Extending Relationships (2)

When I had completed yesterday’s entry, I realised that the results could be generalised further in a remarkable way.

Thus the starting sequence (where n = 2) is,

0, 4, 12, 24, …,  = 4(0, 1, 3, 6, …).

In other words the n^th term of the starting sequence = 4 * n^th term of the triangular sequence, where n = 0, 1, 2, 3, …

Now the n^th term of the triangular sequence = n(n – 1)/2

Therefore the n^th term of our starting sequence = 2n(n – 1) = 2n² – 2n.

Now each further sequence for n = 3, 4, 5, … is obtained by successively adding n² (i.e. the n^th term of the sum of squares) to the previous result.

So for example the n^th term of the octagonal sequence, where n = 3, is 3n² – 2n, and the n^th term of the sequence, where n = 4, is 4n² – 2n, the n^th term of the sequence where n = 5, is 5n² – 2n and so on.

This means for example when n = 11, the n^th term of the corresponding sequence = 11n² – 2n.

Therefore the corresponding sequence is

9, 40, 93, 168, …

The reason for choosing this sequence is that the starting base is therefore 10 with subsequent relevant number bases increasing by 11.

So clearly 9 in base 10 (using 2 digits = 09) and the reverse = 90

And 90 (reverse) = 09 (original number) * 10.

However whereas this is clearly true, the result is not unique in this base as any other digit (from 1 – 8 inclusive) can be used with 0 so that the reverse = original number * 10.

Indeed this non-exclusivity applies to all number bases > 2.

When n = 3 and the starting base = 2. Then 01 is the  2-digit original number that can arise in this base with 10 its reverse so that 10 – 01 = 01. However even here the 0 in 01 is strictly redundant.  

And 10 reverse = 01 (original number) * 2.

Then when we keep inserting 1 (in base 2) between first and last digits the self generating pattern is preserved for higher digit numbers.

So 110 – 011 = 011; 1110 – 0111 = 0111 and so on.

However where n > 3 (and consequent number base > 2) a non-exclusive basis attaches to the original number generated in the starting base.

So again where n = 11 (and starting base = 10) any digit from 1 to 9 can be associated with 0 in the 2-digit case and again any digit from 1 to 9 can be continually inserted as between first and last digits to generate higher digit numbers (where the reverse – original number * 10).

However once we proceed on to the next relevant number base, a unique relationship exists (where 0 is not a starting digit).

For example 40 in base 21 is 1S (where S denotes the number 19).

The reverse S1 = (21*19) + 1 = 400

And 400 (reverse)  = 10 * 40 (original number) in denary terms .

So S1 (reverse) = 10 * 1S (original number) in base 21.

This is then unique as the only 2-digit example with this number property in base 21.

However we can then continually extend this property uniquely to 3-digit, 4-digits, 5-digit, … numbers in base 21 by inserting T (representing units measured in 20’s) as between 1^st and last digits.

So the corresponding 3-digit original number (in base 21) = 1TS = 19 + 20 * 21 + 21* 21 = 880 (in denary terms).

And the reverse ST1 = 1 + 20 * 21 + 19 * 21 * 21 = 8800.

So again as we see in base 21,

ST1 (8800 in denary terms) = 10 * 1TS (880 in denary terms).

We saw when dealing with the octagonal sequence (which offers the most unique self similarity features of number), that it also has a well-defined “shadow” sequence that arises when the nth term of the octagonal i.e. 3n² – 2n is defined for n = 0, – 1, – 2, – 3…

Alternatively this sequence i.e.

0, 5, 16, 33, 56, …,

can be defined, where the nth term = 3n² + 2n for n = 0, 1, 2, 3, …

Now the n^th term of the sequences that we have so far addressed can be defined in general terms as kn² – 2n where k = 2, 3, 4, ….

And the n^th term of the corresponding “shadow” sequences can be defined in general terms as kn² + 2n, where k = 2, 3, 4, …

Therefore the “shadow” sequence to the starting sequence that we have already considered i.e. 0, 4, 12, 24, … has as its n^th term 2n² + 2n , for n = 0, 1, 2, 3, … i.e.

0, 4, 12, 24, 40, …,

which gives us the same starting sequence. Again in the “shadow” case we start with negative base 1, with subsequent bases increasing in negative terms by 2.

Thus 4 in negative base 3 = – 22 i.e. –{(2 * – 3) + 2}

And the reverse is also – 22

Thus – 22 = – 22 * 1

So reverse = original number * 1.

And to give one more example,

12 in negative base 5 = – 33 i.e. –{(3 * – 5) + 3}

And – 33 = – 33 * 1

So again reverse = original number * 1.

Thus we have replicated the same result in the negative bases of the “shadow” sequence as in the original starting sequence for positive bases.

Then when we switch to considering these numbers in corresponding positive bases, the same behaviour occurs.

So 4 in base 3 = 11 with the reverse also 11.

And 12 in base 5 = 22 with the reverse also 22.

So uniquely with the starting sequence, with its shadow is the same sequence, the self replicating feature where the original number = the reverse number (i.e. where both are palindromes) applies to interpretation in  positive and corresponding negative number bases.

Then when we went on to consideration of the “shadow” to the next sequence i.e. octagonal, we found that the same behaviour there ( i.e. where reverse = original number * 2) was replicated in negative number bases 1, 4, 7, …

Thus one interesting feature here is that though the gap between relevant number bases is related to the value of n (= 3 in this case) the starting base for all “shadow” cases starts at 1. This is ultimately due to the fact that we are now defining n with respect to 0, 1, 2, 3, ..., whereas formerly we defined n with respect to 1,2 3, ...  from

Then in the “shadow” case of the octagonal, where we now consider results in positive number bases, palindromes results i.e. where reverse = original number * 1.

And this behaviour universally characterises subsequent sequences.

For example the next “shadow” sequence (where the n^th term is 4n² + 2n) for n = 0, 1, 2, 3, … is

0, 6, 20, 42, … with n = 4 and relevant number bases are 1, 5, 9, 13, …

Thus the 1^st non-trivial result applies to 6 in negative base 5, which is – 24. And the reverse is – 42 (i.e. 18 in denary terms).

Thus – 42 (reverse) = – 24 * 3 (original number)

And this replicates behaviour for its complementary sequence (in positive number bases).

And when we interpret base 3 in positive terms 6 = 11 (a palindrome) so that

11 (reverse) = 11 (original number) * 1.