Cyclic Primes (2)

Wells makes the interesting observation that we partition 142857 into its two halves i.e. 142 and 857 and then divide the 2^nd partition (857) by the 1^st (142) we get the quotient 6, with a remainder of 5.

So the number 6 is 1 less than the cyclic prime involved i.e. 7, while the remainder  5 is 2 less.

This is a property which universally holds.

For example for the next cyclic prime (17) the 1^st and 2^nd partitions are

05882352 and 94117647. And dividing 94117647 by 05882352 we obtain the quotient 16 and the remainder 15!

More complicated - though in their own way equally exciting patterns - can be seen through breaking the repeating full cycle of digits into more than 2 equal partitions.

One intriguing case arises when we divide into 4 equal partitions. This requires the repeating sequence of digits of the cyclic prime to be divisible by 4!

Up to 100 the only candidates that qualify are the cyclic primes 17, 29, 61 and 97.

So starting with 17 we now divide its 16 repeating digits into the 4 equal partitions

0588   2352   9411  7647

Now beginning with the last we divide 7647, 9411 and 2352 respectively by 0588 finding in each case the quotient and remainder.

We then get in the 3 cases respectively

13 and 3

16 and 3

4  and  0

Now these 6 numbers can be represented in general terms by the following grid

a₁₁   a₁₂

a₂₁   a₂₂

a₃₁   a₃₂

The following set of relationships now hold (which universally apply in all such cases):

a₂₁ is always one less than the cyclic prime in question. Therefore in this case of the cyclic prime 17,

a₂₁ is 16.

So,

a₂₁  = 16 (i.e. 1 less than the cyclic prime)

a₁₁ + a₃₁  =  17 (i.e. the cyclic prime)

a₂₁ –  a₂₂   = a₁₁ (i.e. 13)

a₃₁ = a₂₂  + 1 (i.e. 4)

a₁₁ +  a₁₂  = a₂₁  – a₃₂ (i.e. 16)

a₃₁  – a₃₂  = a₁₂ + 1 (i.e. 4)

a₁₂ +  a₃₂  = a₂₂  (i.e. 4)

I then calculated the corresponding grid of numbers with reference to the cyclic primes 29, 61 and 97 respectively.  

These are

  p = 29              p = 61             p = 97

12     7              11     9             22    17

28    16             60    49            96    74

17     9              50    40            75    57    

Then as we can see the value for a₂₁ is always one less than the cyclic prime in question. It then becomes like a fascinating variation on a Suduko puzzle to fill in the remaining values on the grid using the general relationships I have listed (which always hold).

A number of possible incorrect variations however are consistent with the relationships given. This suggests that the values are determined in a simultaneous holistic manner!

However there is another interesting observation in that the ratios of terms in each row seem to approximate closely with each other.  So the closest relationships possible as between the ratios will correspond therefore to the correct entries in the grid.

The next cyclic prime with recurring digits divisible by 4 is 109.

So it would be interesting bearing in mind what I have said to predict the correct grid entries without manually calculating the values.          

A  simpler situation relates to three partitions which requires that the repeating decimal sequence of the cyclic prime be divisible by 3.

The first case therefore arises with the cyclic prime 7.

Again we break the repeating digits into 3 successive partitions, i.e.

14    28  57

Then starting with the last we divide 57 and 28 by 14 respectively, listing again the quotient and remainder in each case.

This time we get 4 and 1 in the first case and 2 and 0 in the second

So listing in a grid with general form

a₁₁   a₁₂

a₂₁   a₂₂

we get,

4              1

2              0

Then for the next applicable cyclic prime 19, the corresponding grid is

7             4

11           6  

Then the cyclic prime 61 the grid is

47              10

   13       2

What is clear in each case is that the sum of quotients i.e.

a₁₁ +  a₂₁  = p – 1

Likewise a₁₁ +  a₁₂  + a₂₂  = p  – 2.