Friday, February 16, 2018

Octagonal and Triangular Numbers (1)

Some time ago when I was exploring the use of holistic (circular) notions for deriving perhaps novel new insights with respect to simple number relationships, I became fascinated with - what I referred to at the time as - self generating numbers.

So a self generating number refers to a number that repeats itself according to some well-defined procedure based on subtraction.

And I basically defined two types of such relationships linear and circular respectively.

So one linear type example (with hierarchy) results from the ordering of the digits in descending (largest digit 1st) and ascending (smallest digit 1st) respectively and then subtracting the latter from the former number. So when the original number results from this operation, we thereby have a self-generating number.

For example if we take the number 495 to illustrate when arranging its digits in descending order, we obtain 954 and then in ascending order 459.

And then when we obtain 954 – 459, the result is 495 (i.e. the original number).

Now in fact 496 represents the only example of a 3-digit number (in the denary system) that is self-generating in this sense, with 6174, the only example of a corresponding 4-digit number.

One other interesting feature of such numbers is that when divided by 9 that a palindrome (or close palindrome) frequently results.

So 495/9 = 55; 6174/9 = 686.

I also looked at the circular equivalent, which is especially well represented through cyclic primes.

Self generation is this sense results when the same circular sequence of digits is preserved through subtraction.

For example 142857 represents the well-known unique cyclic system of digits associated with the reciprocal of 7.

Then if we obtain the difference of any 6-digit numbers, that preserves the same cyclic order of these digits that the result will be a number (that likewise preserves the same cyclic order).

So for example, 428571 – 142857 = 285714 (which preserves the same cyclic order).

Note that if we adopt the initial linear order (already mentioned) that self-generation will not occur
So 875421 – 124578 = 750483 (which is neither self-generating in a linear or circular sense).

I then examined the simpler linear case (without hierarchy) where subtraction takes place with the order of digits simply reversed.

So for example taking 6174 again to illustrate its reverse is 4716

And the difference = 6174 – 4716 = 1458. So self-generation does not take place.

Now, perhaps surprisingly there does not appear to be any numbers (in base 10) that are self-generating (in this non-hierarchical sense).

So we have to look to other number bases for this feature to arise.

Fro example in base 8, if we take the 2-digit number, 275 and then subtract it from its reverse we get 572 – 275 = 275.  So this clearly then is a self-generating number in a linear (non-hierarchical) manner.

I then realised that a particularly interesting case would arise here in the 2-digit case as - by definition - both linear and circular self-generation would be involved. And also the hierarchical and non-hierarchical definitions would likewise coincide.

So the quest was on therefore to find all 2-digit self generating numbers that satisfied these requirements (in all relevant number bases).

And I found that a fascinating solution existed. So in general terms if n represents the number base, the relevant bases where self-generation occurs is given by 2 + 3k (where k = 0, 1, 2, 3, …)

And if a represents the 1st digit of the corresponding self-generating number, its value is given as k with the second digit = 2k + 1.

Therefore when k = 0 the relevant number base = 2. The 1st digit = 0 and the second digit = 1

Therefore in base 2, 01 is a self generating number (in both a linear and cyclical sense).

So 10 – 01 = 01.

Then when k = 1, the next relevant number base = 5. Therefore the 1st digit = 1, and the 2nd = 3.

So in base 5, 13 is a self-generating number in the manner defined.

So 31 – 13 = 13.

When k = 2, the next number base = 8. The 1st digit = 2, and the 2nd digit = 5.

So 52 – 25 = 25.

And giving just one more example to illustrate, the next number base is 11 with the 1st digit = 3 and the 2nd digit = 7.

So 73 – 37 = 37 (in base 11).   

Now if one converts all these self-generating numbers into conventional denary (base 10) format, the resulting sequence arise

i.e. 1, 8, 21, 40, …, which represents the well-known octagonal numbers.

Thus the octagonal numbers in this sense are uniquely related to all these 2-digit self-generating numbers (in other number bases).

One other fascinating feature of these numbers struck me at the time i.e. that they represent the cyclic digit sequence (in their respective number bases) of the reciprocal of 3.

So for example if we were to express 1/3 in base 8, we would obtain .252525…

Thus the uniquely recurring digits here are 25.

In one way, this is a remarkable feature, which connects the octagonal and triangular numbers 1, 3, 6, 10, … in a way that is perhaps not commonly realised.

