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A112310
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Number of terms in lazy Fibonacci representation of n.
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25
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0, 1, 1, 2, 2, 2, 3, 2, 3, 3, 3, 4, 3, 3, 4, 3, 4, 4, 4, 5, 3, 4, 4, 4, 5, 4, 4, 5, 4, 5, 5, 5, 6, 4, 4, 5, 4, 5, 5, 5, 6, 4, 5, 5, 5, 6, 5, 5, 6, 5, 6, 6, 6, 7, 4, 5, 5, 5, 6, 5, 5, 6, 5, 6, 6, 6, 7, 5, 5, 6, 5, 6, 6, 6, 7, 5, 6, 6, 6, 7, 6, 6, 7, 6, 7, 7, 7, 8, 5, 5, 6, 5, 6, 6, 6, 7, 5, 6, 6, 6, 7, 6, 6, 7, 6
(list;
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listen;
history;
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internal format)
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OFFSET
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0,4
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COMMENTS
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Equivalently, the number of ones in the maximal Fibonacci bit-representation (A104326) of n.
Conjecture: if we split the sequence in groups that contain Fibonacci(k) terms like (0), (1), (1, 2), (2, 2, 3), (2, 3, 3, 3, 4), (3, 3, 4, 3, 4, 4, 4, 5) etc, the sums in the groups are the terms of A023610. - Gary W. Adamson, Nov 02 2010
Equivalently, the number of periods in the length-n prefix of the infinite Fibonacci word (A003849). An integer p, 1 <= p <= n, is a period of a length-n word x if x[i] = x[i+p] for 1 <= i <= n-p. - Jeffrey Shallit, May 23 2020
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LINKS
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FORMULA
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EXAMPLE
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a(10) = 3 because A104326(10) = 1110 contains three ones.
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MATHEMATICA
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DeleteCases[IntegerDigits[Range[200], 2], {___, 0, 0, ___}]
A112309 = Map[DeleteCases[Reverse[#] Fibonacci[Range[Length[#]] + 1], 0] &, DeleteCases[IntegerDigits[-1 + Range[200], 2], {___, 0, 0, ___}]]
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PROG
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(Haskell)
a112310 n = a112310_list !! n
a112310_list = concat fss where
fss = [0] : [1] : (map (map (+ 1))) (zipWith (++) fss $ tail fss)
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CROSSREFS
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Number of terms in row n of A112309.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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