189, A277315, A277328 for related sequences with similar graphs.
Cf. A086592 and references therein to other sequences related to Kepler's tree of fractions.

a002487 n = a002487_list !! n
a002487_list = 0 : 1 : stern [1] where
   stern fuscs = fuscs' ++ stern fuscs' where
     fuscs' = interleave fuscs $ zipWith (+) fuscs $ (tail fuscs) ++ [1]
   interleave []     ys = ys
   interleave (x:xs) ys = x : interleave ys xs
-- Reinhard Zumkeller, Aug 23 2011

using Nemo
function A002487List(len)
    a, A = QQ(0), [0,1]
    for n in 1:len
        a = next_calkin_wilf(a)
        push!(A, denominator(a))
    end
A end
A002487List(91) |> println # Peter Luschny, Mar 13 2018

[&+[(Binomial(k, n-k-1) mod 2): k in [0..n]]: n in [0..100]]; // Vincenzo Librandi, Jun 18 2019

A002487 := proc(n) option remember; if n <= 1 then n elif n mod 2 = 0 then procname(n/2); else procname((n-1)/2)+procname((n+1)/2); fi; end: seq(A002487(n),n=0..91);
A002487 := proc(m) local a,b,n; a := 1; b := 0; n := m; while n>0 do if type(n,odd) then b := a+b else a := a+b end if; n := floor(n/2); end do; b; end proc: seq(A002487(n),n=0..91); # Program adapted from E. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232. - Igor Urbiha (urbiha(AT)math.hr), Oct 28 2002. Since A007306(n) = a(2*n+1), this program can be adapted for A007306 by replacing b := 0 by b := 1.
A002487 := proc(n::integer) local k; option remember; if n = 0 then 0 elif n=1 then 1 else add(K(k,n-1-k)*procname(n - k), k = 1 .. n) end if end proc:
K := proc(n::integer, k::integer) local KC; if 0 <= k and k <= n and n-k <= 2 then KC:=1; else KC:= 0; end if; end proc: seq(A002487(n),n=0..91); # Thomas Wieder, Jan 13 2008
# next Maple program:
a:= proc(n) option remember; `if`(n<2, n,
      (q-> a(q)+(n-2*q)*a(n-q))(iquo(n, 2)))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Feb 11 2021
fusc := proc(n) local a, b, c; a := 1; b := 0;
    for c in convert(n, base, 2) do
        if c = 0 then a := a + b else b := a + b fi od;
    b end:
seq(fusc(n), n = 0..91); # Peter Luschny, Nov 09 2022
Stern := proc(n, u) local k, j, b;
    b := j -> nops({seq(Bits:-Xor(k, j-k), k = 0..j)}):
    ifelse(n=0, 1-u, seq(b(j), j = 2^(n-1)-1..2^n-1-u)) end:
seq(print([n], Stern(n, 1)), n = 0..5); # As shown in the comments.
seq(print([n], Stern(n, 0)), n = 0..5); # As shown in the examples. # Peter Luschny, Sep 29 2024

a[0] = 0; a[1] = 1; a[n_] := If[ EvenQ[n], a[n/2], a[(n-1)/2] + a[(n+1)/2]]; Table[ a[n], {n, 0, 100}] (* end of program *)
Onemore[l_] := Transpose[{l, l + RotateLeft[l]}] // Flatten;
NestList[Onemore, {1}, 5] // Flatten  (*gives [a(1), ...]*) (* Takashi Tokita, Mar 09 2003 *)
ToBi[l_] := Table[2^(n - 1), {n, Length[l]}].Reverse[l]; Map[Length,
Split[Sort[Map[ToBi, Table[IntegerDigits[n - 1, 3], {n, 500}]]]]]  (*give [a(1), ...]*) (* Takashi Tokita, Mar 10 2003 *)
A002487[m_] := Module[{a = 1, b = 0, n = m}, While[n > 0, If[OddQ[n], b = a+b, a = a+b]; n = Floor[n/2]]; b]; Table[A002487[n], {n, 0, 100}] (* Jean-François Alcover, Sep 06 2013, translated from 2nd Maple program *)
a[0] = 0; a[1] = 1;
Flatten[Table[{a[2*n] = a[n], a[2*n + 1] = a[n] + a[n + 1]}, {n, 0, 50}]] (* Horst H. Manninger, Jun 09 2021 *)
nmax = 100; CoefficientList[Series[x*Product[(1 + x^(2^k) + x^(2^(k+1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 08 2022 *)

