A287823 a(n) = A287729(n)*A001511(n).

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

Offset: 1

Author

I. V. Serov, Jun 03 2017

Keywords

• nonn
• frac

Comments

For n > 0, define the sequence chf(n) of Christoffel words over an alphabet {-,+} by recursion with chf(1) = '-' and chf(2*n+0) = negate(chf(n)), chf(2*n+1) = negate(concatenate(chf(n),chf(n+1))).

The chf(n) word has length fusc(n) = A002487(n) and splits uniquely into two parent Christoffel words - the left Christoffel word lef(n) of length l-fusc(n) = A288002(n) and the right Christoffel word rig(n) of length r-fusc(n) = A288003(n).

For n > 0, define the lap(n) word to be the chf(n) word taken to the power A001511(n), so that the factor-word chf(n) repeats in lap(n) A001511(n) times.

Consider the odd bisection CHF(n) of the chf(n) sequence. Each CHF(n) word has length A007306(n). Then the sequence lap(n) is obtained by cutting off the word lef(n) of length A288002(n) at the left side of negate(CHF(n)) and negating the first letter of the new word iff lef(n) is empty, i.e., iff A288002(n)==0.

The length of the lap(n) word is A287896(n), the number of '-'-signs in the lap(n) word is a(n) and the number of '+'-signs in the lap(n) word is A287824(n).

There are exactly n words of length n in the sequence lap(n). These n words lap(j) are all different with 1<=a(j)<=n for 1<=j<=n.

a(n)/A287824(n) enumerates all fractions along the tree of fractions.

a(n)/A287896(n) enumerates all proper fractions along the tree of proper fractions.

See the Serov links and the example below.

Examples

			n  chf(n)  A070939(n) A001511(n) A002487(n) lef(n) CHF(n)     lap(n)   a(n)
1  '-'     1          1          1          ''     '-'        '-'      1
2  '+'     2          2          1          ''     '+-'       '++'     0
3  '+-'    2          2          2          '+'    '--+'      '+-'     1
4  '-'     3          1          1          ''     '-++'      '---'    3
5  '--+'   3          3          3          '-'    '+++-'     '--+'    2
6  '-+'    3          1          2          '-'    '++-+-'    '-+-+'   2
7  '-++'   3          1          3          '-+'   '+-+--'    '-++'    1
8  '+'     4          1          1           ''    '+---'     '++++'   0
9  '+++-'  4          4          4          '+'    '----+'    '+++-'   1
10 '++-'   4          1          3          '+'    '---+--+'  '++-++-' 2
11 '++-+-' 4          1          5          '++-'  '--+--+-+' '++-+-'  2
12 '+-'    4          1          2          '+'    '--+-+-+'  '+-+-+-' 3
13 '+-+--' 4          2          5          '+-'   '-+-+-++'  '+-+--'  3
14 '+--'   4          1          3          '+-'   '-+-++-++' '+--+--' 4
15 '+---'  4          1          4          '+--'  '-++-+++'  '+---'   3
16 '-'     5          1          1          ''     '-++++'    '-----'  5
17 '----+' 5          5          5          '-'    '+++++-'   '----+'  4 .
		

Links

• I. V. Serov, Table of n, a(n) for n = 1..8192
• J. Berstel, A. Lauve, C. Reutenauer & F. Saliola, Combinatorics on Words: Christoffel Words and Repetitions in Words, 2008.
• I. V. Serov, OEIS: The tree of all fractions as sequence a(n)/A287824(n)
• I. V. Serov, OEIS: The tree of all proper fractions as sequence a(n)/A287896
• I. V. Serov, CHF Diatomic Algorithm Benchmark

Crossrefs

Cf. A001511, A287729, A287824, A002487, A070939, A287896, A154269.

Programs

Python

def c(n): return 1 if n==1 else s(n/2) if n%2==0 else s((n - 1)/2) + s((n + 1)/2)
def s(n): return 0 if n==1 else c(n/2) if n%2==0 else c((n - 1)/2) + c((n + 1)/2)
def a001511(n): return bin(n)[2:][::-1].index("1") + 1
def a(n): return c(n)*a001511(n) # Indranil Ghosh, Jun 08 2017

Formula

a(n) = A287896(n) - A287824(n).

a(n) = (A287729(n-1) + A287729(n) + A287729(n+1))/2 + A154269(n).

gcd(a(n), A287896(n)) == gcd(a(n), A287824(n)) == A001511(n).

From Yosu Yurramendi, Apr 09 2019: (Start)

For m >= 0, m even, 0 <= k < 2^m, a(2^m+k) = A287896(2^m-k).

For m >= 0, m odd, 0 <= k < 2^m, a(2^m+k) = A287896(k). A287896(0) = 0 is needed. (End)