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

Polygonal Numbers and e (1)

A polygonal number is a number that can be expressed as dots in the shape of a regular polygonal. So we for example the well known triangular numbers,

1, 3, 6, 10, 15, …

For simplicity we can express this series as 3-gonal (where n = 3).

 

Then we have the square numbers

1, 4, 9, 16, 25, …,

which we can express as 4-gonal (where n = 4)

 

And we have the pentagonal numbers

1, 5, 12, 22, 35, …,

which we can express as 5-gonal (where n = 5)

 

In fact the earlier series,

1, 2, 3, 4, 5, …,

can be expressed as 2-gonal (where n = 2)

And we can have continue without limit to 6-gonal 7-gonal, 8-gonal … n-gonal series.

For example the 8-gonal series (where n = 8) is that of the well known octagonal numbers,

1, 8, 21, 40, 65, …

 

Then in general for all polygonal series,

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

where a_1, the 1^st term = 1.

 

So, for example with the simplest 2-gonal series

(1/0! + 2/1! + 3/2! + 4/3! + 5/4! + …)/e   = 2 + 1 – 1 = 2

 

And then with the 3-gonal (Triangular) series

(1/0! + 3/1! + 6/2! + 10/3! + 15/4! + …)/e   = 3 + 1.5 – 1 = 3.5

 

Then more generally

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

 

So with the 3-gonal (Triangular) series,     

1, 3, 6, 10, 15, …, where n = 3,

if k = 4, then a_{k + 1} = 15 (i.e. the 5^th term in the series).

Therefore,

10/0! + 15/1! + 21/2! + 28/3! + …)/e  = 15 + 1.5 – 1 = 15.5

 

For the 8-gonal (Octagonal)  1, 8, 21, 40, 65, 96, …   i.e. n = 8

If k = 1, then a_{k + 1} = 8

Therefore,

1/0! + 8/1! + 21/2! + 40/3! + …)/e  = 8 + 4 – 1 = 11

 

Then if k = 3, a_{k + 1} = 40 (i.e. the 4^th term)

Therefore,

21/0! + 40/1! + 65/2! + 96/3! + …)/e  = 40 + 4 – 1 = 43

 

What is remarkable from all of this is that e can therefore be computed from any polygonal i.e. n-gonal series starting with the 1^st term, or indeed the k^th term (where k = 1, 2, 3, …) using the formula   

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

 

For example when n = 24 we get what are called the icositetragonal numbers,

1, 24, 69, 136, 225, 336, 469, 624, 801, 1000, …

So if, say,  k = 3, then

e = (69/0! + 136/1! + 225/2! + 336/3! + 469/4! + 624/5! + 801/6! + 1000/7!…)/(136 + 12 – 1)

= (69/0! + 136/1! + 225/2! + 336/3! + 469/4! + 624/5! + 801/6! + 1000/7!…)/147

= 2.718044…

So using just the 1st 8 terms, we already can compute the value of e accurate to 3 decimal places.  

Generalised Formulae for e

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + …,

is a special case of

e = {1 + (1 + k)/1! + (1 + 2k)/2! + (1 + 3k)/3! + (1+ 4k)/4! + …}/(1 + k),

where k = 0.

So, when k = 1,

e = {1 + 2/1! + 3/2! + 4/3! + 5/4! + …}/2

And when for example k = 10,

e = {1 + 11/1! + 21/2! + 31/3! + 41/4! + …}/11

This relationship holds for all real values of k (including non-integers) except – 1.

 

The case of k = – 1 is fascinating as it suggests that,

e ={(1 + 0/1! – 1/2! – 2/3! – 3/4! – …)}/0

And the limiting value of the expression inside the brackets as the number of terms

n ⁓ ∞ = 0.

So in this case the suggested limiting value of 0/0 = e?

 

Then,

e^x = x⁰/0! + x¹/1! + x²/2! + x³/3! + x⁴/4! + …

So when x = 1,

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + …

 

And when x = 2,

e² = 2⁰/0! + 2¹/1! + 2²/2! + 2³/3! + 2⁴/4! + …

 

Therefore, combining the two formulae,

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + …,

is a special case of,

e^x(1 + 2k) = x⁰/0! + x¹(1 + k)/1! + x²(1 + 2k)/2! + x³(1 + 3k)/3! + x⁴(1 + 4k)/4! + …

for all x where x = 1 and k = 0.