{a(n) = n=abs(n); if( n<2, n>0, a(n\2) + if( n%2, a(n\2 + 1)))};

fusc(n)=local(a=1,b=0);while(n>0,if(bitand(n,1),b+=a,a+=b);n>>=1);b \\ Charles R Greathouse IV, Oct 05 2008

A002487(n,a=1,b=0)=for(i=0,logint(n,2),if(bittest(n,i),b+=a,a+=b));b \\ M. F. Hasler, Feb 12 2017, updated Feb 14 2019

from functools import lru_cache
@lru_cache(maxsize=None)
def a(n): return n if n<2 else a(n//2) if n%2==0 else a((n - 1)//2) + a((n + 1)//2) # Indranil Ghosh, Jun 08 2017; corrected by Reza K Ghazi, Dec 27 2021

def a(n):
    a, b = 1, 0
    while n > 0:
        if n & 1:
            b += a
        else:
            a += b
        n >>= 1
    return b
# Reza K Ghazi, Dec 29 2021

def A002487(n): return sum(int(not (n-k-1) & ~k) for k in range(n)) # Chai Wah Wu, Jun 19 2022

# (fast way for big vectors)
from math import log, ceil
import numpy
how_many_terms = 2**20  # (Powers of 2 recommended but other integers are also possible.)
A002487, A002487[1]  = numpy.zeros(2**(ce:=ceil(log(how_many_terms,2))), dtype=object), 1
for exponent in range(1,ce):
    L, L2 = 2**exponent, 2**(exponent+1)
    A002487[L2 - 1] = exponent + 1
    A002487[L:L2][::2] = A002487[L >> 1: L]
    A002487[L + 1:L2 - 2][::2] = A002487[L:L2 - 3][::2]  +  A002487[L + 2:L2 - 1][::2]
print(list(A002487[0:100])) # Karl-Heinz Hofmann, Jul 22 2025

N <- 50 # arbitrary
a <- 1
for (n in 1:N)
{
  a[2*n    ] = a[n]
  a[2*n + 1] = a[n] + a[n+1]
  a
}
a
# Yosu Yurramendi, Oct 04 2014

# Given n, compute a(n) by taking into account the binary representation of n
a <- function(n){
  b <- as.numeric(intToBits(n))
  l <- sum(b)
  m <- which(b == 1)-1
  d <- 1
  if(l > 1) for(j in 1:(l-1)) d[j] <- m[j+1]-m[j]+1
  f <- c(0,1)
  if(l > 1) for(j in 3:(l+1)) f[j] <- d[j-2]*f[j-1]-f[j-2]
  return(f[l+1])
} # Yosu Yurramendi, Dec 13 2016

# computes the sequence as a vector A, rather than function a() as above.
A <- c(1,1)
maxlevel <- 5 # by choice
for(m in 1:maxlevel) {
  A[2^(m+1)] <- 1
  for(k in 1:(2^m-1)) {
    r <- m - floor(log2(k)) - 1
    A[2^r*(2*k+1)] <- A[2^r*(2*k)] + A[2^r*(2*k+2)]
}}
A # Yosu Yurramendi, May 08 2018

def A002487(n):
    M = [1, 0]
    for b in n.bits():
        M[b] = M[0] + M[1]
    return M[1]
print([A002487(n) for n in (0..91)])
# For a dual see A174980. Peter Luschny, Nov 28 2017

;; An implementation of memoization-macro definec can be found for example in: http://oeis.org/wiki/Memoization
(definec (A002487 n) (cond ((<= n 1) n) ((even? n) (A002487 (/ n 2))) (else (+ (A002487 (/ (- n 1) 2)) (A002487 (/ (+ n 1) 2))))))
;; Antti Karttunen, Nov 05 2016

