Period Panes

In his Penguin Dictionary of Curious and Interesting Numbers (1986), David Wells says that 142857 is “beloved of all recreational mathematicians”. He then says it’s the decimal period of the reciprocal of the fourth prime: “1/7 = 0·142857142857142…” And the reciprocal has maximum period. There are 6 = 7-1 digits before repetition begins, unlike the earlier prime reciprocals:


1/2 = 0·5
1/3 = 0·333...
1/5 = 0·2
1/7 = 0·142857 142857 142...

In other words, all possible remainders appear when you calculate the decimals of 1/7:


1*10 / 7 = 1 remainder 3 → 0·1
3*10 / 7 = 4 remainder 2 → 0·14
2*10 / 7 = 2 remainder 6 → 0·142
6*10 / 7 = 8 remainder 4 → 0·1428
4*10 / 7 = 5 remainder 5 → 0·14285
5*10 / 7 = 7 remainder 1 → 0·142857
1*10 / 7 = 1 remainder 3 → 0·142857 1
3*10 / 7 = 4 remainder 2 → 0·142857 14
2*10 / 7 = 2 remainder 6 → 0·142857 142...

That happens again with 1/17 and 1/19, but Wells says that “surprisingly, there is no known method of predicting which primes have maximum period.” It’s a simple question that involves some deep mathematics. Looking at prime reciprocals is like peering through a small window into a big room. Some things are easy to see, some are difficult and some are presently impossible.

In his discussion of 142857, Wells mentions one way of peering through a period pane: “The sequence of digits also makes a striking pattern when the digits are arranged around a circle.” Here is the pattern, with ten points around the circle representing the digits 0 to 9:

The digits of 1/7 = 0·142857142…


But I prefer, for further peers through the period-panes, to create the period-panes using remainders rather than digits. That is, the number of points around the circle is determined by the prime itself rather than the base in which the reciprocal is calculated:

The remainders of 1/7 = 1, 3, 2, 6, 4, 5…


Period-panes can look like butterflies or bats or bivalves or spiders or crabs or even angels. Try the remainders of 1/13. This prime reciprocal doesn’t have maximum period: 1/13 = 0·076923 076923 076923… So there are only six remainders, creating this pattern:

remainders(1/13) = 1, 10, 9, 12, 3, 4


The multiple 2/13 has different remainders and creates a different pattern:

remainders(2/13) = 2, 7, 5, 11, 6, 8


But 1/17, 1/19 and 1/23 all have maximum period and yield these period-panes:

remainders(1/17) = 1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12


remainders(1/19) = 1, 10, 5, 12, 6, 3, 11, 15, 17, 18, 9, 14, 7, 13, 16, 8, 4, 2


remainders(1/23) = 1, 10, 8, 11, 18, 19, 6, 14, 2, 20, 16, 22, 13, 15, 12, 5, 4, 17, 9, 21, 3, 7


It gets mixed again with the prime 73, which doesn’t have maximum period and yields a plethora of period-panes (some patterns repeat with different n * 1/73, so I haven’t included them):

remainders(1/73)


remainders(2/73)


remainders(3/73)


remainders(4/73)


remainders(5/73)


remainders(6/73)


remainders(9/73)


remainders(11/73) (identical to pattern of 5/73)


remainders(12/73)


remainders(18/73)


101 yields a plethora of period-panes, but they’re variations on a simple theme. They look like flapping wings in this animated gif:

remainders of n/101 (animated)


The remainders of 137 yield more complex period-panes:

remainders of n/137 (animated)


And what about different bases? Here are period-panes for the remainders of 1/17 in bases 2 to 16:

remainders(1/17) in base 2


remainders(1/17) in b3


remainders(1/17) in b4


remainders(1/17) in b5


remainders(1/17) in b6


remainders(1/17) in b7


remainders(1/17) in b8


remainders(1/17) in b9


remainders(1/17) in b10


remainders(1/17) in b11


remainders(1/17) in b12


remainders(1/17) in b13


remainders(1/17) in b14


remainders(1/17) in b15


remainders(1/17) in b16


remainders(1/17) in bases 2 to 16 (animated)


But the period-panes so far have given a false impression. They’ve all been symmetrical. That isn’t the case with all the period-panes of n/19:

remainders(1/19) in b2


remainders(1/19) in b3


remainders(1/19) in b4 = 1, 4, 16, 7, 9, 17, 11, 6, 5 (asymmetrical)


remainders(1/19) in b5 = 1, 5, 6, 11, 17, 9, 7, 16, 4 (identical pattern to that of b4)


remainders(1/19) in b6


remainders(1/19) in b7


remainders(1/19) in b8


remainders(1/19) in b9


remainders(1/19) in b10 (identical pattern to that of b2)


remainders(1/19) in b11


remainders(1/19) in b12


remainders(1/19) in b13


remainders(1/19) in b14


remainders(1/19) in b15


remainders(1/19) in b16


remainders(1/19) in b17


remainders(1/19) in b18


remainders(1/19) in bases 2 to 18 (animated)


