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.    

More on Octagonal Numbers

We have already considered the octagonal numbers,

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 + 24) – (3 + 5) = 40 (the 4th number in the sequence.

In fact a very interesting pattern emerges: 
                                                        
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 = n² .

And because 1 is not included the sum of 3 + 5 + 7 + …  = n² – 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  –  n² + 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^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. {3n² – 2n}, – 2n + 1. 

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, …) 

Octagonal and Triangular Numbers (2)

The octagonal numbers i.e.

1, 8, 21, 40, 65, 96, …have built in to their numerical structure some very interesting relationships.

1 =   1 * 1,

8 =   2 * 4,

21 = 3 * 7,

40 = 4 * 10

65 = 5 * 13

96 = 6 * 16

      …

So the kth term i.e. t_k = k(3k – 2) = 3k² – 2k

Also when each term is multiplied by 3, it results in a number 1 less than a perfect square.

So 1 * 3 =  3  =  2² – 1

     8 * 3  = 24 = 5²  – 1

   21 * 3  = 63 = 8² – 1

             …

So once again we see the relationship between 3 and the the sequence 2, 5, 8, … (in which bases we have the unique digit sequence of the reciprocal of 3).

So 3 * t_k  = (3k – 1)² – 1; so t_k = {(3k – 1)² – 1}/3 = 3k² – 2k

Also each term can be expressed as the difference of two squares.

So 1 =  1² – 0²

       8  = 3² – 1²

   21  = 5² – 2²

    40  = 7² – 3²

                 …

So again t_k = (2k – 1)² – (k – 1)²  = k(3k – 2) = 3k² – 2k.

We have already shown the connection between the octagonal self-generating numbers (with both non-hierarchical linear and circular features).

Now the coincidence of both linear and circular self-generating features can only apply when numbers have two digits.

However the non-hierarchical linear aspect can be continued into numbers with more than 2 digits.

Quite simply when a 2-digit number is self-generating in this non-hierarchical sense then a n-digit number (where n > 2) is likewise self-generating where the additional digits are 1 less than the base in question and inserted between the 1st and 2^nd digits (of the 2-digit number).

So for the simplest case, in base 2, 01 is self-generating (as 10 – 01 = 01).

This therefore entails that 3-digit number 011 is also self-generating (where the additional digit inserted is 1 less than the corresponding number base 2).

So 110 – 011 = 011  i.e. 3 in denary terms. 

So in the next relevant number base 5, 143 is self generating, where 341 – 143 = 143, i.e. 48 in denary terms.

In the next relevant number base 8, 275 is self-generating, where 572 – 275 = 275, i.e. 189 in denary terms.

Then in the next relevant number base 11, 3A7 is self-generating as 7A3 – 3A7 = 3A7, i.e. 480 in denary terms.

Now if we look at the denary nature of the sequence generated 3, 48, 189, 480, …, we can again show its links to the octagonal numbers

So, 3 = (1 * 3) *  1

    48 = (2 * 3) *  8

  189 = (3 * 3) * 21

  480 = (4 * 3) * 40  

              …

And the sum of the 1^st n terms of the sequence 1, 2, 3, 4, … is the nth triangular number.

So the sum of the first 4 terms i.e. 1 + 2 + 3 + 4 = 10 i.e. the 4^th triangular no.

Then when we move into 4-digit numbers 0111 is self-generating with respect to the base 2 i.e. 7 in denary terms.

Next 1443 is self-generating with respect to base 5 i.e. 248 in denary terms.

Then 2775 is self-generating with respect to base 8 i.e. 1533 in denary terms.

Finally to illustrate 3AA7 is self-generating with respect to base 11 i.e. 5320 in denary terms.

Again a fascinating connection can be shown with the octagonal numbers.

So   7     =     7  * 1

248   =   31 * 8

1533 =   73 * 21

5320 = 133 * 40

                 …

Now the other sequence of numbers 7, 31, 73, 133, … is fascinating in its own right

7 = 6 + 1, 31 = 30 + 1, 73 = 72 + 1 133 = 132 + 1, …

And 6 =  6 *  1

      30 =  6 *  5

      72 =  6 * 12

     132 = 6 * 22

           …

And 1, 5, 12, 22, … are the pentagonal numbers, which in turn are related to the triangular numbers.

The average of the 1^st n pentagonal numbers is the nth triangular no.

So the average for example of the first four pentagonal numbers = 40/4 = 10 (i.e. the 4^th triangular number).