a(n+1) = (2*k+1)*a(n) - a(n-1) where k = floor(a(n-1)/a(n)). - David S. Newman, Mar 04 2001
Let e(n) = A007814(n) = exponent of highest power of 2 dividing n. Then a(n+1) = (2k+1)*a(n)-a(n-1), n > 0, where k = e(n). Moreover, floor(a(n-1)/a(n)) = e(n), in agreement with D. Newman's formula. - Dragutin Svrtan (dsvrtan(AT)math.hr) and Igor Urbiha (urbiha(AT)math.hr), Jan 10 2002
Calkin and Wilf showed 0.9588 <= limsup a(n)/n^(log(phi)/log(2)) <= 1.1709 where phi is the golden mean. Does this supremum limit = 1? - Benoit Cloitre, Jan 18 2004.  Coons and Tyler show the limit is A246765 = 0.9588... - Kevin Ryde, Jan 09 2021
a(n) = Sum_{k=0..floor((n-1)/2)} (binomial(n-k-1, k) mod 2). - Paul Barry, Sep 13 2004
a(n) = Sum_{k=0..n-1} (binomial(k, n-k-1) mod 2). - Paul Barry, Mar 26 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = v^3 + 2*u*v*w - u^2*w. - Michael Somos, May 02 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = u1^3*u6 - 3*u1^2*u2*u6 + 3*u2^3*u6 - u2^3*u3. - Michael Somos, May 02 2005
G.f.: x * Product_{k>=0} (1 + x^(2^k) + x^(2^(k+1))) [Carlitz].
a(n) = a(n-2) + a(n-1) - 2*(a(n-2) mod a(n-1)). - Mike Stay, Nov 06 2006
A079978(n) = (1 + e^(i*Pi*A002487(n)))/2, i=sqrt(-1). - Paul Barry, Jan 14 2005
a(n) = Sum_{k=1..n} K(k, n-k)*a(n - k), where K(n,k) = 1 if 0 <= k AND k <= n AND n-k <= 2 and K(n,k) = 0 else. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, if we drop the condition k <= n in the above definition, then we arrive at A002083 = Narayana-Zidek-Capell numbers.) - Thomas Wieder, Jan 13 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1; a(2^n - k) + a(k) = a(2^(n+1) + k). Both formulas hold for 0 <= k <= 2^n - 1. G.f.: G(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ... Define f(z) = (1 + z + z^2), then G(z) = lim f(z)*f(z^2)*f(z^4)* ... *f(z^(2^n))*... = (1 + z + z^2)*G(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 11 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1 (0 <= k <= 2^n - 1). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
a(2^n + k) = a(2^n - k) + a(k) (0 <= k <= 2^n). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
Let g(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ..., f(z) = 1 + z + z^2. Then g(z) = lim_{n->infinity} f(z)*f(z^2)*f(z^4)*...*f(z^(2^n)), g(z) = f(z)*g(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
For 0 <= k <= 2^n - 1, write k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1) where b(0), b(1), etc. are 0 or 1. Define a 2 X 2 matrix X(m) with entries x(1,1) = x(2,2) = 1, x(1,2) = 1 - b(m), x(2,1) = b(m). Let P(n)= X(0)*X(1)* ... *X(n-1). The entries of the matrix P are members of the sequence: p(1,1) = a(k+1), p(1,2) = a(2^n - (k+1)), p(2,1) = a(k), p(2,2) = a(2^n - k). - Arie Werksma (werksma(AT)tiscali.nl), Apr 20 2008
Let f(x) = A030101(x); if 2^n + 1 <= x <= 2^(n + 1) and y = 2^(n + 1) - f(x - 1) then a(x) = a(y). - Arie Werksma (Werksma(AT)Tiscali.nl), Jul 11 2008