And when we look more closely at the triangular, striking connections can indeed be found with the octagonal numbers.

Now the triangular numbers repeat naturally groups of 3 where the 1st member of the group leaves a remainder of 1 (when divided by 3) while 3 is always a factor of the other two members of the group.

So taking the 1st 3 members 1, 3 and 6 if we now add the 3 members of the group and subtract 1, the total will then be divisible by 3. So {(1 + 3 + 6) – 1}/3 = 9/3 = 3 (which is the 2nd triangular number).

Then we treat the next group of 3 in the same manner, we obtain {(10 + 15 + 21) – 1}/3 = 45/3 = 15 (which is the 5th triangular number).

Then with the next group of 3 we have {(28 + 36 + 45) – 1}/3 = 108/3 = 36 (which is the 8th triangular number).

So the ordinal ranking of the triangular numbers obtained in this manner exactly match the corresponding number bases (where the self generating numbers arise).

And again the octagonal numbers are then the representation of these numbers in base 10.


And a more direct connection can be made.

If we start with the 3rd triangular number and then divide by 3 we obtain 6/3 = 2.

If we now keep increasing the ordinal rank by 6 and then dividing (the number arising) by the corresponding cardinal number we will generate the sequence 2, 5, 8, 11, … which matches in cardinal fashion the ordinal rankings of the number bases in which the self generating numbers arise.

So the 9th triangular number = 45 and 45/9 = 5; the 15th cardinal number = 120 and 120/15 = 8; the 21st triangular number = 231 and 231/21 = 11 and so on.

Therefore in a way it is very striking how such connections can be shown as between number (with respect to both cardinal and ordinal usage) both within the denary base and across other relevant number bases.

And this in turn relates to the very nature of the self-generating numbers (on which these connections are based) where both linear and circular notions coincide.


Again the octagonal numbers are:

1, 8, 21, 40, 65, 96, 133, ....

And the triangular numbers are:

1, 3,  6, 10, 15, 21,  28, ...

And if we subtract each successive term in the triangular from the corresponding term in the octagonal, we obtain,

0, 5, 15, 30, 50, 75, 105

= 5 (1, 3, 6, 10, 15, 21, ...)

And the sequence within the brackets is once again that of the triangular numbers.

Thursday, February 8, 2018

Interesting Log Relationships (3)

In previous entries, Interesting Log Relationships (1) and Interesting Log Relationships (2), I showed how the (infinite) reciprocal sum of the unique number sequences associated with (x – k)2 =  0 and 
(x k)3 = 0, (with k = 2, 3, 4, …) together with (x + k)2 = 0 and (x + k)3 = 0 (with k = 1, 2, 3, …), result from the application of generally applicable formulae resulting in answers of the form a log k + b (where a and b are rational numbers that can take on positive or negative values).

So once again for example for the general case where (x + k)3 = 0, the general formula is
2k{(k + 1)[log(k + 1/k)] – 1}.

Thus when k = 1, we have the polynomial equation (x + 1)3  = 0 i.e. x3 + 3x2 + 3x + 1 = 0, with the unique number sequence 1, – 3, 6, – 10, 15, …

Therefore the corresponding (infinite) sum of reciprocals (i.e. the alternating sum of reciprocals of the triangular nos.) = 1 – 1/3 + 1/6 – 1/10 + 1/15 … 

= 2{2(log 2) – 1}  = 4 log 2 – 2.

However one can continue on both for higher powers of (x – k)n =  0 and (x + k)n = 0, to generate corresponding results for the associated (infinite) sums of reciprocals relating to the unique number sequences occurring.

Now while it appears that the general form of these results does not change (remaining as a log k + b, it is not yet clear to me whether for any given power of n, that a general formula expression can be given to cover all possible results.

However, I did manage to calculate some specific results (which, not surprisingly, become somewhat more difficult to establish as the power of n and value of k increases).

For example in the simplest case where (x – k)2 =  0 (with k = 2), we have (x – 2)4 =  0,
i.e. x4 – 8x3 + 24x2 –  32x + 16 = 0 with the corresponding unique number sequence
1, 8. 40, 160, 560, … (A001789 in OEIS).

And the corresponding (infinite) sum of reciprocals is,

1 + 1/8 + 1/40 + 1/160 + 1/560 + …,  =  1.15888 …   = 6 log 2 – 3.

