Here’s √2 in base 2:

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√2 = 1.01101010000010011110... (base=2)`

And in base 3:

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√2 = 1.10201122122200121221... (base=3)`

And in bases 4, 5, 6, 7, 8, 9 and 10:

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√2 = 1.12220021321212133303... (b=4)`

√2 = 1.20134202041300003420... (b=5)

√2 = 1.22524531420552332143... (b=6)

√2 = 1.26203454521123261061... (b=7)

√2 = 1.32404746317716746220... (b=8)

√2 = 1.36485805578615303608... (b=9)

√2 = 1.41421356237309504880... (b=10)

And here’s π in the same bases:

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π = 11.00100100001111110110... (b=2)`

π = 10.01021101222201021100... (b=3)

π = 03.02100333122220202011... (b=4)

π = 03.03232214303343241124... (b=5)

π = 03.05033005141512410523... (b=6)

π = 03.06636514320361341102... (b=7)

π = 03.11037552421026430215... (b=8)

π = 03.12418812407442788645... (b=9)

π = 03.14159265358979323846... (b=10)

Mathematicians know that in all standard bases, the digits of √2 and π go on for ever, without falling into any regular pattern. These numbers aren’t merely irrational but transcedental. But are they also normal? That is, in each base *b*, do the digits 0 to [*b*-1] occur with the same frequency 1/*b*? (In general, a sequence of length *l* will occur in a normal number with frequency 1/(*b*^*l*).) In base 2, are there as many 1s as 0s in the digits of √2 and π? In base 3, are there as many 2s as 1s and 0s? And so on.

It’s a simple question, but so far it’s proved impossible to answer. Another question starts very simple but quickly gets very difficult. Here are the answers so far at the *Online Encyclopedia of Integer Sequences* (OEIS):

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2, 572, 8410815, 59609420837337474 – A049364`

The sequence is defined as the “Smallest number that is digitally balanced in all bases 2, 3, … n”. In base 2, the number 2 is 10, which has one 1 and one 0. In bases 2 and 3, 572 = 1000111100 and 210012, respectively. 1000111100 has five 1s and five 0s; 210012 has two 2s, two 1s and two 0s. Here are the numbers of A049364 in the necessary bases:

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10 (n=2)`

1000111100, 210012 (n=572)

100000000101011010111111, 120211022110200, 200011122333 (n=8410815)

11010011110001100111001111010010010001101011100110000010, 101201112000102222102011202221201100, 3103301213033102101223212002, 1000001111222333324244344 (n=59609420837337474)

But what number, a(6), satisfies the definition for bases 2, 3, 4, 5 and 6? According to the notes at the OEIS, a(6) > 5^434. That means finding a(6) is way beyond the power of present-day computers. But I assume a quantum computer could crack it. And maybe someone will come up with a short-cut or even an algorithm that supplies a(*b*) for any base *b*. Either way, I think we’ll get there, π and by.