Many of
their ingenious properties are to be found in that wonderful book by David
Wells “The Penguin Dictionary of Curious and Interesting Numbers”.

So I will
concentrate in this entry on some other interesting properties (not addressed
in that book).

Clearly we
operate customarily in the denary system (base 10). So full cyclic primes i.e.
where reciprocals have a digital decimal period one less than the number in
question) are expressed with respect to base 10.

So when we
say that 7 is the first (full) cyclic prime, this is with respect to the base
10 system, where its reciprocal (1/7) has
a repeating decimal sequence of six digits (i.e. 124857).

However the
earliest base number for which 7 for example is also a (full) cyclic prime is in
base 3 with the recurring digits 010212 and then again in base 5 (032412).

Then 5,
which is not a cyclic prime in base 10 is however already fully cyclic in bases 2 (0011),
3 (0121), 7 (1254) and 8 (1463).

So if a
prime number is not fully cyclic in base 10, it will be cyclic in a number of other
bases.

In this
sense a remarkable cyclic nature is shared by all prime numbers!

Incidentally
2, which is the first prime number is (fully cyclic) in base 3 (with just one
repeating digit (i.e. 1). It is then fully cyclic in all odd number bases.

A simple
formula can be given for the sum of digits of any cyclic prime.

If p is the
prime number and n the base, then the sum of digits = (p – 1)(n – 1)/2.

So again 7
is fully cyclic in base 10. Therefore the sum of digits of its recurring period
sequence = (6 * 9)/2 = 27.

Indeed this
is related to another property.

Because the
period (of all prime numbers other than 2) is necessarily even, we can partition
the digits into two equal halves.

So with
142857 we have 142 and 857. When we add we get 999.

And this is
a universal occurrence (i.e. where each digit of the sum of partitions is one less than the number
base).

Now when we
subdivide into smaller groups (which must be a factor) of the original
partition size, the relationship likewise holds (except that we multiply by
the other factor involved).

So if we
partition in this case into single digit groups, this represents 1/3 of the
original partition size. So the factors here are 1 and 3. Therefore when we sum the terms we get 3 * 9 =
27.

This can be
better demonstrated with respect to the recurring digit sequence of the next
(fully) cyclic prime in base 10 i.e. 17.

So this has
a period of 16 with the recurring digits 0588235294117647.

Now when we
partition digits into two equal groups of 8 and add we get 99999999.

However we
could equally partition into equal groups of 4 and add to get 2 * 9999. (The factors here are 4 and 2!)

We could
the partition into equal groups of 2 and add to get 4 * 99. (The factors are now 2 and 4).

Finally we
could partition into groups of 1 digit to get 8 * 9 = 72 which is necessarily
the sum of digits of this cyclic prime! (The factors here are now 1 and 8).

When one subtracts the original partition values an interesting result ensues.

So 94117647
– 05882352 = 88235295.

The first 7 digits here are a recurring sequence within the
original digital sequence with the last digit 5 (representing the sum of the
next 2 digits in the full sequence (i.e. 4 + 1).

And this is
a universal feature of such behaviour.

For example
the 18 digits sequence for the next cyclic prime (19) is

052631578947368421.

Then, partitioning
into two halves of 9 and subtracting we get,

947368421 –
052631578 = 894736843.

So once
again the sub-sequence of 9 now contains 8 digits of the original
sequence with the last digit (3) representing the sum of the next two digits (2
+ 1) in the original sequence!

142857
* 142857 = 20,408,122,449

Wells
points to the interesting fact that if we place 0 in front of the first digit
we can then partition into two equal groups of 6, which when added obtains the
original number 142857 (i.e. a Kaprekar number).

When we
multiply 142857 by its reverse (758241) we get 108,320,034,037.

What is
perhaps more surprising is that when we partition into two equal groups of 6
and add, we once more get 142857.

I say
surprising, because the reverse number here 758241 does not in fact maintain
the same cyclic sequence of digits!

However
once again this is a property that universally holds for all cyclic primes
across all bases (though not necessarily with original cyclic sequence of digits). .

Another
interesting property of cyclic primes is a great regularity in the composition
of digits. All the digits 0 – 9 are included before the pattern repeats.

For cyclic
primes with period of (10n + 2) where n = 0, 1, 2, 3, …., all 10 digits will appear at least n
times with the two extra digits 3 and 6 occurring n + 1 times.

23 for
example is a cyclic prime. This means that all 10 digits (0 – 9) will occur at
least twice in the full period of 22 digits with 3 and 6 occurring 3 times.

Now cyclic
primes in base 10 cannot have a period of 10n + 4 digits as this would imply
that the prime ends in 5!

For cyclic
primes, with period of 10n + 6 digits all 10 digits will again occur at n times
with the 1, 2, 4, 5, 7 and 8 occurring n + 1 times.

Again for
example 17 as we have seen is a cyclic prime. This means that 0, 3, 6 and 9
occur once with 1, 2, 4, 5, 7 and 8 occurring twice.

For cyclic
primes with period of 10n + 8 digits all digits but 0 and 9 will occur n + 1
times as verified in the case of 29 (which is a cyclic prime).

This means
that it is possible to immediately predict the number of times each digit occurs for any cyclic
prime!

For example
47 is a (full) cyclic prime.

Therefore
we know that its unique period of 46 digits contains 0, 3, 6 and 9 four times,
and all the other digits i.e. 1, 2, 4, 5, 7 and 8, five times.

This of course
implies that the sum of digits = (0 + 3 + 6 + 9) * 4 + (1 + 2 + 4 + 5 + 7 + 8)
* 5 = 72 + 135 = 207 = (p – 1)(n – 1)/2 = (46 * 9)/2.

Here is
another interesting fact related to octagonal numbers that I discovered some
years ago!

The prime
number 3 is cyclic in all bases 3n + 2 where n = 0, 1, 2, 3, ….

For example
when n = 2 this means that 3 is cyclic in base 8, with the two digits i.e. 25 continually repeating it its decimal sequence.

Now the
remarkable property of the number 25 is that when we subtract from its reverse
(i.e. 52) we once again obtain 25 (in base 8).

And this
property holds in all the (given) bases for which 3 is cyclic!

In base 2,
the reciprocal of 3 (as cyclic prime) has the recurring digital sequence 01

Then in
base 5, it is 13, in base 8 (as we have seen) 25, in base 11 it is 37 and so
on.

So for example,
to move from the cyclic prime in base 5 to base 8, we multiply the first digit
by 2 then subtracting 1 from the second digit, multiply by 2 and add 1.

Then in
base 11 we multiply the first digit (with respect to base 5 by 3, again
subtracting 1 from second digit and this time multiplying by 3 before adding 1.

Now when we
convert these numbers 01, 13, 25, 37,….to base 10, we get

1, 8, 21,
40, …., which are the octagonal numbers!

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