a(n) = A126606(n + 1) / 2. - Reikku Kulon, Oct 05 2008
Equals infinite convolution product of [1,1,1,0,0,0,0,0,0] aerated A000079 - 1 times, i.e., [1,1,1,0,0,0,0,0,0] * [1,0,1,0,1,0,0,0,0] * [1,0,0,0,1,0,0,0,1]. - Mats Granvik and Gary W. Adamson, Oct 02 2009; corrected by Mats Granvik, Oct 10 2009
a(2^(p+2)*n+2^(p+1)-1) - a(2^(p+1)*n+2^p-1) = A007306(n+1), p >= 0 and n >= 0. - Johannes W. Meijer, Feb 07 2013
a(2*n-1) = A007306(n), n > 0. - Yosu Yurramendi, Jun 23 2014
a(n*2^m) = a(n), m>0, n > 0. - Yosu Yurramendi, Jul 03 2014
a(k+1)*a(2^m+k) - a(k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014
a(2^(m+1)+(k+1))*a(2^m+k) - a(2^(m+1)+k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014
a(5*2^k) = 3. a(7*2^k) = 3. a(9*2^k) = 4. a(11*2^k) = 5. a(13*2^k) = 5. a(15*2^k) = 4. In general: a((2j-1)*2^k) = A007306(j), j > 0, k >= 0 (see Adamchuk's comment). - Yosu Yurramendi, Mar 05 2016
a(2^m+2^m'+k') = a(2^m'+k')*(m-m'+1) - a(k'), m >= 0, m' <= m-1, 0 <= k' < 2^m'. - Yosu Yurramendi, Jul 13 2016
From Yosu Yurramendi, Jul 13 2016: (Start)
Let n be a natural number and [b_m b_(m-1) ... b_1 b_0] its binary expansion with b_m=1.
Let L = Sum_{i=0..m} b_i be the number of binary digits equal to 1 (L >= 1).
Let {m_j: j=1..L} be the set of subindices such that b_m_j = 1, j=1..L, and 0 <= m_1 <= m_2 <= ... <= m_L = m.
If L = 1 then c_1 = 1, otherwise let {c_j: j=1..(L-1)} be the set of coefficients such that c_(j) = m_(j+1) - m_j + 1, 1 <= j <= L-1.
Let f be a function defined on {1..L+1} such that f(1) = 0, f(2) = 1, f(j) = c_(j-2)*f(j-1) - f(j-2), 3 <= j <= L+1.
Then a(n) = f(L+1) (see example). (End)
a(n) = A001222(A260443(n)) = A000120(A277020(n)). Also a(n) = A000120(A101624(n-1)) for n >= 1. - Antti Karttunen, Nov 05 2016
(a(n-1) + a(n+1))/a(n) = A037227(n) for n >= 1. - Peter Bala, Feb 07 2017
a(0) = 0; a(3n) = 2*A000360(3n-1); a(3n+1) = 2*A000360(3n) - 1; a(3n+2) = 2*A000360(3n+1) + 1. - M. Jeremie Lafitte (Levitas), Apr 24 2017
From I. V. Serov, Jun 14 2017: (Start)
a(n) = A287896(n-1) - 1*A288002(n-1) for n > 1;
a(n) = A007306(n-1) - 2*A288002(n-1) for n > 1. (End)
From Yosu Yurramendi, Feb 14 2018: (Start)
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + 2^m + k) = 2*a(k), m >= 0, 0 <= k < 2^m.
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1)       + k) = a(2^m + k), m >= 0, 0 <= k < 2^m.
a(2^m + k) = a(k)*(m - floor(log_2(k)) - 1) + a(2^(floor(log_2(k))+1) + k), m >= 0, 0 < k < 2^m,  a(2^m) = 1, a(0) = 0. (End)
From Yosu Yurramendi, May 08 2018: (Start)
a(2^m) = 1, m >= 0.
a(2^r*(2*k+1)) = a(2^r*(2*k)) + a(2^r*(2*k+2)), r < - m - floor(log_2(k)) - 1, m > 0, 1 <= k < 2^m. (End)
Trow(n) = [card({k XOR (j-k): k=0..j}) for j = 2^(n-1)-1..2^n-2] when regarded as an irregular table (n >= 1). - Peter Luschny, Sep 29 2024
a(n) = A000120(A168081(n)). - Karl-Heinz Hofmann, Jun 16 2025

