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Number of rationals in [0, 1) having at most n preperiodic bits, then at most n periodic bits.
0

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

%S 2,12,72,336,1632,6720,29568,120576,499200,2012160,8214528,32894976,

%T 132882432,532070400,2136637440,8551464960,34282536960,137135652864,

%U 549148164096,2196721631232,8791208755200,35166005231616

%N Number of rationals in [0, 1) having at most n preperiodic bits, then at most n periodic bits.

%F a(n) = 2^n * sum_{j=1..n} 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) = |{ 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

%t Table[2^n Sum[Plus@@((2^Divisors[j]-1)MoebiusMu[j/Divisors[j]]),{j,1,n}],{n,1,22}]

%Y Elementwise product of 2^n (offset 1) and A119917. Also, diagonal of A119919.

%K nonn,base,easy

%O 1,1

%A Brad Chalfan (brad(AT)chalfan.net), May 28, 2006