Then when k = 0,

e ^x = x⁰/0! + x¹/1! + x²/2! + x³/3! + x⁴/4! + …

And when also x = 1,

e = 1⁰ + 1¹/1! + 1²/2! + 1³/3! + 1⁴/4! + …

However, if for example k = 2 and x = 2,

Then 5e² = 1⁰/0! + (3*2¹)/1! + (5*2²)/2! + (7*2³)/3! + (9*2⁴)//4! + …

e ² = {1⁰/0! + (3*2¹)/1! + (5*2²)/2! + (7*2³)/3! + (9*2⁴)//4! + …}/5

so that,

e = [{1⁰/0! + (3*2¹)/1! + (5*2²)/2! + (7*2³)/3! + (9*2⁴)//4! + …}/5]^1/2

Ramanujan's Tau Function and Euler's Pentagonal Number Theorem

Ramanujan found that when the infinite series,

x(1 – x)²⁴(1 – x²) ²⁴(1 – x³)²⁴(1 – x⁴)²⁴ …  is multiplied out we obtain

1x – 24x² + 252x³ – 1472x⁴ + 4830x⁵ – 6048x⁶ – 16744x⁷ + 84480x⁸ – … 

This is now known as the Ramanujan tau function.

Now if we divide by x then

(1 – x)²⁴(1 – x²)²⁴(1 – x³)²⁴(1 – x⁴)²⁴ …  

= {(1 – x)(1 – x²)(1 – x³)(1 – x⁴) …}²⁴  

 = 1 – 24x + 252x² – 1472x³ + 4830x⁴ – 6048x⁵ – 16744x⁶ + 84480x⁷ – …  

So the tau values for each term thereby remain unchanged.

Interestingly, Euler earlier discovered that

(1 – x)(1 – x²)(1 – x³)(1 – x⁴) …

= x⁰ – x¹ – x²  + x⁵ + x⁷ – x¹² – x¹⁵ + x²² + x²⁶ – …

So the powers of the x terms in this infinite series comprise the generalised pentagonal numbers

i.e. 0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, …

This thereby means that we can show a fascinating connection as between these generalised pentagonal numbers and the corresponding Ramanujan tau values.

Thus {x⁰ – x¹ – x²  + x⁵ + x⁷ – x¹² – x¹⁵ + x²² + x²⁶ – …}²⁴

= 1 – 24x + 252x² – 1472x³ + 4830x⁴ – 6048x⁵ – 16744x⁶ + 84480x⁷ – …  

Then a related Euler formula shows that

(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 this latter expression involves (in the sole x term of each numerator) the triangular numbers 1, 3, 6, 10, 15, …

Therefore we can equally show a fascinating connection as between the Ramanujan tau values and the triangular numbers.

Pentagonal Number Theorem (4)

I mentioned in yesterday’s entry the complementary nature of the two product expressions relating to the Riemann zeta function and the new derived formula (based on the pentagonal number theorem).

So on a n * n grid (n rows and n columns respectively) the entries (in the horizontal rows) represent the product terms for the Riemann zeta function ζ(s), (s ≥ 1), whereas the corresponding entries (in the vertical columns) entries represent the reciprocals of the product entries (1/x > 1) based on our derived formula,

(1 – x)(1 – x²)(1 – x³)(1 – x⁴) … = 1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x² – 1)} –
1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)} + 1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)(1/x⁴ – 1)} – …

Now the Riemann zeta function provides important information regarding the collective relationship of the primes to the natural number system.

Therefore in complementary manner the derived formula should provide important information on the individual nature of each prime.

With this in mind I then set about looking for a way of using the formula to test for primality (of individual numbers).

Now when we start with x = 1/ 2 (with 1/x = 2) the formula generates the terms 1, 3, 7, 15, … i.e. 2ⁿ – 1, for n = 1, 2, 3, 4, …

I had earlier shown how this number sequence is in turn related to the unique number sequence associated with (x – 1)(x – 2) = x² – 3x + 2 which is 1, 3, 7, 15, …

So in this case we multiply the unique terms of this sequence * 1 to obtain the corresponding numbers that the formula generates for 1/x = 2.

I had also shown that when x = 1/3 (with 1/x = 3) the formula generates the terms 2, 8, 26, 80 i.e. 3ⁿ – 1, for n = 1, 2, 3, 4, …

And this in turn is related to the unique number sequence associated with (x – 1)(x – 3) = x² – 4x + 3, which is 1, 4, 13, 40, …

So in this case, we multiply the unique terms of the sequence * 2 to obtain the corresponding numbers that the formula generates for 1/x = 3.

And in general when we start with x = 1/k (1/x = k), the formula generates the terms
kⁿ – 1, for n = 1, 2, 3, 4, …

And this in turn is related to the unique number sequence associated with (x – 1)(x – k) = where we multiply the unique terms of the sequence * (k – 1) to obtain the corresponding numbers that the formula generates for 1/x = k.