Additional references and comments from Len Smiley, Joshua Zucker, Rick L. Shepherd and Herbert S. Wilf
Typo in definition corrected by Reinhard Zumkeller, Aug 23 2011
Incorrect formula deleted and text edited by Johannes W. Meijer, Feb 07 2013

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

A020650 Numerators 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, 2, 1, 3, 1, 3, 2, 4, 1, 4, 3, 5, 2, 5, 3, 5, 1, 5, 4, 7, 3, 7, 4, 7, 2, 7, 5, 8, 3, 8, 5, 6, 1, 6, 5, 9, 4, 9, 5, 10, 3, 10, 7, 11, 4, 11, 7, 9, 2, 9, 7, 12, 5, 12, 7, 11, 3, 11, 8, 13, 5, 13, 8, 7, 1, 7, 6, 11, 5, 11, 6, 13, 4, 13, 9, 14, 5, 14, 9, 13, 3, 13, 10, 17, 7, 17, 10, 15, 4, 15, 11, 18, 7, 18

Offset: 1

David W. Wilson

The fractions are given in their reduced form, thus gcd(a(n), A020651(n)) = 1 for all n. - Antti Karttunen, May 26 2004
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,
   2,1,
   3,1,3,2,
   4,1,4,3,5,2,5,3,
   5,1,5,4,7,3,7,4, 7,2, 7,5, 8,3, 8,5,
   6,1,6,5,9,4,9,5,10,3,10,7,11,4,11,7,9,2,9,7,12,5,12,7,11,3,11,8,13,5,13,8,
then the sum of the m-th row is 3^m (m = 0,1,2,), and each column is an arithmetic sequence.
If the rows are written in a right-aligned fashion:
                                                                          1,
                                                                        2,1,
                                                                   3,1, 3,2,
                                                         4,1, 4,3, 5,2, 5,3,
                                      5,1,5,4, 7,3, 7,4, 7,2, 7,5, 8,3, 8,5,
  6,1,6,5,9,4,9,5,10,3,10,7,11,4,11,7,9,2,9,7,12,5,12,7,11,3,11,8,13,5,13,8,
each column k is a Fibonacci sequence. (End)
a(n2^m+1) = a(2n+1), n > 0, m > 0. - Yosu Yurramendi, Jun 04 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) = numerator of r(n).
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) = numerator of gamma_n.
Index entries for fraction trees

Cf. A020651.

Bisection: A086592.

import Data.List (transpose); import Data.Ratio (numerator)
a020650_list = map numerator ks where
   ks = 1 : concat (transpose [map (+ 1) ks, map (recip . (+ 1)) ks])
-- Reinhard Zumkeller, Feb 22 2014

A020650 := n -> `if`((n < 2),n, `if`(type(n,even), A020650(n/2)+A020651(n/2), A020651(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_] := Numerator[f[n]]; Table[a[n], {n, 1, 94}] (* Jean-François Alcover, Nov 22 2011 *)
a[1]=1; a[2]=2; a[3]=1; a[n_] := a[n] = Switch[Mod[n, 4], 0, a[n/2+1] + a[n/2], 1, a[(n-1)/2+1], 2, a[(n-2)/2+1] + a[(n-2)/2], 3, a[(n-3)/2]]; Array[a, 100] (* Jean-François Alcover, Jan 22 2016, after Yosu Yurramendi *)

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

a(1) = 1, a(2n) = a(n) + A020651(n), a(2n+1) = A020651(2n) = A020651(n). - Antti Karttunen, May 26 2004
a(2n) = A020651(2n+1). - Yosu Yurramendi, Jul 17 2014
a((2*n+1)*2^m + 1) = A086592(n), n > 0, m > 0. For n = 0, A086592(0) = 1 is needed. For m = 0, a(2*(n+1)) = A086592(n+1). - Yosu Yurramendi, Feb 19 2017
a(n) = A002487(1+A231551(n)), n > 0. - Yosu Yurramendi, Aug 07 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

A093873 Numerators in Kepler's tree of harmonic fractions.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 3, 1, 3, 2, 3, 1, 4, 3, 4, 2, 5, 3, 5, 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, 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, 6

Offset: 1

N. J. A. Sloane and Reinhard Zumkeller, May 24 2004

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).

			The first few fractions are:
1 1 1 1 2 1 2 1 3 2 3 1 3 2 3 1 4 3 4 2 5 3 5 1 4 3 4 2 5 3 5
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ...
1 2 2 3 3 3 3 4 4 5 5 4 4 5 5 5 5 7 7 7 7 8 8 5 5 7 7 7 7 8 8

R. Zumkeller, Table of n, a(n) for n = 1..10000
Johannes Kepler, Harmonices Mundi, Liber III, see p. 27.
Index entries for fraction trees
Index entries for sequences related to music

The denominators are in A093875. Usually one only considers the left-hand half of the tree, which gives the fractions A020651/A086592. See A086592 for more information, references to Kepler, etc.

See A294442 for another version of Kepler's tree of fractions.

{-# LANGUAGE ViewPatterns #-}
import Data.Ratio((%), numerator, denominator)
rat :: Rational -> (Integer,Integer)
rat r = (numerator r, denominator r)
data Harmony = Harmony Harmony Rational Harmony
rows :: Harmony -> [[Rational]]
rows (Harmony hL r hR) = [r] : zipWith (++) (rows hL) (rows hR)
kepler :: Rational -> Harmony
kepler r = Harmony (kepler (i%(i+j))) r (kepler (j%(i+j)))
           where (rat -> (i,j)) = r
-- Full tree of Kepler's harmonic fractions:
k = rows $ kepler 1 :: [[Rational]] -- as list of lists
h = concat k :: [Rational] -- flattened
a093873 n = numerator $ h !! (n - 1)
a093875 n = denominator $ h !! (n - 1)
a011782 n = numerator $ (map sum k) !! n -- denominator == 1
-- length (k !! n) == 2^n
-- numerator $ (map last k) !! n == fibonacci (n + 1)
-- denominator $ (map last k) !! n == fibonacci (n + 2)
-- numerator $ (map maximum k) !! n == n
-- denominator $ (map maximum k) !! n == n + 1
-- eop.
-- Reinhard Zumkeller, Oct 17 2010

