Pentagonal Number Theorem (1)

I was reading again recently that marvellous little book by Davis Wells “The Penguin Dictionary of Curious and Interesting Numbers” when my attention was drawn to the section on pentagonal numbers.

Now the pentagonal numbers are generally listed as,

1, 5, 12, 22, 35, 51, 70, …, with the n^th pentagonal number given as n(3n – 1)/2 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 – x²)(1 – x³)(1 – x⁴) … he discovered that the first few terms were,

1 – x¹ – x²  + x⁵ + x⁷ – x¹² – x¹⁵ + …

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 – x²)(1 – x³)(1 – x⁴) …  

= 1 – x/{(1 – x)} + x³/{(1 – x)(1 – x²)}  – x⁶/{(1 – x)(1 – x²)( 1 – x³)} + …

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 = 46, and dividing by 9 (while 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).