It then struck me that if we subtract k – 1 from kⁿ – 1 i.e. kⁿ – 1 – k + 1 = kⁿ – k that
(kⁿ – k)/n would offer a good test of primality.

Fro example in the simplest case where k = 2, we have (2ⁿ – 2)/n which invariably results in an integer when n is prime.

For example, when n = 2, (2² – 2)/2 =  1
                                n = 3, (2³ – 2)/3 =  2
                                n = 5, (2⁵ – 2)/5 =  6
                                n = 7, (2⁷ – 2)/7 = 18

Furthermore when n is not prime, (2ⁿ – 2)/n is not an integer.

For example when n = 4, (2⁴ – 2)/4 =  14/4 (which is not an integer)
                     when n = 6, (2⁶ – 2)/6 =  62/6 (which is not an integer)
                     when n = 8, (2⁸ – 2)/8 = 254/8 (which is not an integer)
                     when n = 9, (2⁹ – 2)/9 = 510/9 (which is not an integer).


So in this case, where k = 2, (2ⁿ – 2)/n offers a clear cut test for primality

Thus for any integer value of n if (2ⁿ – 2)/n is perfectly divisible by n (so that n is thereby a factor) then n is prime; if not perfectly divisible then n is not prime.

When k > 3, results are not quite so clearcut.

It still is true that when n is prime, and not a factor of k, (kⁿ – k)/n will result in an integer. Unfortunately however n will no longer be exclusively confined to primes.

For example when k = 3, (3⁶ – 3)/6 = 726/6 = 121. However n = 6 is not prime.
However for all other values where (3ⁿ – 3)/n results in an integer up to but not including 66, n is prime.

Then when k = 4 many non-prime values for n will occur.

For example (4⁴ – 4)/4 = 252/4 = 63 (but 4 is not prime)  

              And (4⁸ – 8)/8 = 65,528/8 = 8191 (but 8 is not prime).

Then when k = 5 again many non-prime values will occur. For example in the first 20, n = 4, 5, 10, 15 are all non-prime.

There are close connections here with Fermat’s Little Theorem. However what is important is the context in which my own investigation has arisen.

Pentagonal Number Theorem (3)

Let us return briefly to the formula given in the previous entry i.e.

(1 – x)(1 – x²)(1 – x³)(1 – x⁴) … = 1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x² – 1)} –
1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)} + 1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)(1/x⁴ – 1)} – …

We have seen that when x = 1/2, this results in the relationship,

1/2 * 3/4 * 7/8 * 15/16 * …  = 1 – 1/(1) + 1/(1 * 3) – 1/(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) – …

Then when x = 1/3, the following relationship results,

2/3 * 8/9 * 26/27 * 80/81 … = 1 – 1/(2) + 1/(2 * 8) – 1/(2 * 8 * 26) + 1/(2 * 8 * 26 * 80) – …

When x = 1/5, we get,

4/5 * 24/25 * 124/125 * 624/625 … = 1 – 1/(4) + 1/(4 * 24) – 1/(4 * 24 * 124) +

1/(4 * 24 * 124 * 624) – …

And when x = 1/7, we get,

6/7 * 48/49 * 342/343 * 2400/2401 … = 1 – 1/(6) + 1/(6 * 48) – 1/(4 * 48 * 342) +

1/(4 * 48 * 342 * 2400) – …

                                                                                                                                        _∞

If we now look at the product over primes version of the Riemann Zeta Function for ζ(s)

                                                                                                                                       ^s=1

we get the following

ζ(1) = 2/1 * 3/2 * 5/4 * 7/6 * …

ζ(2) = 4/3 * 9/8 * 25/24 * 49/48 * …

ζ(3) = 8/7 * 27/26 * 125/124 * 343/342 * …

ζ(4) = 16/15 * 81/80 * 625/624 * 2401/2400 * …

                   …

What is remarkable here is that what represent rows of numbers with respect to the Riemann Zeta function, represent similar columns with respect to the reciprocals of the products generated through our formula i.e.

(1 – x)(1 – x²)(1 – x³)(1 – x⁴) … = 1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x² – 1)} –
1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)} + 1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)(1/x⁴ – 1)} – …
So for example the 1st row with respect to the Riemann Zeta function is,

2/1 * 3/2 * 5/4 * 7/6 * …

And the corresponding 1^st column with respect to the products arising from the formula is

1/2 * 2/3 * 4/5 * 6/7 * …

This suggests therefore that there is an intimate relationship here as between the stated formula (that generates the product results) and the corresponding formula that generates the product results for the Riemann zeta function.

