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A287823 a(n) = A287729(n)*A001511(n). 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
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.
LINKS
J. Berstel, A. Lauve, C. Reutenauer & F. Saliola, Combinatorics on Words: Christoffel Words and Repetitions in Words, 2008.
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)
EXAMPLE
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 .
PROG
(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
CROSSREFS
Sequence in context: A155917 A291770 A343229 * A143378 A131961 A276426
KEYWORD
nonn,frac
AUTHOR
I. V. Serov, Jun 03 2017
STATUS
approved

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Last modified March 28 18:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)