The quest to be normal =======Warren D Smith====Aug 2015======== http://mathworld.wolfram.com/ChampernowneConstant.html constructs a stupid but normal (in radix 10) real number: concatenate all the positive integers in order to get its decimal expansion. I say that construction is "stupid" because it occurs to me that there is a much simpler-to-prove-normal construction: 1. Let K=1. 2. Concatenate all the K-digit numbers in order. (Including numbers starting with 0, e.g. "02" is a "2-digit number.") 3. Double K. 4. Go back to step 2. The radix-10 normality of my number is trivial to see. For Champernowne's number it is much less trivial. Incidentally, sum(n=0..B-1)of n*B^(-n-1) = B^(-B) * (B^B-B^2+B-1) / (B-1)^2 so that if we write L_N = sum(n=0..N-1)of 2^n * B^(2^n) with L_0 = 0, then my number which is normal to radix B is given by MyNumber = sum(k=0,1,2,...)of B^(-B-L_k) * (B^B-B^2+B-1) / (B-1)^2. NEW(?) PROBLEM: Construct a number which is normal to every radix B>=2 simultaneously. (It is easy to see an uncountable infinity of such numbers exist, but I'm unaware of anybody actually constructing any example.) Here is a candidate, constructed as a mixed-radix number to radix B where B increases. 1. Let K=1 and B=2 and N=2. 2. Concatenate all the K-digit radix-B numbers in order. (Including numbers starting with 0, e.g. "01" is a "2-digit number.") 3. Multiply K by N. 4. Increment N. 5. Replace B by B^(N^2). 6. Multiply B by the Nth prime. 7. Go back to step 2. I think this number is normal simultaneously in every radix of the form (2*3*5*7*11*13*17*...*P)^k with P prime and k>0. Is it, and if so is that enough to prove normality to all radices B>=2? end.