In the next case where (x – 3)4 =  0, i.e. x4 – 12x3 + 54x2 – 108x + 81 = 0, the unique numbers associated are 1, 12, 90, 540, 2835, … (A036216 in OEIS).   

Thus the corresponding (infinite) sum of reciprocals is,

1 + 1/12 + 1/90  + 1/540 + 1/2835 + …,   = 1.09674… = {72 log (3/2) – 27}/2.

Then in the next case, where (x – 4)4 =  0, i.e. x4 – 16x3 + 96x2 – 256x + 256 = 0,
the unique associated number sequence is

1, 16, 160, 1280, 8960, … (i.e. A038846 in OEIS).

Thus the corresponding sum of reciprocals

= 1 + 1/16 + 1/160 + 1/1280 + 1/8960 + … =  1.06966…  = 108 log (4/3) – 30


Then in the case where (x + 1)4 = 0, i.e. x4 + 4x3 + 6x2 + 4x + 1 = 0, the unique associated number sequence is 1, – 4, 10, – 20, 35, … 

So the corresponding (infinite) sum of reciprocals (i.e. sum of alternating reciprocal terms of the tetrahedral numbers) is,

1 – 1/4 + 1/10 – 1/20 + 1/35, …,  = .817766…    = (24 log 2 – 15)/2.   

Then in the case where (x + 2)4 = 0, i.e. x4 + 8x3 + 24x2 + 32x + 16 = 0, the unique associated number sequence is 1, – 8, 40, – 160, 560, …

So the corresponding (infinite) alternating sum of reciprocals,

= 1 – 1/8 + 1/40 – 1/160 + 1/560 – … = .89511…   54 log (3/2) – 21.

And in the case where (x – 5)4 = 0 i.e. x4 – 20x3 + 150x2 – 500x + 625 = 0, the unique digit sequence is 0, 0, 0, 1, 20, 250, 2500, ...

So the sum of reciprocals = 1 + 1/20 + 1/250 + 1/2500 + ... ,(A081143 in OEIS).

= (480 log 5/4  – 105)/2.

Thursday, February 1, 2018

Interesting Log Relationships (2)

In a former entry, “Interesting Log Relationships 1” I showed how the infinite sums of reciprocals of the unique numbers associated with simple polynomial equations of the form (x – k)2 = 0, where k = 2, 3, 4, … are related to simple log type relationships.

So, again for example, when k = 2, the equation is (x – 2)2 = 0, i.e. x2 – 4x + 4 = 0.

Then by the same process by which the unique numbers forming the Fibonacci sequence is obtained from the equation x2 – x – 1 = 0, in like manner a unique number sequence can be obtained for x2 – 4x + 4 = 0, i.e. 1, 4, 12, 32, 80, … (listed as A001787 in the On-line Encyclopedia of Integer Sequences).

Then the infinite sum of reciprocals of this number sequence,

= 1 + 1/4 + 1/12 + 1/32 + 1/80 + … = 2 log 2.

And in that piece, I provided a general formula to cover the infinite sum of reciprocals for all cases entailing unique number sequences of form (x – k)2= 0,

i.e. k{log k/(k – 1)}.

So for example when k = 6 we have (x – 6)2= x2 – 12x + 36 = 0.
And the unique sequence of numbers associated with this equation – again derived in a similar manner to the Fibonacci sequence – is 1, 12, 108, 864, (i.e. A053469 in the OEIS).

Therefore the infinite sum of reciprocals of this sequence

= 1 + 1/12 + 1/108 + 1/864 + …  = 6{log (6 – 1)/5} = 6 log 6/5 = 1.0939…


I also provided in that entry a general formula to cover all cases where (x + k)2 = 0, with k = 2, 3, 4, …

i.e. k{log (k + 1)/k}

In this case the same sequence of numbers is obtained as for (x – k)2 = 0, except that they alternate as between positive and negative values.

So the corresponding infinite sum of associated reciprocals

= 1 – 1/12 + 1/108 – 1/864 + …   = 6{log (6 + 1)/6} = 6 log 7/6 = .9249…

So interestingly the sum of the alternating sum of reciprocals can be directly obtained from the (positive) sum of reciprocals by increasing both numerator and denominator of the log expression in the latter formula by 1.


