Polygonal Numbers and e (2)

The general formula for the nth term of an n-gonal number series is given by,

{(n  – 2)k² – (n – 4)k}/2

So therefore if for example one wishes to calculate the 6^th term of the octagonal number series, 

then n = 8 and k = 6,

Thus {(n  – 2)k² – (n – 4)k}/2

= {(6 * 6²) – (4.6)}/2 = (216 – 24)/2 = 192/2 = 96

An easy empirical way of obtaining all terms of polygonal series is given in the following manner

Let us start with the 2-gonal natural number series

1, 2, 3, 4, 5, 6, …

Then to get the 1^st term of successive n-gonal series keep adding 0 to the next term (starting with 1).

So the 1^st term of all n-gonal series = 1

To get the 2^nd term, keep adding 1 to the next term (starting with 2).

So the 2^nd term of the 3-gonal (triangular) series is 3, the 2^nd term of the 4-gonal (square) series is 4, the 2^nd term of the 5-gonal (pentagonal) series is 5 and so on.

 

To get the 3^rd term, keep adding 3 to the 3^rd term (starting with 3)

So the 3^rd term of the 3-gonal series is 6, the 3^rd term of the 4-gonal is 9, the 3^rd term of the 5-gonal, 12 and so on.

 

Then to get the 4^th term, keep adding 6 to the 4^th term (starting with 4).

So the 4^th term of the 3-gonal series is 10, the 4^th term of the 4-gonal is 16, the 4^th term of the 5-gonal, 22 and so on.

 

Now the terms that we keep successively adding on to generate the k^th term of each sequence are the successive terms of the 3-gonal (triangular) series i.e.

0, 1, 3, 6, 10, 15, …

 

To get the 5^th term of each polygonal series, keep adding 10 to the 5^th term (starting with 5).

Therefore the 5^th term of the 3-gonal series is 15, the 5^th term of the 4-gonal series 25, the 5^th term of the 5-gonal, 35 and so on.

 

To obtain the 6^th term keep adding 15 to the 6^th term (starting with 6).

Therefore the 6^th term of the 3-gonal series is 21, the 6^th term of the 4-gonal is 36, the 6^th term of the 5-gonal, 51 and so on.

 

In all the polygon series so far the k^th term has been given a positive value. However k equally can take on the value 0 and negative values – 1, – 2, – 3, …

So the 0^th term of all polygonal series = 0.

 

Then if take the pentagonal series to illustrate (where n = 5) we can derive successive terms where k =  – 1, – 2, – 3, …

So when k = – 1, {(n  – 2)k² – (n – 4)k}/2

=  (3 * (– 1)² – (– 1}/2 = 2

 

When k = – 2, {(n  – 2)k² – (n – 4)k}/2

= {(3 * (– 2)² – (– 2}/2 = 7

 

When k = – 3, {(n  – 2)k² – (n – 4)k}/2

= {(3 * (– 3)² – (– 3}/2 = 12

 

In this way we generate a new shadow pentagonal series

2, 7, 15, 26, 40, 57, 77, 100, …

 

And the same general formula applies as given in the previous entry,

(a_k/0! + a_{k + 1}/1! + a_{k + 2}/2! + …)/e = a_k+ 1 + n/2 – 1

 

Therefore in this case when k = 1,

(a₁/0! + a₂/1! + a₃/2! + …)/e = a₂ + n/2 – 1

So (2/0! + 7/1! + 15/2! + …)/e = 7 + 2.5 – 1 = 8.5

 

Then the generalised pentagonal number series, with numbers written in ascending order, containing the original sequence (for positive values of k and the shadow sequence (for negative values) is

1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 92, 100, ..,

Euler was able to use this sequence to great effect is establishing a link to the divisors of numbers and to partitions of numbers.

 

However for the formula to e to work these two sequences must be joined separately with sequences unfolding in opposite directions (with 0 in the middle common to both).

i.e.  … 15, 7, 2, 0, 1, 5, 12, 22, …

So for example with respect to this new conjoined sequence if a_k = 15, a_{k + 1} = 7

 

Then

(a_k/0! + a_{k + 1}/1! + a_{k + 2}/2! + …)/e =

= 15/0! + 7/1! + 2/2! + 0/3! + 1/4! + 5/5! + 12/6! + 22/7! + … = 7 + 2.5 – 1 = 8.5