Hi, I've been looking at a "flagstone problem," where no two adjacent n-tuples can be identical. I solved the problem with Icon using an array of stacks and was going to explore how to do it in Haskell when I saw another way to do it explained in the same text. Just count the ones between the zeros in a(n) to get b(i), a non-repeating sequence of zeros, ones, and twos. An a(n) is 0 if the number of one bits in the binary representation of n is even, otherwise odd, and the initial a(n) must be even. n a(n) b(i) 20 = 010100 0 21 = 010101 1 2 22 = 010110 1 23 = 010111 0 0 24 = 011000 0 25 = 011001 1 2 26 = 011010 1 27 = 011011 0 1 28 = 011100 1 29 = 011101 0 0 30 = 011110 0 31 = 011111 1 2 32 = 100000 1 33 = 100001 0 0 34 = 100010 0 1 35 = 100011 1 36 = 100100 0 ========= Here's my Haskell code ======== import Data.Bits import Data.List flagstone n = foldl (++) "" (take n (map show (f $ group [if even y then 0 else 1 | y <- [bitcount x | x <- [20..]]]))) bitcount :: (Integral t) => t -> t bitcount 0 = 0 bitcount n = let (q,r) = quotRem n 2 in bitcount q + r f [] = [] f ([0]:xs) = f xs f ([0,0]:xs) = 0 : f xs f ([1]:xs) = 1 : f xs f ([1,1]:xs) = 2 : f xs ========= My question ======== A further exercise in the text: "Exercise 5.-(a) Define a(n) as the sum of the binary digits in the binary representation of n. Define b(i) as the number of a's between successive zeros as before. Then T = b(1) b(2) b(3) b(4) ... gives an infinite sequence of *seven* symbols with no repeats. (b) Write a routine to generate a sequence for seven colors of beads on a string with no repeats." I may be misreading, but does this make any sense? There's a reference to The American Mathematical Monthly, June-July 1963, p. 675, by C. H. Braunholtz. Michael