Q: 0 01 01011 0101101011011 0101101011011010110101101101011011 etc.
Each string is formed from the previous string by substituting '01' for '0' and '011' for '1' simultaneously at each occurance. Notice that each string is an initial substring of the previous string so that we may consider them all as initial substrings of an infinite string. The puzzle then is, given n, determine if the nth digit is 0 or 1 without having to construct all the previous digits. That is, give a non-recursive formula for the nth digit.
S: Let G equal the limit string generated by the above process and define the string F by
where floor(x) is the greatest integer =< x and phi = (1 + \/5)/2; I claim that F = G.
I will try to motivate my solution. Let g0?="0" and define gn+1? to be the string that results from replacing "0" in gn? with "01" and "1" with "011"; furthermore, let s(n) and t(n) be the number of "0"'s and "1"'s in gn?, respectively. Note that we have the following recursive formulas : s(n+1) = s(n) + t(n) and t(n+1) = s(n) + 2t(n). I claim that s(n) = Fib(2n-1) and t(n) = Fib(2n), where Fib(m) is the mth Fibonacci number (defined by Fib(-1) = 1, Fib(0) = 0, Fib(n+1) = Fib(n) + Fib(n-1) for n>=0); this is easily established by induction. Now noting that Fib(2n)/Fib(2n-1) -> phi as n -> infinity, we see that if the density of the "0"'s and "1"'s exists, they must be be 1/phi^2 and 1/phi, respectively. What is the simplest generating sequence which has this property? Answer: the one given above.
Proof: We start with
Beatty's Theorem: if a and b are positive irrational numbers such that 1/a + 1/b = 1, then every positive integer has a representation of the form floor(am) or floor(bm) (m a positive integer), and this representation is unique.
This shows that F is well-defined. I now claim that
Lemma: If S(n) and T(n) (yes, two more functions; apparently today's the day that functions have their picnic) represent the number of "0"'s and "1"'s in the initial string of F of length n, then S(n) = ceil(n/phi^2) and T(n) = floor(n/phi) (ceil(x) is the smallest integer >= x).
T(n-1) + 1. Now note that if Fn-1?"1" ==> n-1 = floor(phi*m) for some positive integer m and since phi*m-1 < floor(phi*m) < phi*m ==> m-1/phi < (n-1)/phi < m ==> T(n) = T(n-1) + 1. To finish, note that if Fn-1?"0" ==> n-1 = floor(phi^2*m) for some positive integer m and since phi^2*m-1 < floor(phi^2*m) < phi^2*m ==> m-1/phi^2 < (n-1)/phi^2 < m ==> S(n) = S(n-1) + 1. Q.E.D.
I will now show that F is invariant under the operation of replacing "0" with "01" and "1" with "011"; it will then follow that F=G. Note that this is equivalent to showing that F2S(n) + 3T(n)? = "0", F2S(n) + 3T(n) + 1? = "1", and that if n = phi*m? for some positive integer m, then F2S(n) + 3T(n) + 2? = "1". One could waste hours trying to prove some fiendish identities; watch how I sidestep this trap. For the first part, note that by the above lemma F2S(n) + 3T(n)? = F2*ceil(n/phi^2) + 3*floor(n/phi)? = F2n + floor(n/phi)? = F2n + floor(n*phi-n)? = Ffloor(phi*n+n)? = Ffloor(phi^2*n)? ==> F2S(n) + 3T(n)? = "0". For the second, it is easy to see that since phi^2>2, if Fm?"0" ==> Fm?"1" hence the first part implies the second part. Finally, note that if n = phi*m? for some positive integer m, then F2S(n) + 3T(n) + 3? = F2S(n+1) + 3T(n+1)? = "0", hence by the same reasoning as above F2S(n) + 3T(n) + 2? = "1".