

A122632


Table T(n,k) = number of initial segments of Beatty sequences for numbers > 1 of length k, cutting sequence so that all terms are < n.


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1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 3, 2, 3, 1, 1, 3, 2, 2, 3, 1, 1, 4, 3, 2, 3, 4, 1, 1, 4, 3, 3, 3, 3, 4, 1, 1, 5, 3, 4, 2, 4, 3, 5, 1, 1, 5, 4, 3, 3, 3, 3, 4, 5, 1, 1, 6, 4, 4, 5, 2, 5, 4, 4, 6, 1, 1, 6, 4, 4, 4, 4, 4, 4, 4, 4, 6, 1, 1, 7, 5, 5, 4, 6, 2, 6, 4, 5, 5, 7, 1, 1, 7, 5, 5, 4, 6, 4, 4, 6, 4, 5, 5, 7, 1
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OFFSET

1,5


COMMENTS

Enumerate all rational numbers q in [0,1) with denominator <= n. T(n,k) is the number of these with floor(n*q) = k1. Problem suggested by David W. Wilson.


LINKS



EXAMPLE

T(6,3) = 2; the sequences for n=6, k=3 are 0,2,4 and 0,2,5. The sequence 0,1,3 is not counted because the next term of a Beatty sequence beginning 0,1,3 must be 4 or 5, so 0,1,3 is not a Beatty sequence truncated to numbers less than 6.


MATHEMATICA

Flatten@Table[Count[Select[Union@Flatten@Outer[Divide, Range[n + 1]  1, Range[n]] , # <= 1 &], _?(Floor[n #] == k &)], {n, 12}, {k, n}] (* Birkas Gyorgy *)


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AUTHOR



STATUS

approved