Thus is we were to generate a n * n grid (i.e. with n rows and n columns) with respect to the product entries for both our formula and the Riemann zeta function, the combined products of products (whereby the values of all rows and then values of all columns are multiplied together) in either case would represent the reciprocal of the alternative result. However we would need here to exclude the 1^st row and 1^st column to ensure a finite result!

And it has to be remembered in this context that likewise that each product over primes expression can be given an alternative sum over the natural numbers expression.

So in the case of the Riemann zeta function, the corresponding sum over the natural numbers expression for the corresponding product over primes expression is the harmonic series i.e.

1/2 * 3/2 * 5/4 * 7/6    =   1/1¹ + 1/2¹ + 1/3¹ + 1/4¹ + … (which in this one particular case is divergent).

Thus strictly speaking we are obtaining here the sum over the natural numbers for the base aspect of number where by contrast the dimensional aspect is fixed in each case to just one of the natural numbers (in this case 1).

Then with respect to our formula the sum over natural numbers expression again is given as 

1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x² – 1)} –
1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)} + 1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)(1/x⁴ – 1)} – …

So here in complementary fashion it is the dimensional aspect that varies over the natural numbers with the base aspect in each given case fixed.

Thus when x = 1/2 with 1/x = 2, we have 

1/2 * 3/4 * 7/ 8 * 15/16 * …  =  1 – 1 + 1/3 – 1/21 + 1/315 – …


We have seen already how the Zeta 2 function can be used to provide a similar vertical (column) readout of the entries for the Riemann Zeta function where terms are added (rather than multiplied)

Now excluding values for s = 1, the sum of 1^st column will be 1/4 + 1/8 + 1/16 + 1/32 +

 = (1 + 1/4 + 1/8 + 1/16 + 1/32 + …)  – 1.

And this corresponds to the Zeta 2 function (1 + x + x² + x³ + …) – 1, with x = (1/p² – 1) (where p = 2) = 1/2.

                                                                                  _∞

And we used this earlier relationship to prove that ∑{ζ(s) – 1)} = 1.

                                                                                                                   ^s=2 

Pentagonal Number Theorem (2)

In yesterday’s entry, I mentioned in connection with the Pentagonal Number Theorem a related Euler formula i.e.

(1 – x)(1 – x²)(1 – x³)(1 – x⁴) …      =  1 – {x/(1 – x)} + x³/{(1 – x)(1 – x²)}
– x⁶/{(1 – x)(1 – x²)( 1 – x³)} + …

I then decided to use this in testing this expression by letting x = 1/2, to come up with this interesting result i.e.

1/2 * 3/4 * 7/8 * 15/16  = 1/(1 * 3) – 1(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + … = .288788…

In fact fully expressed the RHS,

 = 1 – 1/(1)  + 1/(1 * 3) – 1(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + …

I then realised that this provided just one especially interesting application of a general formula that can be written as follows

(1 – x)(1 – x²)(1 – x³)(1 – x⁴) … = 1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x² – 1)} –

1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)} + 1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)(1/x⁴ – 1)} – …

And this formula will converge where x < 1.

So for example, where x = 1/3 we obtain,

2/3 * 8/9 * 26/27 * 80/81 * …  = 1 – 1/(2) + 1/(2 * 8) – 1/(2 * 8 * 26) +

1/(2 * 8 * 26 * 80) – … = .56012 …

I also soon discovered the closely related general formula that can be written as follows

(1 + x)(1 + x²)(1 + x³)(1 + x⁴) … = 1 + 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x² – 1)} +

1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)} + 1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)(1/x⁴ – 1)} + …

So once again, illustrating where x = 1/2, we obtain,

3/2 * 5/4 * 9/8 * 17/16 * …   = 1 + 1/(1) + 1/(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + …

Therefore once again, when for each term on the LHS we decrease the numerator by 2 we obtain on the RHS the alternating version of the same terms with

1/2 * 3/4 * 7/8 * 15/16  = 1/(1 * 3) – 1/(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + …

There is a fascinating connection here with e and  1/e respectively.

In other work on this blog I have shown how we can associate a unique infinite sequence of terms with every regular polynomial equation.

Perhaps the best known sequence is the Fibonacci which arises from the equation

x² – x – 1.

