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A263991 a(n) is the number of uniform consecutive subintervals of the unit interval each of size 2^(-ceiling(log_2(n))) that are completely covered by one of the n uniform consecutive subintervals (of size 1/n each) of the unit interval. 0

%I

%S 1,2,2,4,4,4,2,8,8,8,6,8,4,4,2,16,16,16,14,16,12,12,10,16,8,8,6,8,4,4,

%T 2,32,32,32,30,32,28,28,26,32,24,24,22,24,20,20,18,32,16,16,14,16,12,

%U 12,10,16,8,8,6,8,4,4,2,64,64,64,62,64,60,60,58,64,56,56,54,56,52,52,50,64,48,48,46,48,44,44,42,48,40,40,38,40,36,36,34,64,32,32,30

%N a(n) is the number of uniform consecutive subintervals of the unit interval each of size 2^(-ceiling(log_2(n))) that are completely covered by one of the n uniform consecutive subintervals (of size 1/n each) of the unit interval.

%C For n > 1, we know that 2 <= a(n) <= n. The lower bound of 2 is because [0, 2^(-ceiling(log_2(n)))] will always be covered by [0, 1/n] and [1 - 2^(-ceiling(log_2(n))), 1] will always be covered by [1 - 1/n, 1]. The upper bound of n is because each of the n consecutive subintervals (of size 1/n each) can only completely cover up to 1 of the 2^(ceiling(log_2(n))) consecutive subintervals (of size 2^(-ceiling(log_2(n))) each). In the case that n is a power of 2, a(n) = n.

%e For n=5, the 4 solutions are these intervals: [0, 1/8] is covered completely by [0, 1/5], [1/4, 3/8] is covered completely by [1/5, 2/5], [5/8, 3/4] is covered completely by [3/5, 4/5], and [7/8, 1] is covered completely by [4/5, 1].

%o (Python)

%o def getNumber(n):

%o y = int(math.ceil(math.log(n, 2)))

%o smallIncrement = 0.5 ** y

%o marker = 0

%o nMarker = 0

%o toReturn = 0

%o while marker < 1:

%o newMarker = marker + smallIncrement

%o if (nMarker + 1) / n < newMarker:

%o nMarker += 1

%o if nMarker / n <= marker and (nMarker + 1) / n >= newMarker:

%o toReturn += 1

%o marker = newMarker

%o return toReturn

%Y Cf. A029837, A062383.

%K nonn

%O 1,2

%A _Apoorv Khandelwal_, Oct 31 2015

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Last modified August 12 17:17 EDT 2020. Contains 336439 sequences. (Running on oeis4.)