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_{ }

^{k}C

_{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

^{2}C_{2 }= 1
And subsequent numbers are

^{3}C_{2 }= 3,^{4}C_{2 }= 6,^{5}C_{2 }= 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/

^{k}C_{n - 1 }= sum of reciprocals of unique digit sequence associated with (x – 1)^{n}^{k = n - 1}

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

_{ ∞}

∑1/

^{ k}C_{2 }= sum of reciprocals of unique digit sequence associated with (x – 1)^{3}^{ k = 2}

= 1/2C

_{2 }+_{ }1/3C_{2 }+ 1/4C_{2 }+ 1/5C_{2 }+ …
= 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 n

^{3 }and when n is composite, not divisible by n^{3}(where p ≥ 5).
Where p = 3, the {(n + 1)th term – 1} i.e. 9, is divisible
by n

^{2}(i.e. 3^{2}).
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}/n^{3}, to see whether an integer (without remainder) results. Alternatively, we check to see if n^{3 }is a factor.
So for example to test whether n = 29 is prime, we compute

(

^{57}C_{28 }– 1)/n^{3 }= [{57!/(28! * 29)!} – 1]/24389 = 616410400171 (exactly).
Therefore 29 is prime.

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

In all these cases for primes, where n

^{3}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 n

^{3}, 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 3

^{rd}cycle (i.e. 11^{th}term) is 1001.
Then when we subtract 1, we obtain 1000, which is divisible
by n

^{3}=125.