We can however proceed to obtain further log expressions for the infinite sums of reciprocals associated with the unique numbers associated with the polynomial equations of the form (x – k)3 = 0, where k = 1, 2, 3, 4, …    

For example, when k = 2, we have (x – 2)3 = x3 – 6x2  + 12x – 8 = 0.

And the unique sequence of numbers associated with this equation is 1, 8, 40, 160, 560, … (A001789 in OEIS).

So the corresponding sum of reciprocals of this sequence is given as

1 + 1/8 + 1/40 + 1/160 + 1/560 + …  = 4(1 – log 2).

Once again a general formula can be given to cover all such cases,

i.e. 2k{1 – (k – 1)[log k/(k – 1)]}

So for example when k = 3, we have (x – 3)3 = x3 – 9x2 + 27x – 27 = 0.

And the unique sequence of digits associated with this equation is 1, 9, 54, 270, 1215, …, (A027472  in OEIS).

Therefore the corresponding infinite sum of reciprocals

= 1 + 1/9 + 1/54 + 1/270 + 1/1215 + …

= 6{1 – [(3 – 1)log 3/(3 – 1)]} = 6{1 – 2 log (3/2)} = 1.1344 …

And as before, a related general formula can be given to cover the infinite sums of reciprocals of the unique number sequences associated with (x + k)3 = 0,

i.e. 2k{(k + 1)[log(k + 1/k)] – 1}.

So once again this provides the alternating case to the corresponding positive sum of reciprocals

Thus again for k = 3, the infinite sum of reciprocals of the unique number sequence associated with (x + 3)3

 = 1 – 1/9 + 1/54 – 1/270 + 1/1215 – …

Therefore from the formula the required sum = 6{[(3 + 1)log 3 + 1/(3)] – 1}

= 6{4 log (4/3) – 1} = .9043 …

A special case with respect to (x – k)3 = 0, occurs where k = 1 i.e. (x – 1)3
= x3 – 3x2 + 3x – 1 = 0.

And the unique sequence of numbers associated with this sequence is the set of triangular nos. i.e. 1, 3, 6, 10, 15, …

So the infinite sum of reciprocals of this sequence, i.e.

1 + 1/3 + 1/6 + 1/10 + 1/15 + … = 2{1 – (1 – 1)[log 1/(1 – 1)]} = 2{1 – 0[log (1/0)]}

= 2.

And the alternating sum of reciprocals, corresponding to the unique sequence of numbers associated with (x + 1)3 = 0, i.e.

1 – 1/3 + 1/6 – 1/10 + 1/15 – …


= 2{(1 + 1)[log(1 + 1/1)] – 1} = 2(2 log 2 – 1) = .772588 …

Friday, January 12, 2018

Testing for Primes

As we have seen a unique (infinite) number sequence is associated with the polynomial equation (x – 1)= 0.

For example when n = 3, the sequence relates to the triangular nos.,

1, 3, 6, 10, 15, 21, 28, 36, …

There is however a fascinating alternative way of obtaining this general sequence.

So again when (x – 1)= 0, each number in the sequence corresponds to

     
 kCn - 1,   =  k!/{(n – 1)!( k – n + 1)!}, where k = n – 1, n, n + 1, n + 2, …

Therefore, when the starting value of n = 3, the first number in sequence is 2C2  = 1

And subsequent numbers are 3C= 3, 4C= 6, 5C= 10, …

Thus these successive numbers represent (n – 1) combinations with respect to 2, 3, 4, 5, … , items respectively.

So when n = 3, successive numbers represent just 2 combinations with respect to 2, 3, 4, 5, …, items respectively.  


For the above general series, in the first number cycle of n, the sum of last (n – 1) digits
= (n + 1)th term – 1

So again, for example when n = 3, the 1st cycle comprises the 3 numbers 1, 3 and 6 respectively.

And the sum of the last two of these numbers = 9, which is 1 less than the 4th term in the sequence (i.e. 10).


This relationship holds for all values of n.

So for example when n = 4, the unique associated (infinite) sequence of digits is,

1, 4, 10, 20, 35, 56, 84, …

These now represent 3 combinations of 3, 4, 5, 6, ... items respectively. 

Thus, the 1st cycle comprises the 4 numbers 1, 4, 10 and 20. And the sum of the last three of these numbers = 34, which now is 1 less than the 5th number in the sequence (i.e. 35).

