This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A288003 #44 Apr 28 2021 16:27:30
%S A288003 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,
%T A288003 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,
%U A288003 9,4,4,3,11,7,7,3,10,3,3,2,11,8,8,5,13,5
%N 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).
%C A288003 Define a sequence chf(n) of Christoffel words over an alphabet {-,+}:
%C A288003 chf(1) = '-',
%C A288003 chf(2*n+0) = negate(chf(n)),
%C A288003 chf(2*n+1) = negate(concatenate(chf(n),chf(n+1))).
%C A288003 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.
%H A288003 Charles R Greathouse IV, <a href="/A288003/b288003.txt">Table of n, a(n) for n = 1..10000</a>
%F A288003 a(n) = A002487(n) - A288002(n). [r-fusc(n) = fusc(n) - l-fusc(n).]
%F A288003 gcd(a(n),A288002(n)) = gcd(a(n),A002487(n)) = 1.
%e A288003 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.
%e A288003 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.
%e A288003 n chf(n) A070939(n) A002487(n) rig(n) a(n) CHF(n)
%e A288003 fusc(n) r-fusc(n) bisection of chf(n)
%e A288003 1 '-' 1 1 '+' 1 '-'
%e A288003 2 '+' 2 1 '-' 1 '+-'
%e A288003 3 '+-' 2 2 '-' 1 '--+'
%e A288003 4 '-' 3 1 '+' 1 '-++'
%e A288003 5 '--+' 3 3 '-+' 2 '+++-'
%e A288003 6 '-+' 3 2 '+' 1 '++-+-'
%e A288003 7 '-++' 3 3 '+' 1 '+-+--'
%e A288003 8 '+' 4 1 '-' 1 '+---'
%e A288003 9 '+++-' 4 4 '++-' 3 '----+'
%e A288003 10 '++-' 4 3 '+-' 2 '---+--+'
%e A288003 11 '++-+-' 4 5 '+-' 2 '--+--+-+'
%e A288003 12 '+-' 4 2 '-' 1 '--+-+-+'
%e A288003 13 '+-+--' 4 5 '+--' 3 '-+-+-++'
%e A288003 14 '+--' 4 3 '-' 1 '-+-++-++'
%e A288003 15 '+---' 4 4 '-' 1 '-++-+++'
%e A288003 16 '-' 5 1 '+' 1 '-++++'
%e A288003 17 '----+' 5 5 '---+' 4 '+++++-'
%o A288003 (Python)
%o A288003 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)
%o A288003 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)
%o A288003 print([r(n) for n in range(1, 151)]) # _Indranil Ghosh_, Jun 11 2017
%o A288003 (PARI) l(n)=if(n%2, if(n==1, 0, l(n\2) + a(n\2)), l(n/2))
%o A288003 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
%Y A288003 Cf. A002487, A070939, A287729, A287730, A288002.
%K A288003 nonn
%O A288003 1,5
%A A288003 _I. V. Serov_, Jun 10 2017