

A082524


a(1)=1, a(2)=2, then use the rule when a(n) is the end of a run, n appears a(n) times.


0



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, 21, 21, 21, 21, 21, 21, 21, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55
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OFFSET

1,2


COMMENTS

All Fibonacci numbers >=1 occur. For k>=4, the kth Fibonacci number occurs F(k1) times. Sequence na(n) consists of (0,0) union successive runs 1,2,...,F(k) for k>=1.
Beginning with a(3), this is the index sequence of the blockfractal sequence A003849; see A280511 for definitions.  Clark Kimberling, Jan 06 2017


LINKS

Table of n, a(n) for n=1..74.


FORMULA

(n1)/tau < a(n) < n where tau is the golden ratio; k>=3 a(F(k))=F(k1) where F(k) is the kth Fibonacci number.


EXAMPLE

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
From Clark Kimberling, Jan 06 2017: (Start)
Connection of this sequence to the infinite Fibonacci word A003849 (see Comments):
A003849 = (0,1,0,0,1,0,1,0,0,1,0,0,1,...) = (s(0), s(1), ... ).
(initial block #1) = (0) first repeats at s(2), so that a(3) = 2;
(initial block #2) = (0,1) first repeats at s(3), so that a(4) = 3;
(initial block #3) = (0,1,0) first repeats at s(3), so that a(5) = 3. (End)


CROSSREFS

Sequence in context: A045767 A108221 A169682 * A099961 A286107 A285735
Adjacent sequences: A082521 A082522 A082523 * A082525 A082526 A082527


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Apr 30 2003


STATUS

approved



