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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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%I #8 Nov 30 2016 23:38:43

%S 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,

%T 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,

%U 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

%N 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.

%C Enumerate all rational numbers q in [0,1) with denominator <= n. T(n,k) is the number of these with floor(n*q) = k-1. Problem suggested by _David W. Wilson_.

%e 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.

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

%Y Cf. A002088 (row sums), A006842/A006843 (Farey fractions).

%K nonn,tabl

%O 1,5

%A _Franklin T. Adams-Watters_, Oct 20 2006