Normality is sufficient, but I do not think it is necessary. Normality requires that if the number is represented in any base b then each each of the b possible digits occure with density 1/b. All that is needed for every work to appear is that each digit occur at least once. They do not have to occure with equal density.
Normality is not necessary. Counter-counter example is the digits of pi in groups of 1, 2, 3... with strings of zeros in between:
3.0140015900026530000....
this number is obviously nor normal since 0 occurs with greater frequency than any other digit. But if pi is normal, then this sequence contains all finite sequences too.
Each base 10 digit occurs at least once in your number. It's not obvious to me that each base b digit occurs at least once for all bases b in your number.
Comments
Normality is sufficient, but I do not think it is necessary. Normality requires that if the number is represented in any base b then each each of the b possible digits occure with density 1/b. All that is needed for every work to appear is that each digit occur at least once. They do not have to occure with equal density.
Nope, normality is necessary.
Counter example would be a=0,123456789010011000111000011110000011111....
Normality is not necessary. Counter-counter example is the digits of pi in groups of 1, 2, 3... with strings of zeros in between:
this number is obviously nor normal since 0 occurs with greater frequency than any other digit. But if pi is normal, then this sequence contains all finite sequences too.Each base 10 digit occurs at least once in your number. It's not obvious to me that each base b digit occurs at least once for all bases b in your number.