4 ms·
Nope, normality is necessary. >>All that is needed for every work to appear is that each digit occur at least once. Counter example would be a=0,1234567890100
by ithinkso 12y ago
Nope, normality is necessary.
>>All that is needed for every work to appear is that each digit occur at least once.
Counter example would be a=0,123456789010011000111000011110000011111....
- millstone 12y agoNormality 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.
- tzs 12y agoEach 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.