M:= 8: # to get a(1) .. a(2^M-1)
gen[1]:= [1];
for n from 2 to M do
  gen[n]:= map(t -> (numer(t)/(numer(t)+denom(t)),
      denom(t)/(numer(t)+denom(t))), gen[n-1]);
od:
seq(op(map(numer,gen[i])),i=1..M): # Robert Israel, Jan 11 2016

num[1] = num[2] = 1; den[1] = 1; den[2] = 2; num[n_?EvenQ] := num[n] = num[n/2]; den[n_?EvenQ] := den[n] = num[n/2] + den[n/2]; num[n_?OddQ] := num[n] = den[(n-1)/2]; den[n_?OddQ] := den[n] = num[(n-1)/2] + den[(n-1)/2]; A093873 = Table[num[n], {n, 1, 97}] (* Jean-François Alcover, Dec 16 2011 *)

a(n) = a([n/2])*(1 - n mod 2) + A093875([n/2])*(n mod 2).
a(A029744(n-1)) = 1; a(A070875(n-1)) = 2; a(A123760(n-1)) = 3. - Reinhard Zumkeller, Oct 13 2006
A011782(k) = SUM(a(n)/A093875(n): 2^k<=n<2^(k+1)), k>=0. [Reinhard Zumkeller, Oct 17 2010]
a(1) =  1. For all n>0  a(2n) =  a(n), a(2n+1) =  A093875(n). - Yosu Yurramendi, Jan 09 2016
a(4n+3) = a(4n+1), a(4n+2) = a(4n+1) - a(4n), a(4n+1) = A071585(n). - Yosu Yurramendi, Jan 11 2016
G.f. G(x) satisfies G(x) = x + (1+x) G(x^2) + Sum_{k>=2} x (1+x^(2^(k-1))) G(x^(2^k)). - Robert Israel, Jan 11 2016
a(2^(m+1)+k) = a(2^(m+1)+2^m+k) = A020651(2^m+k), m>=0, 0<=k<2^m. - Yosu Yurramendi, May 18 2016
a(k) = A020651(2^(m+1)+k) - A020651(2^m+k), m>0, 0Yosu Yurramendi, Jun 01 2016
a(2^(m+1)+k) - a(2^m+k) = a(k) , m >=0, 0 <= k < 2^m. For k=0 a(0)=0 is needed. - Yosu Yurramendi, Jul 22 2016
a(2^(m+2)-1-k) = a(2^(m+1)-1-k) + a(2^m-1-k), m >= 1, 0 <= k < 2^m. - Yosu Yurramendi, Jul 22 2016
a(2^m-1-(2^r -1)) = A000045(m-r), m >= 1, 0 <= r <= m-1. - Yosu Yurramendi, Jul 22 2016
a(2^m+2^r) = m-r, , m >= 1, 0 <= r <= m-1 ; a(2^m+2^r+2^(r-1)) = m-(r-1), m >= 2, 0 <= r <= m-1. - Yosu Yurramendi, Jul 22 2016
A093875(2n) - a(2n) = A093875(n), n > 0; A093875(2n+1) - a(2n+1) = a(n), n > 0. - Yosu Yurramendi, Jul 23 2016

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 *)

A093875 Denominators in Kepler's tree of harmonic fractions.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane and Reinhard Zumkeller, May 24 2004

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).

It appears that A071585 is a bisection of this sequence, which itself is a bisection of A093873. - Yosu Yurramendi, Jan 09 2016

			The first few fractions are:
1 1 1 1 2 1 2 1 3 2 3 1 3 2 3 1 4 3 4 2 5 3 5 1 4 3 4 2 5 3 5
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ...
1 2 2 3 3 3 3 4 4 5 5 4 4 5 5 5 5 7 7 7 7 8 8 5 5 7 7 7 7 8 8

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for fraction trees

The numerators are in A093873. Usually one only considers the left-hand half of the tree, which gives the fractions A020651/A086592. See A086592 for more information, references to Kepler, etc.

