Octagonal and Triangular Numbers (3)

We have seen that the nth term of the octagonal series, i.e.

1, 8, 21, 40, 65, …,

is given by 3n² – 2n where n = 1, 2, 3, …

However what is interesting is that a “shadow” series (comprising positive valued terms) can be obtained by letting n = 0, – 1, – 2, …, i.e.

0, 5, 16, 33, 56, …

And again the same neat numerical features that characterise the octagonal series, also characterise its “shadow” referred to in OEIS as A045944 (The Rhombic Matchstick Numbers).

So when we multiply each of the numbers of this sequence by 3, we obtain a number that is 1 less than a perfect square.

So 0 * 3 = 0 = 1² – 1; 5 * 3 = 15 = 4² – 1; 16 * 3 = 48 = 7² – 1 and so on.

Likewise each of these numbers can be expressed as the difference of two squares.

So 0 = 1² – 1²; 5 = 3² – 2² ; 16 = 5² – 3² and so on.

And                     Also in the case of the octagonal series

0 =   0 * 2            1 =  1 * 1

5 =   1 * 5            8 =  2 * 4

16 = 2 * 8           21 = 3 * 7

33 = 3 * 11         40 = 4 * 10

     …                         …

So we can see that the number bases that were so relevant in the case of self generating numbers and the octagonal series i.e. 2, 5, 8, 11, …, now naturally appear with this “shadow” sequence.

Likewise in complementary fashion the number bases applicable to self generating numbers in the “shadow” sequence i.e. 1, 4, 7, 10, …naturally appear in the octagonal series.

In fact because the value of the respective numbers bases is directly related to the value of k it is the negative number bases that are relevant with respect to the “shadow” sequence.

In other words the self generating numbers that arise in the negative number bases i.e.

– 1, – 4, – 7, – 10,…, in fact are the respective terms 0, 5, 16, 33, … of the shadow sequence.

The 1st case is trivial for 00 in base – 1  is 00 and 00 – 00 = 00.

However it is somewhat trickier in subsequent number bases.

In fact in base – 4, 5 would be written as – 23.

Now if we look at the right hand digit this indeed represent 3 units. However the 2 in base – 4 = 2 * (– 4) = – 8.

Therefore 23 in base – 4 =  – 8 + 3 =   – 5 (in denary terms).

Therefore 5 (in base 10) = – 23 (in base – 4).

And the reverse of – 23 = – 32 (in base – 4).

The difference of the two = – 32 – (– 23) = – 32 + 23

And  32 = 3 * (– 4) + 2 = – 12 + 2 = – 10 (in denary terms)

Therefore – 32 = 10 and 23 = – 5.

So – 32  + 23 = 10 – 5 = 5.

Therefore when we subtract – 23  from its reverse in base – 4,  we get – 23.

So – 23 is self-generating (in both cyclical and linear non-hierarchical terms) in this base.

Then next case arises in base – 7, where – 35 (= 16 in base 10) is now the relevant number.

And – 53 – (– 35 ) in base – 7,  =  – 53 + 35 = 32 – 16 = 16 (in base 10).

So – 35 is self-generating in base – 7.

And just as the two digits of the self-generating numbers in the positive bases 2, 5, 8, 11, … represent the unique digit sequence of the reciprocal of the cyclic prime 3 in these bases, likewise the two digits in the negative bases – 1, – 4, – 7, – 10 again represent the unique digit sequence of the reciprocal of 3.

Now if we line up the octagonal and its corresponding “shadow” sequence as follows, we have,

1, 8, 21, 40, 65, …

0, 5, 16, 33, 56, …

Then subtracting each term of the latter “shadow” from the corresponding term of the former octagonal sequence we get,

1, 3, 5, 7, 9, …

In other words, we have the sequence of odd-numbered natural integers.

This for example using Leibniz’s well- known formula, we can express π in terms of the octagonal and shadow sequences.

So π/4 = 1/(1 – 0) – 1/(8 – 5) + 1/(21 – 16) – 1/(40 – 33) + …

It is also fascinating how we can derive the natural number sequence, through dividing each successive term of one sequence by the corresponding successive number base of the complementary sequence.

Thus with respect to the “shadow” sequence the terms are 0, 5, 16, 33, ..

and the corresponding number bases (related to self-generating numbers) of the octagonal sequence are 2, 5, 8, 11, …

And 0/2 = 0; 5/5 = 1; 16/8 = 2; 33/11 = 3 and so on.

Then with respect to the octagonal sequence, the terms are 1, 8, 21, 40, … and the corresponding number bases (related to self-generating numbers) of the “shadow” sequence are – 1, – 4, – 7, – 10, …

And 1/– 1 = – 1; 8/– 4 = – 2; 21/– 7 = – 3; 40/– 10 = – 4 and so on.

If we now combine the positive number bases i.e. 2, 5, 8, 11, … directly with the corresponding terms of the octagonal series, a fascinating pattern emerges.

So  2 *  1  =  2

  5 *  8  =  40

  8 *  21 = 168 

 11 * 40 = 440

      …

Now the sum of the first (i.e. one) term = 2              =  1 * 2

The sum of the first two terms  = 2 + 40 = 42           =  6 * 7

The sum of the first 3 terms = 42 + 168 = 210          = 14 * 15

The sum of the first 4 terms = 210 = 440 = 650        = 25 * 26

So the cumulative total of terms always can be expressed as the product of two consecutive numbers.

This also applies to the shadow sequence, multiplied by the number bases,  1, 4, 7, 10, …

So  1 *  0 =      0

    4  * 5 =     20

    7 * 16 =   112

  10 * 33 =   330.

So the cumulative total of 1 = 0 =     0 *  1

The cumulative total of 2 =    20   =  4 *  5 

The cumulative total of 3 =    132 = 11 * 12

The cumulative total of 4 =    462 = 21 * 22

Notice how the increase of each number in the consecutive sequence is related to the number bases. So each number increases by 4, then 7, then 10 and so on whereas with octagonal numbers each number (in the consecutive sequence).

Finally, we can again make a connection with the triangular nos. in the following way.

 

Again, if we line up the octagonal and its corresponding “shadow” sequence as follows, we have,

1, 8, 21, 40, 65, …

0, 5, 16, 33, 56, …

Then adding each respective term we get,

1, 13, 37, 73, 121, …

Then subtracting 1 we get

0, 12, 36, 72, 120, …

= 12 (0, 1, 3, 6, 10, …)