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

Author

I. V. Serov, Jun 10 2017

Keywords

• nonn

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.

Examples

			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                           '+++++-'
		

Links

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

Crossrefs

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

Programs

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

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

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.