Jim asks: << Can anyone find an explicit construction of an infinite sequence S of 0's and 1's with the property that for any finite bit-string s, the relative frequency with which s occurs in the first N bits of S converges to (1/2)^(length of s) with error that falls like (log N)/N?

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I have an idea for a strategy that may result in a solution. In the cubical torus T^n = R^n/Z^n one may wish to find a vector v in R^n such that stepping along T^n in equal steps each equal to [v] (= the image of v in T^n by the quotient map R^n -> T^n) gets you dense in T^n as fast as possible. I.e., define R_v(n) := the radius of the largest open ball in T^n lying in the complement of {[v], [2v], . . ., [nv]}. We want to find v such that R_v(n) decreases as fast as possible. A folk theorem holds that the vector defined by V_n := (phi,phi^2,...,phi^n), where p is the golden ratio, can't be beaten, at least in the limit. (This is based on the non-folk theorem that stepping along the circumference-1 circle T = R/Z in steps of equal length gets you dense in T as fast as possible by using steps of length phi.) THE PRESENT PROBLEM may be thought of as similar to a series of such torus problems on tori T^n, where a point [x_1, . . .,x_n] of T^n is mapped to the 0-1 sequence (*) floor{2*x_1}, . . ., floor{2*x_n} where {y} denotes the fractional part of y. The idea being, to get a 0-1 sequence with all the right probabilities having the least error, PERHAPS it will work to let n = 1, 2, 3, . . . for L(n) consecutive steps, and stepping through T^n as in the above paragraph, and *concatenating* all the length-n 0-1 sequences obtained from each step via (*). I suspect L(n) should be something of the order of L(n) = 2^n. The step itself, while stepping through T^n, should perhaps be the vector V_n as above. QUESTION: What if this whole thing were collapsed down to the 1-dimensional version of just stepping through T = R/Z with a constant steplength = phi, and writing down the sequence e_n = floor(2*n*phi) --Dan