|
%I #20 Feb 27 2017 06:13:51
%S 1,2,2,3,3,5,5,5,8,8,8,8,8,13,13,13,13,13,13,13,13,21,21,21,21,21,21,
%T 21,21,21,21,21,21,21,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,
%U 34,34,34,34,34,55,55,55,55,55,55,55,55,55,55,55,55,55,55,55,55,55,55,55
%N a(1)=1, a(2)=2, then use the rule when a(n) is the end of a run, n appears a(n) times.
%C All Fibonacci numbers >=1 occur. For k>=4, the k-th Fibonacci number occurs F(k-1) times. Sequence n-a(n) consists of (0,0) union successive runs 1,2,...,F(k) for k>=1.
%C Beginning with a(3), this is the index sequence of the block-fractal sequence A003849; see A280511 for definitions. - _Clark Kimberling_, Jan 06 2017
%F (n-1)/tau < a(n) < n where tau is the golden ratio; k>=3 a(F(k))=F(k-1) where F(k) is the k-th Fibonacci number.
%e Sequence begins 1,2,2: a(3)=2 is the end of the second run, hence 3 will appear twice and sequence continues: 1,2,2,3,3. Now a(5)=3 is the end of the third run, hence 5 appears 3 times and sequence continues: 1,2,2,3,3,5,5,5. - _Labos Elemer_
%e From _Clark Kimberling_, Jan 06 2017: (Start)
%e Connection of this sequence to the infinite Fibonacci word A003849 (see Comments):
%e A003849 = (0,1,0,0,1,0,1,0,0,1,0,0,1,...) = (s(0), s(1), ... ).
%e (initial block #1) = (0) first repeats at s(2), so that a(3) = 2;
%e (initial block #2) = (0,1) first repeats at s(3), so that a(4) = 3;
%e (initial block #3) = (0,1,0) first repeats at s(3), so that a(5) = 3. (End)
%K nonn
%O 1,2
%A _Benoit Cloitre_, Apr 30 2003
|
