1, 8, 21, 40, 65, 96, ..., in the respective bases, 2, 5, 8, 11, 14, ..., where they are self generating to that when each number, in its respective number base is subtracted from its reverse, the same number re-occurs.

So for example the appropriate number base to consider 21, is base 8.

And 21 (in denary terms) = 25 (in base 8). Thus the reverse of 25 = 52.

And 52 – 25 = 25.

Now another way of expressing this result is that 25 = 52 * 1/2 (in base 8).

Equally we considered the "shadow" counterpart of the octagonal numbers

0, 5, 16, 33, 56, 85, ..., in the respective negative bases – 1, – 4, – 7, – 10, – 13, ... , where they are equally self-generating so that again when each number is subtracted from its reverse, the same number re-occurs.

So for example the appropriate number base to consider for 16 is base – 7.

And in base – 7, 16 (in denary terms) = – 35. Once again 5 here relates to + 5, while 3 = – 7 * 3 =

– 21. So 35 = – 21 + 5 = – 16.

Therefore 16 (in denary terms) – 35 (in base – 7).

And the reverse of – 35 = – 53 = – {(– 7 * 5) + 3} = – (– 35 + 3) = 32.

So – 53 – (– 35 ) = – 35 i.e. 32 – 16 = 16 (in denary terms).

And again another way of expressing this result is – 35 = – 53 * 1/2 (in base – 7).

In this context, it is therefore interesting to see what happens when we now - with reference to our original octagonal sequence - change the respective number bases to their negative counterparts, and then - with reference to the "shadow" sequence - change the respective negative number bases to their positive counterparts.

So again using 21 in the original sequence to illustrate now in base – 8, this would be written as

– 33 i.e. – {(– 8 * 3) + 3},

So the number here is palindromic so that when subtracted from its reverse 0 will result.

However another way of expressing this result is – 33 = – 33 * 1 (in base – 8),

So in the case of its respective positive number base each term of the octagonal sequence is equal to half of its reverse number; however in the case of the corresponding negative number base, each term is now equal to its reverse number.

This is also borne out with reference to the "shadow" octagonal sequence, now considered for respective positive number bases.

So 16 in base 7 = 22 (which of course being a palindrome is equal to its reverse.

So 16 (i.e. – 35 in base – 7) is half of its reverse (– 53).

However 16 (i.e. 22 in base 7) is equal to its reverse (22).

These results can then be extended for both sequences in the respective number bases to 3, 4, ..., digit numbers where a digit (1 less than the number base in question) is continually inserted between 1st and last digits..

Mind you it is very tricky showing this in negative number bases.

For example we can extend this in the case of our example of – 35 (in base – 7)

So we insert 6 (which 1 less than the positive base 7) to get – 365 However this now calculates as – {(5 + (6 * 7) – (3 * 49)} = – (5 + 42 – 147) = 100 (in denary terms).

Then the reverse – 563 calculates as – {(3 + (6 * 7) – (5 * 49)} = – (5 + 42 – 247) = 200 (in denary terms).

So again the starting number is half of its reverse.

Then for the positive number base 7, 100 (in denary terms) = 202 (which is a palindrome)

So now the starting number is equal to its reverse in this number base.

Interesting further connections can be shown as between the octagonal number sequence and its "shadow" sequence.

If one looks at the entry on Octagonal Numbers at Mathworld one can see how geometrically the octagonal sequence can be represented.

So in the 1st diagram we have 1 point (the 1st number in the sequence).

In the 2nd, we have now 8 points (the 2nd number in the sequence).

Now each new circle drawn from the same starting point includes 8 additional points.

So in the 3rd diagram the inner circle contains 8 points and the outer circle 16 points. However 3 of these points are common to both circles. So the total no. of points (where each point is counted once) = 8 + 16 – 3 = 21 (the 3rd number in sequence).

Then in the next diagram a further circle is added this time containing 24 points. However with respect to the 1st circle 3 points repeat and with respect to the second 5 points.

So the total of non-repeating points = (8 + 16

In fact a very interesting pattern emerges

1

So we have 1 * 8 = 8

3 * 8 (– 3) = 21

6 * 8 (– 3 – 5) = 40

10 * 8 (– 3 – 5 – 7 ) = 65

So in fact, each (n + 1)th term of the octagonal can be represented as the product of the corresponding nth term of the triangular series * 8 (minus an adjustment based on sum of the odd integers except 1).

The sum of
1st n odd integers = n

^{2 }.
And because
1 is not included the sum of 3 + 5 + 7 + …
= n

^{2 }– 1.The nth term of the triangular sequence = n(n + 1)/2.

Therefore
for example the (n + 1)

^{th}term of the octagonal,
= {n(n + 1)/2} * 8 – n

Then when n = 5 (i.e. 6th term of octagonal sequence) we thereby obtain,

^{2 }+ 1.Then when n = 5 (i.e. 6th term of octagonal sequence) we thereby obtain,

(15 *
8) – 25 + 1 = 120 – 24 = 96.

A geometric basis can also be found for the “shadow” octagonal system relating
to non-overlapping points within the circles.

For
example in the first circle (of 8) there are no overlapping points.

So the
first number in the “shadow” sequence = 0.

Then
in the next representation, there 3 over-lapping points on the inner circle
(leaving 5 points that do not overlap)

So the
second number in the “shadow” sequence = 5.

Then
in the next diagram there are 5 additional over-lapping points on the 2

^{nd}circle (of 16 points).
Thus the
total number that do not overlap with respect to both circles = 5 + 11 = 16

So the third number in the “shadow” sequence = 16.

Then
in the final diagram there are an additional 7 over-lapping points with respect
to the 3

^{rd}circle (of 24 points).
Thus
the total number of points that do overlap with respect to the 3 circles = 5 +
11 + 17 = 33.

And the
4

^{th}number in the “shadow” sequence = 33.
Therefore, the (n + 1)

^{th}term of the shadow” sequence = {n(n + 1)/2 * 8} – (n + 1)^{2 }+ 1}
Thus
the 5

Incidentally one simple implication of the above is that the nth term of the "shadow" octagonal series = nth term of octagonal series i.e. {3n

^{th}term = (10 * 8) – 25 + 1 = 80 – 24 = 56.Incidentally one simple implication of the above is that the nth term of the "shadow" octagonal series = nth term of octagonal series i.e. {3n

^{2}– 2n}, – 2n + 1.
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