However
this in fact represents but a specific example of a more general number
phenomenon.

Now when we
return to the octagonal numbers, we may recall that the two digits that occur
in the respective number bases 2, 5, 8, … (when converting each term to its
appropriate number base) represent the unique 2-digit recurring sequence of the
reciprocal of 3 in each of these bases.

So again
for example, 8 in base 5 = 13 (with 31 – 13 = 13). 13 then equally
represents the unique 2-digit sequence of 1/3 in this base (.131313…)

Thus the number in question here (to which 1st reciprocal
relates) is 3.

And the 1st relevant number base (through which the
respective terms of the octagonal sequence are expressed) = 2 (i.e. 3 – 1). And
then subsequent number bases keep increasing by 3.

So we now can express a more general number phenomenon in
these terms.

Let n represent a number. Then when we obtain the unique digit
sequence of its reciprocal (1/n) in the respective number bases n – 1, 2n – 1,
3n – 1, …, a unique recurring connection will characterise the
relationship between each resulting 2-digit number and its reverse (in its
respective number base).

The simplest case occurs when n = 2, Therefore the
appropriate number bases here to express the unique digit sequence of 1/2 are 1, 3, 5, …

Now 1/2 expressed in base 1 does not have a meaningful
expression. But 1/2 in base 3 is .111… Though there is only one recurring digit in this case, we will preserve the first two digits as 2 unique digits will arise when n> 2. So the digit sequence here is 11.

And in this case 11 is equal to its reverse. So 11 (reverse)
= 11 (original number) * 1. And 11 in base 10 = 4.

And this is equally the case for all subsequent number
bases. For example in base 5, the reciprocal of 2 = .222… So 22 (reverse) = 22
(original number) * 1. And 22 in base 10 = 12

So when the number (to which the reciprocal relates) is 2
with relevant number bases 3, 5, 7, …,

reverse = original number * 1

Now when expressed in base 10, these numbers in
the respective number bases leads to a unique sequence i.e. the triangular numbers * 4,

0, 4, 12, 24, …

Then as we have seen in the next case, where 3 is the number
to which the reciprocal relates (bases 2, 5, 8, …), a unique 2-digit sequences
arise.

And in all these cases (as for example 31 and 13 in base 5),

reverse = original number * 2.

And these numbers in their respective number bases,
expressed in base 10, lead to the octagonal sequence of numbers,

1, 8, 21, 40, …

Then when we subtract each term of the original sequence from the octagonal sequence we get

1, 4, 9, 16, ... (i.e. the sum of squares).

1, 4, 9, 16, ... (i.e. the sum of squares).

And continuing on, in the next case, where 4 is the starting
number to which the reciprocal relates, the relevant number bases (for
expressing its unique 2-digit sequence) are 3, 7, 11, …

So for example in base 3, 1/4 = .0202…

Thus the unique 2-digit sequence = 02.

Then when we subtract 02 from its reverse we get 20 – 02 =
11

So in denary terms the reverse = 6 and the original number =
2.

Then in base 7, 1/4 =
.1515…

Thus the unique 2-digit sequence is 15.

And 51 in denary terms = 36 and 15 = 12

So generalising for all such cases related to starting
number 4,

reverse = original number * 3

Once again a unique sequence is associated with each of
these original numbers, when expressed in a denary manner i.e.

2, 12, 30, 56, …

In fact we can see a discernible pattern emerging as we move to higher
number bases.

Thus the original number in base 3 = 02, in base 7 = 15, in
base 11 = 28, base 15 = 3A and so on.

And 2, 12, 30, 56 = 2(1, 6, 15, 28, …)

The terms inside the brackets constitute the hexagonal
sequence (which comprises the odd numbered terms of the triangular sequence).

It is fascinating in this context that if now subtract each
term of the previous octagonal sequence from each corresponding term of this
new sequence, we again obtain

1, 4, 9, 16, … (i.e. the squares of the natural numbers).

So using just one final case for illustration, when the
starting number is 5, we consider 1/5 in bases 4, 9, 14, …

1/5 in base 4 = .0303…

Therefore the unique 2-digit sequence = 03 which now
constitutes our original number in base 4.

And the 30 = 12 (in denary terms)

So 30 = 03 * 4 (in base 4).

In base 9, 1/5 = .1717…

Therefore the unique 2-digit sequence = 17, which now
constitutes our original number in base 9.

And 17 = 16 and 71 (the reverse) = 64 (in denary terms).

So 71 = 17 * 4 (in base 9).

So generalising for all such cases related to starting
number 5,

reverse = original number * 4.

Then the corresponding unique associated number sequence (in
base 10) is

3, 16, 39, 72, … (A147874
in OEIS)

Once again when we subtract each terms of the previous
sequence from the corresponding terms of the new sequence, we obtain,

1, 4, 9, 16,…(the squares of the natural numbers)

Expressed in even more general terms when the starting
number (to which the reciprocal 1/n relates in number bases n – 1, 2n – 1, 3n –
1, … then with respect to the original numbers (based on the unique digit
sequence of the reciprocal)

reverse = original number * (n – 1)

Thus when the starting number is 6

reverse = original number * 5

We can readily confirm this for 1/6 in base 5 = .0404…

Therefore the unique digit sequence = 04

So the relevant original number (in base 5) = 04 and reverse
40 (i.e. 4 and 20 in denary terms)

Thus 40 = 04 * 5.

Incidentally the sequence (in base 10) associated with these
numbers is

4, 20, 48, 88, … = 4
(1, 5, 12, 22, …)

And the terms inside the brackets comprises the pentagonal
sequence.

Once again when we subtract each term of the previous
sequence from each corresponding term of the present sequence we obtain

1, 4, 9, 16, … (i.e. the squares of the natural numbers). And
this appears to be universally the case.

We can therefore write down immediately the unique
digit sequence in the next case(for n = 7) by adding each of these sum of square terms to the
respective term in the previous sequence, to obtain

5, 24, 57, 104, …

So 5 in base 6 = 05

And 50 = 05 * 6 (in base 6)

Likewise 24 in base 13 = 1B

And its reverse B1 = 11* 13 + 1 = 144

And 144 (reverse) = 24 (original number * 6).

So starting with the original sequence (where n = 2), we can directly
generate all further sequences by simply adding each term of the original
sequence to the corresponding term of the sum of squares.system.

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