Here are a few more period-panes in different bases:

remainders(1/11) in b2


remainders(1/11) in b7


remainders(1/13) in b6


remainders(1/43) in b6


remainders in b2 for reciprocals of 29, 37, 53, 59, 61, 67, 83, 101, 107, 131, 139, 149 (animated)


And finally, to performativize the pun of “period pane”, here are some period-panes for 1/29, whose maximum period will be 28 (NASA says that the “Moon takes about one month to orbit Earth … 27.3 days to complete a revolution, but 29.5 days to change from New Moon to New Moon”):

remainders(1/29) in b4


remainders(1/29) in b5


remainders(1/29) in b8


remainders(1/29) in b9


remainders(1/29) in b11


remainders(1/29) in b13


remainders(1/29) in b14


remainders(1/29) in various bases (animated)


Period Panes

In The Penguin Dictionary of Curious and Interesting Numbers (1987), David Wells remarks that 142857 is “a number beloved of all recreational mathematicians”. He then explains that it’s “the decimal period of 1/7: 1/7 = 0·142857142857142…” and “the first decimal reciprocal to have maximum period, that is, the length of its period is only one less than the number itself.”

Why does this happen? Because when you’re calculating 1/n, the remainders can only be less than n. In the case of 1/7, you get remainders for all integers less than 7, i.e. there are 6 distinct remainders and 6 = 7-1:

(1*10) / 7 = 1 remainder 3, therefore 1/7 = 0·1...
(3*10) / 7 = 4 remainder 2, therefore 1/7 = 0·14...
(2*10) / 7 = 2 remainder 6, therefore 1/7 = 0·142...
(6*10) / 7 = 8 remainder 4, therefore 1/7 = 0·1428...
(4*10) / 7 = 5 remainder 5, therefore 1/7 = 0·14285...
(5*10) / 7 = 7 remainder 1, therefore 1/7 = 0·142857...
(1*10) / 7 = 1 remainder 3, therefore 1/7 = 0·1428571...
(3*10) / 7 = 4 remainder 2, therefore 1/7 = 0·14285714...
(2*10) / 7 = 2 remainder 6, therefore 1/7 = 0·142857142...

Mathematicians know that reciprocals with maximum period can only be prime reciprocals and with a little effort you can work out whether a prime will yield a maximum period in a particular base. For example, 1/7 has maximum period in bases 3, 5, 10, 12 and 17:

1/21 = 0·010212010212010212... in base 3
1/12 = 0·032412032412032412... in base 5
1/7 =  0·142857142857142857... in base 10
1/7 =  0·186A35186A35186A35... in base 12
1/7 =  0·274E9C274E9C274E9C... in base 17

To see where else 1/7 has maximum period, have a look at this graph:

Period pane for primes 3..251 and bases 2..39


I call it a “period pane”, because it’s a kind of window into the behavior of prime reciprocals. But what is it, exactly? It’s a graph where the x-axis represents primes from 3 upward and the y-axis represents bases from 2 upward. The red squares along the bottom aren’t part of the graph proper, but indicate primes that first occur after a power of two: 5 after 4=2^2; 11 after 8=2^3; 17 after 16=2^4; 37 after 32=2^5; 67 after 64=2^6; and so on.

If a prime reciprocal has maximum period in a particular base, the graph has a solid colored square. Accordingly, the purple square at the bottom left represents 1/7 in base 10. And as though to signal the approval of the goddess of mathematics, the graph contains a lower-case b-for-base, which I’ve marked in green. Here are more period panes in higher resolution (open the images in a new window to see them more clearly):

Period pane for primes 3..587 and bases 2..77


Period pane for primes 3..1303 and bases 2..152


An interesting pattern has begun to appear: note the empty lanes, free of reciprocals with maximum period, that stretch horizontally across the period panes. These lanes are empty because there are no prime reciprocals with maximum period in square bases, that is, bases like 4, 9, 25 and 36, where 4 = 2*2, 9 = 3*3, 25 = 5*5 and 36 = 6*6. I don’t know why square bases don’t have max-period prime reciprocals, but it’s probably obvious to anyone with more mathematical nous than me.

Period pane for primes 3..2939 and bases 2..302


Period pane for primes 3..6553 and bases 2..602


Like the Ulam spiral, other and more mysterious patterns appear in the period panes, hinting at the hidden regularities in the primes.