

A112310


Number of terms in lazy Fibonacci representation of n.


24



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
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OFFSET

0,4


COMMENTS

Equivalently, the number of ones in the maximal Fibonacci bitrepresentation (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 lengthn prefix of the infinite Fibonacci word (A003849). An integer p, 1 <= p <= n, is a period of a lengthn word x if x[i] = x[i+p] for 1 <= i <= np.  Jeffrey Shallit, May 23 2020


LINKS



FORMULA



EXAMPLE

a(10) = 3 because A104326(10) = 1110 contains three ones.


MATHEMATICA

DeleteCases[IntegerDigits[Range[200], 2], {___, 0, 0, ___}]
A112309 = Map[DeleteCases[Reverse[#] Fibonacci[Range[Length[#]] + 1], 0] &, DeleteCases[IntegerDigits[1 + Range[200], 2], {___, 0, 0, ___}]]


PROG

(Haskell)
a112310 n = a112310_list !! n
a112310_list = concat fss where
fss = [0] : [1] : (map (map (+ 1))) (zipWith (++) fss $ tail fss)


CROSSREFS

Number of terms in row n of A112309.


KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



