A288003 R-fusc, sequence r of the mutual diatomic recurrence pair: l(1)=0, r(1)=1, l(2n) = l(n), r(2n) = r(n), l(2n+1) = l(n)+r(n), r(2n+1) = l(n+1)+r(n+1), where l(n) = A288002(n).

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

Offset

1,5

Comments

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))).

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

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Formula

a(n) = A002487(n) - A288002(n). [r-fusc(n) = fusc(n) - l-fusc(n).]

gcd(a(n),A288002(n)) = gcd(a(n),A002487(n)) = 1.

Example

The odd bisection CHF(n) of the chf(n) sequence shifted rightwards by A288002(n) determines the longest overlap of the words CHF(n) and CHF(n+1). Note that the first overlapping letters differ for n == 2^k or equivalently when A288002(n)==0.
To construct the word CHF(n+1) from the word CHF(n): cut off the word negate(lef(n)) of length A288002(n) at the left side of CHF(n), add the word negate(rig(n)) of length a(n) at the right side of CHF(n) and negate the first letter of the new word iff A288002(n)==0.
n chf(n)  A070939(n) A002487(n) rig(n) a(n)       CHF(n)
                     fusc(n)           r-fusc(n)  bisection of chf(n)
1  '-'     1          1         '+'     1        '-'
2  '+'     2          1         '-'     1        '+-'
3  '+-'    2          2         '-'     1        '--+'
4  '-'     3          1         '+'     1         '-++'
5  '--+'   3          3        '-+'     2         '+++-'
6  '-+'    3          2         '+'     1          '++-+-'
7  '-++'   3          3         '+'     1           '+-+--'
8  '+'     4          1         '-'     1             '+---'
9  '+++-'  4          4       '++-'     3             '----+'
10 '++-'   4          3        '+-'     2              '---+--+'
11 '++-+-' 4          5        '+-'     2               '--+--+-+'
12 '+-'    4          2         '-'     1                  '--+-+-+'
13 '+-+--' 4          5       '+--'     3                   '-+-+-++'
14 '+--'   4          3         '-'     1                     '-+-++-++'
15 '+---'  4          4         '-'     1                       '-++-+++'
16 '-'     5          1         '+'     1                          '-++++'
17 '----+' 5          5      '---+'     4                           '+++++-'

Prog

(Python)
def l(n): return 0 if n==1 else l(n//2) if n%2==0 else l((n - 1)//2) + r((n - 1)//2)
def r(n): return 1 if n==1 else r(n//2) if n%2==0 else l((n + 1)//2) + r((n + 1)//2)
print([r(n) for n in range(1, 151)]) # Indranil Ghosh, Jun 11 2017
(PARI) l(n)=if(n%2, if(n==1, 0, l(n\2) + a(n\2)), l(n/2))
a(n)=if(n%2, if(n==1, 1, l(n\2+1) + a(n\2+1)), a(n/2)) \\ Charles R Greathouse IV, Jun 11 2017

Crossrefs

Cf. A002487, A070939, A287729, A287730, A288002.

Sequence in context: A107249 A062842 A126805 * A304382 A304717 A110963

Adjacent sequences: A288000 A288001 A288002 * A288004 A288005 A288006

Keyword

nonn

Author

I. V. Serov, Jun 10 2017

Status

approved