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) + 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) + 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 …

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) + p(n – 15) – ...

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).

Fascinating Palindrome Connection (2)

The approach in the last entry can be generalised for all self-generating numbers, where in the appropriate number bases, the original starting number * k (where k = 2, 3, 4, 5, …) = its reverse number.

When n = 4, a unique repeating 2-digit sequence applies to 1/n in bases 2n – 1, 3n – 1, 4n – 1, …, so
that the original starting number (using these 2 digits) * 3 = its reverse number.

And just as the k^th term in the previous case “Fascinating Palindrome Connection 1”, where n = 3, is given by 3n² – 2n, the k^th term (where n = 4) is given by 4n² – 2n, resulting in the series,

2, 12, 30, 56, 90, …

So the relevant number bases, where this form of self-generating behaviour applies is in the number bases 7, 11, 15, … (Once again we omit the number base where n – 1 = 3, applying to the 1^st term of the series i.e. 2, as this results in a redundant 1^st digit of 0 in the original starting number).

However the 2^nd term, applying to base 7 is fully valid.

And 12 in base 7 = 15.

Therefore, 15 * 3 = 51 (in base 7).

However this can equally be expressed as 2 * (2n – 2) = 2 * 6 = 12 (in base 10), i.e. 15 in base 7.

So the subsequent 2-digit starting terms are given by 3 * 10 (= 28 in base 11), 4 * 14 (= 3B in base 15), 5 * 18 (= 4E in base 19) and so on.

Thus  28 * 3 = 82 (in base 11)

         3B * 3 = B3 (in base 15)

         4E * 3 = E4 (in base 19)

                  …

Then one 3-digit term in each case is obtained from 22 * 6 in base 7, 33 * 10 (in base 11), 44 * 14 (in base 15), 55 * 18 (in base 19) and so on.

And as we saw in “Fascinating Palindrome Connection 1” one can then use the formulas k/2(k/2 + 1), where k is odd and k/2 * k/2 where k is even, to calculate the total number of self-generating numbers (of a particular type) up to (and including) k digits.

So for example the total collection of self-generating numbers (where the original starting number * 3 = its reverse) -up to an including 4 digits - in each relevant number base, is    

4/2 * 4/2 = 4.

And in base 7, these are 15, 165, 1515 and 1665 respectively.

Fascinating Palindrome Connection (1)

I wish to correct a recent assertion regarding the frequency of self generating numbers in different number bases.

What I mean in this context is a number that when multiplied by a positive integer (> 1) = its reverse.

So for example in base 8, 25 (original starting number) * 2 = 52 (i.e. its reverse number).

Now I claimed that where such a 2-digit number exists (in an appropriate base), that one example will then occur for every higher digit number (in the same number base).

However I have since discovered, through an interesting number connection, that this is not in fact strictly true.

Once again when we start with a positive integer n (> 1), then the reciprocal of n will result in a unique repeating 2-digit decimal sequence in the corresponding number bases

2n – 1, 3n – 1, 4n – 1, …    

So again for example when n = 3, then 1/n results in this unique 2-digit sequence in base 5, 8, 11, …

And then when we take these 2 digits as the original starting number and subtract it from its reverse, the same starting number will result.

Alternatively, when we multiply the starting number by 2, we obtain its reverse in these number bases.

Thus in base 5, 13 * 2 = 31; in base 8, 25 * 2 = 52; in base 11, 37 * 3 = 73, and so on.

However there is another revealing perspective with respect to these numbers.

Thus when we divide each of these numbers by the number that is one less than the number base in question i.e. 2n – 2, 3n – 2, 4n – 2, … respectively we obtain 2, 3, 4, …

Now once again we are excluding the 2-digit number in base 2 i.e. 01, as the first digit as a 0 is strictly redundant.

So 2 * 4 = 13 (in base 5); 3 * 7 = 25 (in base 8); 4 * 10 = 37 (in base 11) and so on.

Now a corresponding higher digit numbers (where starting number * 2 = its reverse number) can now be created with respect to the original starting number provided that its palindrome nature is preserved and where the inclusion of one or more zeros is allowed (between first and last digits).

So when the palindrome has 2 digits, then only one possibility arises.

Thus in base 5, 22 * 4 = 143 and 143 (starting number) * 2 = 341 (its reverse number).

In like manner, in base 8,  33 * 7 = 275 and 275 (starting number) * 2 = 572 (its reverse number).

And in base 11, 44 * 10 = 3A7 and 3A7 (starting number) * 2 = 7A3 (its reverse number).

Thus there is indeed only one example of a self-generating number (of this type) with respect to all 3-digit numbers in bases 5, 8, 11, …

However when the palindrome now has 3 digits, two possibilities exist

So with respect for example to base 8, we can have 333 or 303.

Thus in base 8, we have as our possible starting numbers 333 * 7 = 2775 and 2775 (starting number) * 2 = 5772 (its reverse number).

However, we also have 303 * 7 = 2525 and 2525 (starting number) * 2 = 5252 (its reverse number).

Thus for 4-digit numbers in bases 5, 8, 11, …, we have 2 examples of such self-generating numbers (i.e. where the starting number * 2 = its reverse number).

Now when the initial palindrome has 4 digits, again illustrating with respect to base 8, we have just two possibilities i.e. 3333 or 3003.

And in base 8, 3333 * 7 = 27775 and 27775 (starting number) * 2 = 57772 (its reverse).

However, we also have 3003 * 7 = 25025 and 25025 (starting number) * 2 = 52052 (its reverse).

So for 4 and 5-digit numbers, we have two examples in each case of such self-generating numbers.

Then when we go to 6 and 7-digit numbers, we have three examples in each case, with 8 and 9-digit numbers 4 examples, with 10 and 11-digit numbers five examples and so on.

Thus when k is an even integer we have k/2 examples of such numbers in the relevant number bases.

And when k is an odd integer, we have (k – 1)/2 examples.

And the cumulative number of such numbers up to and including k digits is k/2(k/2 + 1), where k is odd and k/2 * k/2 where k is even.

Therefore, for example the total number of such self generating numbers up to and including 6 digits (i.e. where the reverse is twice the original starting number), in the appropriate number bases  5, 8, 11, ..., = 3 * 3 = 9.