Other examples [of substitution systems] (a) (Perioddoubling sequence) After t steps, there are a total of 2^^{t} elements, and the sequence is given by Nest[MapAt[1#&, Join[#, #], 1]&, {0}, t]. It contains a total of Round[2^^{t}/3] black elements, and if the last element is dropped, it forms a palindrome. The n^^{th} element is given by Mod[IntegerExponent[n,2],2]. As discussed on page 885, the sequence appears in a vertical column of cellular automaton rule 150. The ThueMorse sequence discussed on page 890 can be obtained from it by applying 1Mod[Flatten[Partition[FoldList[Plus,0,list],1,2]],2] (b) The n^^{th} element is simply Mod[n,2]. (c) Same as (a), after the replacement 1>{1,1} in each sequence. Note that the spectra of (a) and (c) are nevertheless different, as discussed on page 1080. (d) The length of the sequence at step t satisfies a[t] ==2a[t  1] + a[t  2], so that a[t]=Round[(1 + Sqrt[2])^(t  1)/2] for t>1. The number of white elements at step t is then Round[a[t]/Sqrt[2]]. Much like example (c) on page 83 there are m+1 distinct blocks of length m, and with f = Floor[(1  1/Sqrt[2])(# + 1/Sqrt[2])] & the n^^{th} element of the sequence is given by f[n + 1]  f[n] (see page 903). (e) For large t the number of elements increases like λ^^{t} with λ=(Sqrt[13]+1)/2; there are always λ times as many white elements as black ones. (f) The number of elements at step t is Round[(1+Sqrt[2])^^{t}/2], and the n^^{th} element is given by Floor[Sqrt[2] (n+1)]Floor[Sqrt[2] n] (see page 903). (g) The number of elements is the same as in (f). (h) The number of black elements is 2^(t1); the total number of elements is 2^(t2) (t+1). (i) and (j) The total number of elements is 3^(t1).