 ∞
 ∑1/kCn - 1 = sum of reciprocals of unique digit sequence associated 
k = n - 1

with (x – 1)n

So when for example, n = 3 (and n – 1 = 2),

    ∞
  ∑1/ kC= sum of reciprocals of unique digit sequence associated 
  k = 2

with (x – 1)3

= 1/2C2  +  1/3C2  + 1/4C2  + 1/5C2  + … 

= 1 + 1/3 + 1/6 + 1/10 + …   = 2


Remarkably when n is prime the {(n + 1)th term – 1} appears to be always divisible by n3 and when n is composite, not divisible by n3 (where p ≥ 5).

Where p = 3, the {(n + 1)th term – 1} i.e. 9, is divisible by n2 (i.e. 32).

And when p = 2, the {(n + 1)th term – 1} i.e. 2, is divisible by n (i.e. 2).


In principle this provides a fascinating test for whether a number is prime.

Thus in general terms, we can compute,

{(2n – 1Cn – 1) – 1}/n3, to see whether an integer (without remainder) results. Alternatively, we check to see if n3 is a factor.

So for example to test whether n = 29 is prime, we compute

(57C28  – 1)/n3 = [{57!/(28! * 29)!} – 1]/24389     =  616410400171 (exactly).

Therefore 29 is prime. 

I have tested for all primes from 5 to 61 (inclusive) and found no exceptions.

In all these cases for primes, where n3 was indeed a factor, in no case was a higher power of n a factor.

It would equally appear to be the case that if we continued on to further cycles - where once again a cycle contains the same number of terms, as the number in question (n) - that the 1st number of the next cycle less 1, will again be divisible by n3, where n is prime.

However in some cases this number will also be divisible by higher powers of n.

For example when n = 5, the first number in the 5th cycle (i.e. 11th term) is 10626.

Then when we subtract 1, we obtain 10625, which is divisible by n4 = 625.  

Thursday, November 30, 2017

Further Interesting Connections

As we have seen the triangular nos. i.e. 0, 1, 3, 6, 10, 15, ... represent the unique digit sequence associated with the simple polynomial equation (x – 1)3 = 0.

And this number sequence is intimately related to the corresponding number sequence associated with the Zeta 1 (Riemann) function where s = – 2, i.e. ζ1( – 2), i.e. 1, 4, 9, 16, 25, ...

Then when we look at the numbers in this later sequence, we can see that they are related to the  corresponding numbers of the former sequence in the following manner

1 = 0 + 1; 4 = 1 + 3; 9 = 3 + 6; 16 = 6 + 10; 25 = 10 + 15 and so on.

So in more general terms, tk – 1 + tk (with respect to the former infinite sequence) = tk (with respect to the latter).


With respect to the former sequence, 



1 – 0   = 1  and  1 + 0  = 1 (i.e. 12);

3 – 1    = 2  and  3 + 1   = 4 (i.e. 22);

6 – 3   = 3  and   6 + 3 =   9 (i.e. 32);

10 – 6 = 4  and 10 + 6 = 16 (i.e. 42);

So in general terms the tth – (t – 1)th term = t;   and tth + (t – 1)th term = t2


 
Again with respect to the former sequence, 

0 + 1   = 1    and 02 + 12        = 1       (i.e. the 1st term);  

1 + 3    = 4    and 12 + 32        = 10     (i.e. the 4th term);


3 + 6   = 9    and 32 + 62       = 45     (i.e. the 9th term);


6 + 10 = 16  and 62 + 102   = 136    (i.e. the 16th term);


Thus in general terms (tk – 1)2 + (tk 2) = the (tk – 1 + tk)th  term of the infinite sequence.


In like manner, with respect to the same sequence,

1 – 0    = 1 and  12 –  0     = 1 (i.e. 13)
 
– 1    = 2 and 32 –  1    = 8 (i.e. 23)

– 3    = 3 and 62 –  3    = 27 (i.e. 33)

10 – 6 = 4 and 102 –  4   = 64 (i.e. 43)

So once more in general terms,   

(tk)2 – (tk – 1)2 = (tk tk – 1)3 for the infinite sequence.


