Testing for Primes

As we have seen a unique (infinite) number sequence is associated with the polynomial equation (x – 1)ⁿ  = 0.

For example when n = 3, the sequence relates to the triangular nos.,

1, 3, 6, 10, 15, 21, 28, 36, …

There is however a fascinating alternative way of obtaining this general sequence.

So again when (x – 1)ⁿ  = 0, each number in the sequence corresponds to

     

 ^kC_{n - 1},   =  k!/{(n – 1)!( k – n + 1)!}, where k = n – 1, n, n + 1, n + 2, …

Therefore, when the starting value of n = 3, the first number in sequence is ²C₂  = 1

And subsequent numbers are ³C₂  = 3, ⁴C₂  = 6, ⁵C₂  = 10, …

Thus these successive numbers represent (n – 1) combinations with respect to 2, 3, 4, 5, … , items respectively.

So when n = 3, successive numbers represent just 2 combinations with respect to 2, 3, 4, 5, …, items respectively.  

For the above general series, in the first number cycle of n, the sum of last (n – 1) digits

= (n + 1)^th term – 1

So again, for example when n = 3, the 1st cycle comprises the 3 numbers 1, 3 and 6 respectively.

And the sum of the last two of these numbers = 9, which is 1 less than the 4^th term in the sequence (i.e. 10).

This relationship holds for all values of n.

So for example when n = 4, the unique associated (infinite) sequence of digits is,

1, 4, 10, 20, 35, 56, 84, …

These now represent 3 combinations of 3, 4, 5, 6, ... items respectively. 

Thus, the 1st cycle comprises the 4 numbers 1, 4, 10 and 20. And the sum of the last three of these numbers = 34, which now is 1 less than the 5^th number in the sequence (i.e. 35).

 _∞

 ∑1/^kC_{n - 1} = sum of reciprocals of unique digit sequence associated 

^{k = n - 1}

with (x – 1)ⁿ

So when for example, n = 3 (and n – 1 = 2),

    _∞

  ∑1/ ^kC₂  = sum of reciprocals of unique digit sequence associated 

  ^{k = 2}

with (x – 1)³

= 1/2C₂  +  1/3C₂  + 1/4C₂  + 1/5C₂  + … 

= 1 + 1/3 + 1/6 + 1/10 + …   = 2

Remarkably when n is prime the {(n + 1)th term – 1} appears to be always divisible by n³ and when n is composite, not divisible by n³ (where p ≥ 5).

Where p = 3, the {(n + 1)th term – 1} i.e. 9, is divisible by n² (i.e. 3²).

And when p = 2, the {(n + 1)th term – 1} i.e. 2, is divisible by n (i.e. 2).

In principle this provides a fascinating test for whether a number is prime.

Thus in general terms, we can compute,

{(^{2n – 1}C_{n – 1}) – 1}/n³, to see whether an integer (without remainder) results. Alternatively, we check to see if n³ is a factor.

So for example to test whether n = 29 is prime, we compute

{⁵⁷C₂₈  – 1)/29³}

And because there is an integer result, 29 is therefore prime. 

I have tested for all primes from 5 to 61 (inclusive) and found no exceptions.

In all these cases for primes, where n³ was indeed a factor, in no case was a higher power of n a factor.

It would equally appear to be the case that if we continued on to further cycles - where once again a cycle contains the same number of terms, as the number in question (n) - that the 1st number of the next cycle less 1, will again be divisible by n³, where n is prime.

However in some cases this number will also be divisible by higher powers of n.

For example when n = 5, the first number in the 5^th cycle (i.e. 11^th term) is 10626.

Then when we subtract 1, we obtain 10625, which is divisible by n⁴ = 625.