## Wednesday, November 16, 2016

### A walk in base-three pi

Looking at the output of my previous post, it would be natural to believe that A039624 is infinite. What about a base-three analogue? For how many of the initial digits in the ternary representation of π are there solutions where the number of zeros, ones, and twos are all exactly equal? In base three, π = 10.010211012222010211... After a preliminary no-digits, the first 15 and 18 digits also clearly qualify. Then what?

I wanted to explore this graphically, so I converted π's ternary digits to vectors in a Cartesian coordinate system. I let the digit 1 be the vector (0,1); the digit 2, (1,0); and the digit 0, (-1,-1). If we start at (0,0), every time we revisit the origin we should have an exactly equal number of zeros, ones, and twos:

1    (0,0)  +  (0,1)  =  (0,1)
2    (0,1)  + (-1,-1) = (-1,0)
3   (-1,0)  + (-1,-1) = (-2,-1)
4   (-2,-1) +  (0,1)  = (-2,0)
5   (-2,0)  + (-1,-1) = (-3,-1)
6   (-3,-1) +  (1,0)  = (-2,-1)
7   (-2,-1) +  (0,1)  = (-2,0)
8   (-2,0)  +  (0,1)  = (-2,1)
9   (-2,1)  + (-1,-1) = (-3,0)
10   (-3,0)  +  (0,1)  = (-3,1)
11   (-3,1)  +  (1,0)  = (-2,1)
12   (-2,1)  +  (1,0)  = (-1,1)
13   (-1,1)  +  (1,0)  =  (0,1)
14    (0,1)  +  (1,0)  =  (1,1)
15    (1,1)  + (-1,-1) =  (0,0)
16    (0,0)  +  (0,1)  =  (0,1)
17    (0,1)  + (-1,-1) = (-1,0)
18   (-1,0)  +  (1,0)  =  (0,0)
19    (0,0)  +  (0,1)  =  (0,1)
20    (0,1)  +  (0,1)  =  (0,2)
...

And indeed, we re-arrive at the origin at steps 15 and 18. Then what?

The graph shows 25 million steps, with the initial (0,0) and final (950,2678) points shown in red. Yes, we should get back to the origin eventually but it's not as easy to believe as it is for base-two. It should be more understandable then that providing an exactly-equal number of all base-b digits in some initial terms of base-b π with b > 3 will be difficult. Regardless, we have in base-four, the first 4 digits (3.021) and in base-five, the first 75 digits!