With respect to the sequence  1, 3, 6, 10, 15, 21, 28, 36, 45, 55,  ..., the 1st term of each successive group of 3 leaves a remainder of 1 when divided by 9

So, 1 divided by 9  = 0 (+ remainder of 1);


     10 divided by 9 = 1(+ remainder of 1);

     28 divided by 9 = 3 (+ remainder of 1);

     55 divided by 9 = 6 (+ remainder  of 1); 


Thus continuing on, when we ignore the reminder of 1, we obtain the same series 0, 1, 3, 6,...

We have already noted the remarkable fact that with respect to the Zeta 1 (Riemann) function,


 
∑{ζ1(s) – 1} = .64493... + .20205... + .08232.. + .03692... + ...
s = 2

= 1/2 + 1/6 + 1/12 + 1/20 + …  = 1/2( 1 + 1/3 + 1/6 + 1/10 + …)  = 1.


And once again 1 + 1/3 + 1/6 + 1/10 + ... represents the sum of reciprocals of the unique number sequence associated with (x – 1)3 = 0.

Then 1/2 (1 + 1/3 + 1/6 + 1/10 + ...) +
         
         1/6 (1 + 1/4 + 1/10 + 1/20 + ...) +

        1/12 (1 + 1/5 + 1/15 + 1/35 + ... ) +

       1/20 (1 + 1/6 + 1/21 + 1/42 + ...) +
                          ...

  = 1/2(2/1) + 1/6(3/2) + 1/12(4/3) + 1/20(5/4) + ...                     ...

= 1 + 1/4 + 1/9 + 1/16 + ...

In other words by combining each successive term of 1/2 + 1/6 + 1/12 + 1/20 + ...
with the sum of reciprocals of the unique number sequences associated with (x – 1)n = 0, where n = 3, 4, 5, 6, ..., we obtain the Zeta 1(Riemann) function ζ1(s), where s = – 2.

Thursday, November 9, 2017

Connections with Symmetry

We have already seen how the unique number sequences associated with the simple polynomial equation, 
(x – 1)n = 0, are intimately related in geometrical terms with perfectly symmetrical objects, based on each face representing an equilateral triangle.

So again, the unique numbers associated with (x – 1)3 = 0, are the triangular numbers 1, 3, 6, 10, 15, …

So for example, starting with one point (see Mathworld), when we then place a point at the vertices of a symmetrical i.e. equilateral triangle, we get 3 points in all.

Then when we place another point at the middle of each side (at an equal distance from the end points) we get 6 points in all. Then with additional points places at an equal distance from each other the total increases in accordance with the numbers in our unique number sequence for (x – 1)3 = 0.

And again the unique numbers associated with (x – 1)4 = 0 are the tetrahedral numbers 1, 4, 10, 20, 35, … (Note that the nth term of this sequence represents the sum of the first n terms of the previous sequence)!

So again starting with 1 point (see again Mathworld) the successive terms in the series represent the equally placed points of the fully symmetrical 3-dimensional solid i.e. tetrahedron (as the number of equidistant points - initially with respect to each side of the object - increases.

Thus in the simplest case where we have one point at the vertex of each side of the tetrahedron (with 4 sides in all), we have 4 points.     

And we then extended this thinking more generally for n dimensions in space so that the unique number sequences associated with (x – 1)n = 0, represent the no. of equally spaced points for a “hyper” tetrahedron in (n – 1) space dimensions.

And such a simplex “hyper” tetrahedron (containing n sides) represents the least number of sides required to construct a fully symmetrical object in (n – 1) space dimensions.

Now, as we have seen, all these number sequences are intimately related to - what I refer to as - the Alt Zeta 2 function, which uses the reciprocals of these number sequences, which in turn complements the better known Zeta 1 (Riemann) function.

So again for example, the Zeta 1 (Riemann) function i.e. ζ(s), where s = 2 is

1 + 1/22 + 1/32 + 1/42 + ….     = 4/3 * 9/8 * 25/24 * …      = π2/6.

And the individual terms of the Zeta 1 can then be expressed in terms of infinite series based representing the Alt Zeta 2 function.

For example the 1st term in the product over primes expression above for ζ(2) is 4/3.
And when s = 2 with respect to the Zeta 1, n = 22 + 1 = 5 with respect to the Alt Zeta 2.

Therefore, the relevant Alt Zeta 2 expression represents the (infinite) sum of reciprocal terms based on the unique number sequences associated with (x – 1)5 = 0, i.e.

