%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
