222 A.D. was the year in which the Emperor Heliogabalus was assassinated by his own soldiers. Exactly 1666 years later, the Anglo-Dutch classicist Sir Lawrence Alma-Tadema exhibited his painting *The Roses of Heliogabalus* (1888). I suggested in “Roses Are Golden” that Alma-Tadema must have chosen the year as deliberately as he chose the dimensions of his canvas, which, at 52″ x 84 ^{1}/_{8}“, is an excellent approximation to the golden ratio.

But did Alma-Tadema know that lines at 0º and 222º divide a circle in the golden ratio? He could easily have done, just as he could easily have known that 222 precedes the 48th prime, 223. But it is highly unlikely that he knew that 223 yields a magic square whose columns, rows and diagonals all sum to 222. To create the square, simply list the 222 multiples of the reciprocal 1/223 in base 3, or ternary. The digits of the reciprocal repeat after exactly 222 digits and its multiples begin and end like this:

001/223 = 0.00001002102101021212111012022211122022... in base 3

002/223 = 0.00002011211202120201222101122200021121...

003/223 = 0.00010021021010212121110120222111220221...

004/223 = 0.00011100200112011110221210022100120020...

005/223 = 0.00012110002220110100102222122012012120...

[...]

`218/223 = 0.22210112220002112122120000100210210102... in base 3
219/223 = 0.22211122022110211112001012200122102202...
220/223 = 0.22212201201212010101112102000111002001...
221/223 = 0.22220211011020102021000121100022201101...
222/223 = 0.22221220120121201010111210200011100200...
`

Each column, row and diagonal of ternary digits sums to 222. Here is the full n/223 square represented with 0s in grey, 1s in white and 2s in red:

(Click for larger)

It isn’t difficult to see that the white squares are mirror-symmetrical on a horizontal axis. Here is the symmetrical pattern rotated by 90º:

(Click for larger)

But why should the 1s be symmetrical? This isn’t something special to 1/223, because it happens with prime reciprocals like 1/7 too:

1/7 = 0.010212... in base 3

2/7 = 0.021201...

3/7 = 0.102120...

4/7 = 0.120102...

5/7 = 0.201021...

6/7 = 0.212010...

And you can notice something else: 0s mirror 2s and 2s mirror 0s. A related pattern appears in base 10:

1/7 = 0.142857...

2/7 = 0.285714...

3/7 = 0.428571...

4/7 = 0.571428...

5/7 = 0.714285...

6/7 = 0.857142...

The digit 1 in the decimal digits of n/7 corresponds to the digit 8 in the decimal digits of (7-n)/7; 4 corresponds to 5; 2 corresponds to 7; 8 corresponds to 1; 5 corresponds to 4; and 7 corresponds to 2. In short, if you’re given the digits d1 of n/7, you know the digits d2 of (n-7)/7 by the rule d2 = 9-d1.

Why does that happen? Examine these sums:

1/7 = 0.142857142857142857142857142857142857142857...

+6/7 = 0.857142857142857142857142857142857142857142...

7/7 = 0.999999999999999999999999999999999999999999... = 1.0

2/7 = 0.285714285714285714285714285714285714285714...

+5/7 = 0.714285714285714285714285714285714285714285...

7/7 = 0.999999999999999999999999999999999999999999... = 1.0

` 3/7 = 0.428571428571428571428571428571428571428571...
+4/7 = 0.571428571428571428571428571428571428571428...
7/7 = 0.999999999999999999999999999999999999999999... = 1.0
`

And here are the same sums in ternary (where the first seven integers are 1, 2, 10, 11, 12, 20, 21):

1/21 = 0.010212010212010212010212010212010212010212...

+20/21 = 0.212010212010212010212010212010212010212010...

21/21 = 0.222222222222222222222222222222222222222222... = 1.0

1/21 = 0.010212010212010212010212010212010212010212...

+20/21 = 0.212010212010212010212010212010212010212010...

21/21 = 0.222222222222222222222222222222222222222222... = 1.0

2/21 = 0.021201021201021201021201021201021201021201...

+12/21 = 0.201021201021201021201021201021201021201021...

21/21 = 0.222222222222222222222222222222222222222222... = 1.0

` 10/21 = 0.102120102120102120102120102120102120102120...
+11/21 = 0.120102120102120102120102120102120102120102...
21/21 = 0.222222222222222222222222222222222222222222... = 1.0
`

Accordingly, in base b with the prime p, the digits d1 of n/p correspond to the digits (p-n)/p by the rule d2 = (b-1)-d1. This explains why the 1s mirror themselves in ternary: 1 = 2-1 = (3-1)-1. In base 5, the 2s mirror themselves by the rule 2 = 4-2 = (5-1) – 2. In all odd bases, some digit will mirror itself; in all even bases, no digit will. The mirror-digit will be equal to (b-1)/2, which is always an integer when b is odd, but never an integer when b is even.

Here are some more examples of the symmetrical patterns found in odd bases:

Patterns of 1s in 1/19 in base 3

Patterns of 6s in 1/19 in base 13

Patterns of 7s in 1/19 in base 15

Elsewhere other-posted:

• Roses Are Golden — more on *The Roses of Heliogabalus* (1888)

• Three Is The Key — more on the 1/223 square