A047679 Denominators in full Stern-Brocot tree.

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

Offset: 0

Author

Wouter Meeussen

Keywords

• nonn
• easy
• frac
• nice
• tabf
• look

Comments

Numerators are A007305.

Write n in binary; list run lengths; add 1 to last run length; make into continued fraction. Sequence gives denominator of fraction obtained.

From Reinhard Zumkeller, Dec 22 2008: (Start)

For n > 1: a(n) = if A025480(n-1) != 0 and A025480(n) != 0 then = a(A025480(n-1)) + a(A025480(n)) else if A025480(n)=0 then a(A025480(n-1))+0 else 1+a(A025480(n-1));

a(n) = A007305(A054429(n)+2) and a(A054429(n)) = A007305(n+2);

A153036(n+1) = floor(A007305(n+2)/a(n)). (End)

From Yosu Yurramendi, Jun 25 2014 and Jun 30 2014: (Start)

If the terms are written as an array a(m, k) = a(2^(m-1)-1+k) with m >= 1 and k = 0, 1, ..., 2^(m-1)-1:

1,

2,1,

3,3, 2, 1,

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

5,7, 8, 7, 7, 8, 7,5,4, 5, 5, 4, 3, 3,2,1,

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

then the sum of the m-th row is 3^(m-1), and each column is an arithmetic sequence. The differences of these arithmetic sequences give the sequence A007306(k+1). The first terms of columns are 1 for k = 0 and a(k-1) for k >= 1.

In a row reversed version A(m, k) = a(m, m-(k+1)):

1

1,2

1,2,3,3,

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

1,2,3,3,4,5,5,4,5,7,8,7,7,8,7,5

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

each column k >= 0 is constant, namely A007306(k+1).

This row reversed version coincides with the array for A007305 (see the Jun 25 2014 comment there). (End)

Looking at the plot, the sequence clearly shows a fractal structure. (The repeating pattern oddly resembles the [first completed] facade of the Sagrada Familia!) - Daniel Forgues, Nov 15 2019

Examples

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

Links

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

Crossrefs

Cf. A007305, A007306, A054424, A152976.

Programs

Mathematica

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

PARI

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

R

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

Formula

a(n) = SternBrocotTreeDen(n) # n starting from 1.

From Yosu Yurramendi, Jul 02 2014: (Start)

For m >0 and 0 <= k < 2^(m-1), with a(0)=1, a(1)=2:

a(2^m+k-1) = a(2^(m-1)+k-1) + a((2^m-1)-k-1);

a(2^m+2^(m-1)+k-1) = a(2^(m-1)+k-1). (End)

a(2^m-2^q ) = q+1, q >= 0, m > q

a(2^m-2^q-1) = q+2, q >= 0, m > q+1. - Yosu Yurramendi, Jan 01 2015

a(2^(m+1)-1-k) = A007306(k+1), m >= 0, 0 <= k <= 2^m. - Yosu Yurramendi, May 20 2019

a(n) = A002487(1+A059893(n)), n > 0. - Yosu Yurramendi, Sep 29 2021

Extensions

Edited by Wolfdieter Lang, Mar 31 2015