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A268696
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a(1)=1; thereafter a(n+1) = floor(m/2), where m = number of occurrences of a(n) in [a(1),...,a(n)].
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3
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1, 0, 0, 1, 1, 1, 2, 0, 1, 2, 1, 3, 0, 2, 1, 3, 1, 4, 0, 2, 2, 2, 3, 1, 4, 1, 5, 0, 3, 2, 3, 2, 4, 1, 5, 1, 6, 0, 3, 3, 3, 4, 2, 4, 2, 5, 1, 6, 1, 7, 0, 4, 3, 4, 3, 5, 2, 5, 2, 6, 1, 7, 1, 8, 0, 4, 4, 4, 5, 3, 5, 3, 6, 2, 6, 2, 7, 1, 8, 1, 9, 0, 5, 4, 5, 4, 6, 3, 6, 3, 7, 2, 7, 2, 8, 1, 9, 1, 10, 0
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OFFSET
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1,7
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COMMENTS
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It appears that a(n) <= ceiling(sqrt(n)).
In fact it appears that a(n) <= floor(sqrt(n)) except when n belongs to the sequence S := [99, 120, 142, 167, 193, 222, 252, 285, 319, ...], which has second differences 1,3,1,3,1,3,... and is the sequence {99; A035608(k)+21*k+120, k>=0}. For these values of n it appears that a(n) = ceiling(sqrt(n)). The first example is a(99) = 10 = ceiling(sqrt(99)).
The zeros occur at positions [2, 3, 8, 13, 19, 28, 38, 51, 65, 82, 100, 121, 143, 168, 194, 223, 253, 286, 320, ...], which apart from the initial terms appears to be S+1.
Without the division by 2 in the definition (that is, if a(n+1)=m), we get A158416. (End)
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LINKS
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EXAMPLE
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a(2) is equal to the number of times a(1) = 1 appears in the sequence before, divided by two, rounding down. Since 1 appears once before, a(2) = floor(1/2) = 0.
a(3) is equal to the number of times 0 appears in the sequence before, which is again once, divided by two, rounding down. So a(3) = floor(1/2) = 0.
a(4) is the number of times 0 appears before (twice) divided by two, which gives us 1.
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MATHEMATICA
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a = {1}; Do[AppendTo[a, Floor[Count[a, n_ /; n == a[[k - 1]]]/2]], {k, 2, 120}]; a (* Michael De Vlieger, Feb 11 2016 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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