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A082524
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a(1)=1, a(2)=2, then use the rule when a(n) is the end of a run, n appears a(n) times.
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0
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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
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1,2
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COMMENTS
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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.
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LINKS
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FORMULA
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(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.
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EXAMPLE
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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
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)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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