Cf. A071585, A093873

num[1] = num[2] = 1; den[1] = 1; den[2] = 2; num[n_?EvenQ] := num[n] = num[n/2]; den[n_?EvenQ] := den[n] = num[n/2] + den[n/2]; num[n_?OddQ] := num[n] = den[(n-1)/2]; den[n_?OddQ] := den[n] = num[(n-1)/2] + den[(n-1)/2]; A093875 = Table[den[n], {n, 1, 83}] (* Jean-François Alcover, Dec 16 2011 *)

a(n) = a([n/2]) + A093873([n/2]).
Conjecture of the comment in detail: a(2n+1) = a(2n), n > 0;  a(2n+1) = A071585(n), n >= 0; a(2n) = A071585(n), n > 0. - Yosu Yurramendi, Jun 22 2016
a(2n) - A093873(2n) = a(n), n > 0; a(2n+1) - A093873(2n+1) = A093873(n), n > 0. - Yosu Yurramendi, Jul 23 2016
From Yosu Yurramendi, Jul 25 2016: (Start)
a(2^m)  = m+1, m >= 0; a(2^m + 2) = 2m - 1, m >= 1; a(2^m - 1) = A000045(m+2), m >= 1.
a(2^(m+1) + k) - a(2^m + k) = a(k),   m > 0, 0 <= k < 2^m. For k=0, a(0) = 1 is needed.
a(2^(m+2) - k - 1) = a(2^(m+1) - k - 1) + a(2^m - k - 1), m >= 0, 0 <= k < 2^m. (End)

A268087 a(n) = A162909(n) + A162910(n).

Original entry on oeis.org

2, 3, 3, 5, 4, 4, 5, 8, 7, 5, 7, 7, 5, 7, 8, 13, 11, 9, 12, 9, 6, 10, 11, 11, 10, 6, 9, 12, 9, 11, 13, 21, 18, 14, 19, 16, 11, 17, 19, 14, 13, 7, 11, 17, 13, 15, 18, 18, 15, 13, 17, 11, 7, 13, 14, 19, 17, 11, 16, 19, 14, 18, 21, 34, 29, 23, 31, 25, 17, 27, 30, 25, 23, 13, 20, 29, 22, 26, 31, 23, 19, 17, 22, 13, 8, 16, 17, 27

Offset: 1

Yosu Yurramendi, Jan 26 2016

If the terms (n>0) are written as an array (in a left-aligned fashion) with rows of length 2^m, m >= 0:
2,
3, 3,
5, 4, 4, 5,
8, 7, 5, 7, 7, 5, 7, 8,
13,11, 9,12, 9, 6,10,11,11,10,6, 9,12, 9,11,13,
21,18,14,19,16,11,17,19,14,13,7,11,17,13,15,18,18,15,13,17,11,7,13,14,19,17,11,16, ...
a(n) is palindromic in each level m >= 0 (ranks between 2^m and 2^(m+1)-1), because in each level m >= 0 A162910 is the reverse of A162909:
a(2^m + k) = a(2^(m+1) - 1 - k), m >= 0, 0 <= k < 2^m.
All columns have the Fibonacci sequence property: a(2^(m+2) + k) = a(2^(m+1) + k) + a(2^m + k), m >= 0, 0 <= k < 2^m (empirical observations).
a(2^m + k) = A162909(2^(m+2) + k), a(2^m + k) = A162909(2^(m+1)+ 2^m + k), a(2^m + k) = A162910(2^(m+1) + k), m >= 0, 0 <= k < 2^m (empirical observations).
a(n) = A162911(n) + A162912(n), where A162911(n)/A162912(n) is the bit reversal permutation of A162909(n)/A162910(n) in each level m >= 0 (empirical observations).
a(n) = A162911(2n+1), a(n) = A162912(2n) for n > 0 (empirical observations). n > 1 occurs in this sequence phi(n) = A000010(n) times, as it occurs in A007306 (Franklin T. Adams-Watters's comment), which is the sequence obtained by adding numerator and denominator in the Calkin-Wilf enumeration system of positive rationals. A162909(n)/A162910(n) is also an enumeration system of all positive rationals (Bird 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 A086592 (A020650+A020651).

			m = 3, k = 6: a(38) = 17, a(22) = 10, a(14) = 7.

