Extending Relationships for Octagonal Numbers

We have been looking at the octagonal sequence (and its “shadow” sequence) showing how its terms when expressed in appropriate number bases leads naturally to one important form of self-generating numbers i.e. where the same number results, when subtracted from its reverse.

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).

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.