Now the
pentagonal numbers are generally listed as,

1, 5, 12,
22, 35, 51, 70, …, with the n

^{th}pentagonal number given as n/2{3n – 1} for n = 1, 2, 3, …
However when we allow n to also take on values for n = – 1, –
2, – 3, … we get the generalised pentagonal numbers, which arranged in
ascending order are,

1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100,
…

Euler discovered an important and somewhat surprising
relationship entailing the generalised pentagonal numbers.

In
multiplying out the infinite series

(1 –
x)(1 – x

^{2})(1 – x^{3})(1 – x^{4}) … he discovered that the first few terms were,
1 –
x

^{1 }– x^{2 }+ x^{5 }+ x^{7 }– x^{12 }– x^{15 }+ …
So the powers of the x terms naturally unfolding in this manner were seemingly the
(generalised) pentagonal numbers.

Euler was then able to prove from this finding - now known
as The Pentagonal Number Theorem - two remarkable results.

Firstly he showed that if σ(n) is the sum of the divisors of
n then,

σ(n) = σ(n – 1) + σ(n – 2) – σ(n – 5) – σ(n – 7) + σ(n – 12) + σ(n – 15) – σ(n – 22) – …

So if n =
12, then

σ(12) = σ(11) + σ(10) – σ(7) – σ(5) + σ(0)

The divisors
of 11 are 1, 11; the divisors of 10 are 1, 2, 5, 10; the divisors of 7 are 1, 7;
the divisors of 5 are 1, 5; then when n = 0 the value is given as n, which in
this case = 12.

Therefore σ(12) = 12 + 18 – 8 – 6
+ 12 = 28.

And when we
manually check, the divisors of 12 are 1 + 2 + 3 + 4 + 6 + 12 = 28.

Secondly he
was able to relate his theorem to partitions.

Now p(n)
represents the number of partitions of a number i.e. the number of distinct
ways of expressing a number.

For example
4 has 5 partitions. So if 4 represented a group of pebbles, we could
place all the pebbles in a group (as one possible partition). We could equally place
2 pebbles in one group and 2 in another (representing a second partition). We
could also have 3 pebbles in one group and just one in another (representing a
third partition).

We could
then have 2 pebbles in one group and 1 each in two other groups (as a fourth
partition). Finally, we could place the 4 items in four separate groups (as the
fifth and final partition).

Therefore 4
has 5 possible partitions.

Thus if we
wanted now to find p(n) where for example n = 5, we can use Euler’s formula,
i.e.

p(n) = p(n – 1) + p(n – 2) – p(n – 5) – p(n – 7) + p(n – 12).

Thus p(5) = p(4) + p(3) – p(0). In this case where n = 0, p(0)
= 1.

So, p(5) = 5 + 3 – 1
= 7.

However it
is another feature of the pentagonal theorem that I wish to concentrate on in
this entry.

In other
related work, Euler came up with the following formula,

(1 –
x)(1 – x

^{2})(1 – x^{3})(1 – x^{4}) …
= 1 – {x/(1 – x)} + x

^{3}/{(1 – x)(1 – x^{2})} – x^{6}/{(1 – x)(1 – x^{2})( 1 – x^{3})} + …
So we can
see here how the powers x in the numerator terms follow the pattern of the
triangular numbers, i.e.

1, 3, 6,
10, 15, 21, 28, 36, 45, 55, 66, …

In fact
there are many fascinating connections as between the pentagonal and triangular
numbers.

For example,
when a pentagonal number is multiplied by 3, a triangular number results.

So 12 (the 5

^{th}generalised pentagonal number) * 3 = 36 (the 8^{th}triangular number).
Also when
one combines the powers of x in the pentagonal number theorem in couplets (as
they naturally occur with two positive terms alternating with the two negative
terms), and then divide by 3, then the absolute value represents a triangular number.

So – 1 – 2 = – 3 and 3/3 = 1 (the 1

^{st}triangular number)
– 1 – 2 + 5 + 7 = 9 and 9/3 = 3
(the 2

^{nd}triangular number)
– 1 – 2 + 5 + 7 – 12 – 15 = – 18 and 18/3 = 6 (the 3

^{rd}triangular number)
– 1 – 2 + 5 + 7 – 12 – 15 + 22 + 26 = 30 and 30/3 = 10 (the
4

^{th}triangular number).
And the full sequence of triangular numbers can be generated
in this fashion.

In a reverse manner, if we take the triangular numbers in
groups of 3, a direct connection with the pentagonal numbers can be shown.

So 1 + 3 + 6 = 10. Then dividing by 9 (and ignoring the
remainder of 1) we get 1 (the 1

^{st}of the regular pentagonal sequence).
Then for the next group of 3 triangular numbers 10 + 15 + 21
= 45. then again dividing by 9 (and ignoring remainder of 1) we get 5 (the 2

^{nd}of the pentagonal numbers).
For the next group 28 + 36 + 45 = 109. Again dividing by 9 (and
ignoring the remainder of 1) we get 12
(the 3

^{rd}pentagonal number)
Finally to illustrate for the next group 55 + 66 + 78 = 198.
And dividing by 9 (and ignoring remainder of 1) we get 22 (the 4

^{th}pentagonal number).
So the regular pentagonal number series 1, 5, 12, 22, … can
be generated in this manner.

The “shadow” pentagonal sequence (that is generated when n
takes on negative values)

can also be generated from successive groupings of 3 triangular
numbers.

Here we start with the 2

^{nd}triangular number with the 3 numbers in sequence 3, 6 and 10 respectively.
So the sum = 19 which when divided by 9 (again ignoring the
remainder of 1) = 2.

The next 3 triangular numbers in sequence gives 15 + 21 + 28
= 64. Then dividing by 9 (and ignoring remainder) we obtain 7.

Again finally to illustrate the next grouping of 3 triangular
numbers = 36 + 45 + 55 = 136. Then dividing by 9 (and ignoring remainder) we
get 15.

So we have now generated the first 3 members of the “shadow”
pentagonal series 2, 7, 15, …

Then the generalised pentagonal sequence combines the
regular sequence with its shadow in ascending order of magnitude.

Again with respect to the powers of x in the pentagonal
number theorem, another fascinating observation can be made.

This time we successively take each natural couplet (of
positive and negative terms) and divide by 3 concentrating on the absolute
value of the result.

So the first two negative powers are 1 and 2 and 1 + 2 = 3
with 3/3 = 1. So this is the 1

^{st}term of our new series.
The next couplet of positive powers is 5 and 7 and 5 + 7 =
12 with 12/3 = 4. So this is the 2

^{nd}term of the series.
Then the next two negative powers are 12 and 15 and 12 + 15
= 27 with 27/3 = 9. So 9 is the 3

^{rd}term of the series.
Finally to illustrate the next two positive powers are 22
and 26 and 22 + 26 = 48 with 48/3 = 16. So 16 is the 4

^{th}terms of our series.
So the new series that can be successively generated in this
manner is 1, 4, 9, 16, … (i.e. the squares of the natural numbers).

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