Cf. A162909, A162910, A162911, A162912, A007306, A071585, A086592.

a(n) = my(x=1, y=1); for(i=0, logint(n, 2), if(bittest(n, i), [x, y]=[x+y, x], [x, y]=[y, x+y])); x \\ Mikhail Kurkov, Mar 10 2023

a(2^(m+2)+k) = a(2^(m+1)+k) + a(2^m+k) with m = 0, 1, 2, ... and 0 <= k < 2^m (empirical observation).
a(A059893(n)) = a(n) for n > 0. - Yosu Yurramendi, May 30 2017
From Yosu Yurramendi, May 14 2019: (Start)
Take the smallest m > 0 such that 0 <= k < 2^(m-1), and choose any M >= m,
a((1/3)*(  A016921(2^(m-1)+k)*4^(M-m)-1)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k  ).
a((1/3)*(2*A016921(2^(m-1)+k)*4^(M-m)-2)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k  ) + a(2^(m-1)+k).
a((1/3)*(  A016969(2^(m-1)+k)*4^(M-m)-2)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k+1).
a((1/3)*(2*A016969(2^(m-1)+k)*4^(M-m)-1)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k+1) + a(2^(m-1)+k). (End)
a(n) = A007306(A258996(n)), n > 0. - Yosu Yurramendi, Jun 23 2021

A273493 a(n) = A245327(n) + A245328(n).

Original entry on oeis.org

2, 3, 3, 5, 5, 4, 4, 8, 8, 7, 7, 7, 7, 5, 5, 13, 13, 11, 11, 12, 12, 9, 9, 11, 11, 10, 10, 9, 9, 6, 6, 21, 21, 18, 18, 19, 19, 14, 14, 19, 19, 17, 17, 16, 16, 11, 11, 18, 18, 15, 15, 17, 17, 13, 13, 14, 14, 13, 13, 11, 11, 7, 7, 34, 34, 29, 29, 31, 31, 23, 23, 30, 30, 27, 27, 25, 25, 17, 17, 31, 31, 26, 26, 29, 29, 22, 22, 25

Offset: 1

Yosu Yurramendi, May 23 2016

nonn

The terms (n>0) may be written as a left-justified array with rows of length 2^m, m >= 0:
2,
3, 3,
5, 5, 4, 4,
8, 8, 7, 7, 7, 7, 5, 5,
13,13,11,11,12,12, 9, 9,11,11,10,10, 9, 9, 6, 6,
21,21,18,18,19,19,14,14,19,19,17,17,16,16,11,11,18,18,15,15,17,17,13,13,14,14,...
All columns have the Fibonacci sequence property: a(2^(m+2) + k) = a(2^(m+1) + k) + a(2^m + k), m >= 0,  0 <= k < 2^m (empirical observations).
The terms (n>0) may also be written as a right-justified array with rows of length 2^m, m >= 0:
                                                                              2,
                                                                           3, 3,
                                                                     5, 5, 4, 4,
                                                         8, 8, 7, 7, 7, 7, 5, 5,
                                13,13,11,11,12,12, 9, 9,11,11,10,10, 9, 9, 6, 6,
    ...,19,19,17,17,16,16,11,11,18,18,15,15,17,17,13,13,14,14,13,13,11,11, 7, 7,
Each column is an arithmetic sequence. The differences of the arithmetic sequences repeat the sequence A071585: a(2^(m+2) -1 - 2k) - a(2^(m+1) -1 - 2k) = A071585(k-1),    m > 0,  0 <= k < 2^m ; a(2^(m+2) -1 - 2k - 1) - a(2^(m+1) -1 - 2k - 1) = A071585(k-1), m > 0,  0 <= k < 2^m .
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. A245327(n)/A245328(n) is also an enumeration system of all positive rationals, 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.
a(n) = A273494(A059893(n)), a(A059893(n)) = A273494(n), n > 0. - Yosu Yurramendi, May 30 2017

Yosu Yurramendi, Table of n, a(n) for n = 1..4099

Cf. A007306, A268087, A071585, A086592, A273494

b(n) = my(b=binary(n)); fromdigits(concat(b[1], Vecrev(vector(#b-1, k, b[k+1]))), 2); \\ from A059893
a(n) = my(n=b(n), x=1, y=1); for(i=0, logint(n, 2), if(bittest(n, i), [x, y]=[x+y, y], [x, y]=[y, x+y])); x \\ Mikhail Kurkov, Mar 11 2023

a(n) = A007306(A284459(n)), n > 0. - Yosu Yurramendi, Aug 23 2021

cp's OEIS Frontend

A086593 Bisection of A086592, denominators of the left-hand half of Kepler's tree of fractions.

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A002487 Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1).

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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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A020650 Numerators 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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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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A002487 Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2n) = a(n), a(2n+1) = a(n) + a(n+1).