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Table read by antidiagonals: number of rationals in [0, 1) having at most n preperiodic bits, then at most k periodic bits (read up antidiagonals).
1

%I #1 Sep 29 2006 03:00:00

%S 1,2,3,4,6,9,8,12,18,21,16,24,36,42,51,32,48,72,84,102,105,64,96,144,

%T 168,204,210,231,128,192,288,336,408,420,462,471,256,384,576,672,816,

%U 840,924,942,975,512,768,1152,1344,1632,1680,1848,1884,1950,1965,1024

%N Table read by antidiagonals: number of rationals in [0, 1) having at most n preperiodic bits, then at most k periodic bits (read up antidiagonals).

%F a(n, k) = 2^n * sum_{j=1..k} sum_{d|j} (2^d - 1) * mu(j/d)

%e The binary expansion of 7/24 = 0.010(01)... has 3 preperiodic bits (to the right of the binary point) followed by 2 periodic (i.e., repeating) bits, while 1/2 = 0.1(0)... has one bit of each type. The preperiodic and periodic parts are both chosen to be as short as possible.

%e a(2, 2) = |{ 0/1 = 0.(0)..., 1/3 = 0.(01)..., 2/3 = 0.(10)..., 1/2 = 0.1(0)..., 1/6 = 0.0(01)..., 5/6 = 0.1(10)..., 1/4 = 0.01(0)..., 3/4 = 0.11(0)..., 1/12 = 0.00(01)..., 5/12 = 0.01(10)..., 7/12 = 0.10(01)..., 11/12 = 0.11(10)...}| = 12

%e Table begins:

%e 1 3 9 21

%e 2 6 18 42

%e 4 12 36 84

%e 8 24 72 168

%t Table[2^n Sum[Plus@@((2^Divisors[j]-1)MoebiusMu[j/Divisors[j]]),{j,1,k}],{n,0,10},{ k,1,10}]

%Y Outer product of 2^n (offset 0) and A119917. Also, partial (double) sums of A119918.

%K nonn,base,easy,tabl

%O 1,2

%A Brad Chalfan (brad(AT)chalfan.net), May 29, 2006