Then starting with 0, 1 we multiply the coefficient of the x term (i.e. – 1) by – 1 then multiply by the 2^nd of the starting digits (i.e. 1 and then likewise multiply the coefficient of the 3^rd term (i.e. – 1) again by – 1 before multiplying the result by the 1^st of the starting digits (i.e. 0). Then combining the 2 answers we get 1 + 0 = 1, which is then the next digit in the Fibonacci sequence.

And by concentrating on the two most recent digits generated we can continue to generate subsequent terms in the sequence.

However the same general procedure can in principle be applied to all polynomial equations (though it becomes more cumbersome for higher degree equations).

So the Fibonacci sequence is 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, …

And the significance of the terms in the sequence is that they can be used dividing the

(n + 1)^th by the n^th to approximate the real roots for the associated equation x² – x – 1 = 0.

So for example 233/144 = 1.618055 already provides a good approximation of the positive real root (i.e. phi = 1.618033…).

Now if we consider (x – 1)( x – 1) we obtain the polynomial equation x² – 2x + 1

And using the same procedure for generating subsequent terms (as for the Fibonacci sequence) we obtain 0, 1, 2, 3, 4, 5, …

In other words, the unique digit sequence associated with (x – 1)(x – 1) = (x – 1)² 

is the natural number sequence.

And of course we can use this sequence corresponding to x² – 2x + 1 = 0, to approximate the positive real root (i.e. 1) of the equation.

So as the n^th term = the (n + 1)^th/n ^th term = (n + 1)/n ~ 1..

Now the well known expression for e is as follows

e = 1 + 1/(1) + 1/(1 * 2) + 1/(1 * 2 * 3) + 1/(1 * 2 * 3 * 4) + …

1/e = 1 –  1/(1) + 1/(1 * 2) + 1/(1 * 2 * 3) –  1/(1 * 2 * 3 * 4) + …

So the factorial expressions in the denominator entail the successive products of the natural numbers.

Let us now look at the unique digit sequence associated with (x – 1)(x – 2) = x² – 3x + 2

which is 0, 1, 3, 7, 15, 31, …

And once again the (n + 1)^th/n ^th term e.g. 31/15, will approximate the principle real valued root of the associated equation i.e. x² – 3x + 2 = 0, = 2.

If we look again at the two related expressions for 3/2 * 5/4 * 9/8 * 17/16 * …   and 1/2 * 3/4 * 7/8 * 15/16 , i.e.

1 + 1/(1) + 1(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + …  and

1 – 1/ (1) + 1/(1 * 3) – 1(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + …,

we can see how they readily match the corresponding expressions for e and 1/e respectively.

Whereas in the expressions for e and (i/e) the successive products of  the unique digit sequence associated with (x – 1)(x – 1) are used, in the latter case in the corresponding expressions for 3/2 * 5/4 * 9/8 * 17/16 * …   and 1/2 * 3/4 * 7/8 * 15/16,  the successive products of the unique digit sequence associated with (x – 1)(x – 2) are now used.

So in our two related formulae,

(1 + x)(1 + x²)(1 + x³)(1 + x⁴) … = 1 + 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x² – 1)} +

1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)} + 1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)(1/x⁴ – 1)} + …

and

(1 – x)(1 – x²)(1 – x³)(1 – x⁴) … = 1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x² – 1)} –

1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)} + 1/{(1/x – 1)(1/x² – 1)(1/x³ – 1)(1/x⁴ – 1)} – …,

1/x = 2.

And the numbers in the product denominator sequence for the RHS of the expression correspond here with the unique numbers associated with (x – 1)(x – 2).

And where 1/x = n (as integer), the numbers in the product denominator sequence for the RHS of the expression correspond to (x – 1)(x – n). Strictly each number in the product sequence relates to the corresponding unique numbers - associated with (x – 1)(x – n) - multiplied by 1/x –  1. So, as we have seen, when 1/x = 2 the unique number sequence associated with (x – 1)(x – 2) directly applies!

For example when 1/x = 3, we now use the unique digits associated with (x – 1)(x – 3) = x² – 4x + 3, which are,

0, 1, 4, 13, 40, 121, …

Therefore the appropriate denominator sequence applying entails each of these terms * 2

i.e. 2, 8, 26, 80, 242, …

And as we have already seen when 1/x = 3,

2/3 * 8/9 * 26/27 * 80/81 * …  = 1 – 1/(2) + 1/(2 * 8) – 1/(2 * 8 * 26) +

1/(2 * 8 * 26 * 80) – … = .56012 …

So likewise,

4/3 * 10/9 * 28/27 * 82/81 * …  = 1 + 1/(2) + 1/(2 * 8) + 1/(2 * 8 * 26) +

1/(2 * 8 * 26 * 80) + … = 1.56493 …