1 + 1/5 + 1/15 + 1/35 + 1/70 + …   = 4/3.

So then the next term in the product over primes expression for ζ(2)  i.e. 9/8 now corresponds to the infinite sum of reciprocals of the unique number sequences, where n = 32 + 1 = 10,  i.e. (x – 1)10 = 0,

i.e. 1 + 1/10 + 1/55 + 1/220 +  1/715 + …  = 9/8, and so on.


So again each (individual) term of the Zeta 1 (Riemann) function can be expressed in an ordered fashion, through the appropriate Alt Zeta 2 function. Though the Alt Zeta 2 function is more directly related to the product over primes expression of the Zeta 1 (Riemann) function (with s an integer > 1), it can equally be associated with the sum over natural numbers expression (through the subtraction of 1).

So for example the 2nd term of ζ1(2) in the sum over natural numbers expression = 1/4. This can in turn be related the Alt 2 expression (where n = s2 + 2 = 6).

So the infinite series of reciprocals based on the unique number sequence associated with (x – 1)6 = 0, is

1 + 1/6 + 1/21 + 1/56 + 1/126 + …  = 5/4.

And when we subtract 1 we obtain 1/4, which is the required term in the Zeta 1 (Riemann) function.


However what is fascinating in this context is that the denominator numbers associated with the Zeta 1 (Riemann) function can themselves be directly related to fully symmetrical objects.

Thus in 2-dimensional space we have the square - as earlier the equilateral triangle - as another symmetrical object (with respect to angular rotation). (Again see Mathworld).

So again when s = 2, the denominator numbers associated with ζ1(2) are 1, 4, 9, 16, 25, …

Starting with 1 point, we then move to the 2 dimensional symmetrical object of the square. So when we place a point at each vertex of the square we have 4 (i.e. 22) points in all.

And then when we start by placing 3 equidistant points on each equidistant line of the square, the total number of points = 9 (i.e. 32).

When we place 4 equidistant points on each equidistant line of the square, the total no. of points = 16 (i.e. 42). 

And in general terms when we place s equidistant points on each equidistant line of the square, the total no. of points = s2.

So what we have involved here in 2-dimensional space are the successive denominator terms of
ζ1(2).


And then in 3-dimensional space for the symmetrical object of the cube, we use the successive denominator terms of ζ1(3).
Thus starting with 1, the simplest case in 3 dimensions entails placing a single point at each vertex of the cube. And the total no. of points thereby involved = 8 (i.e. 23)..

And then when we place 3 equidistant points on each side and line through the cube, the total no. of points = 27 (i.e. 33).

And in general terms when we place s equidistant points on each equidistant line (outside and inside the cube) the total no. of points = s3.

Though impossible to properly visualise, we can then extend this into 4 space dimensions to obtain a hypercube (i.e. tesseract) where we now use the successive denominator terms of ζ1(4). So when we now place a point at each vertex (i.e. 2 points at the end of each line) of the tesseract, we obtain 16 (i.e. 24) in all.

And with 3 equidistant points on each line (outside and inside the tesseract) we obtain 81 (i.e. 34) in all.

And in general terms for s equidistant points we obtain s4 in all.


And in even more general terms for s equidistant points with respect to each line of the n-dimensional hyper cube in space, we obtain sn points in all. 


There is also an intimate connection here with the Zeta 2 function i.e. ζ2(s).

So for example where 2 = 1/2, the denominator terms i.e. 1, 4, 8 represent the simplest case of just one point at each vertex, where starting with 1 point, the total no. of points involved as one moves from the 2-dimensional (4) to the 3-dimensional cube (8) to the 4-dimensional tesseract (16) and so on into progressively higher dimensions.

Thursday, October 26, 2017

Symmetrical Geometrical Objects

I have been repeatedly referring to the (infinite) reciprocal sums of the unique number sequences associated with (x – k)n = 0 and especially (x – 1)n = 0.

And I refer to these reciprocal sums associated with (x – 1)n = 0 as the Alternative Zeta 2 function i.e. Alt ζ2(n).

However it is possible now to provide a more general formula.

This is based on the fact that the nth term of these unique number sequences follow a definite pattern.

Then, for (x – 1)2, the unique number sequence is 1, 2, 3, 4 with the kth = k and the
                                                                                                               
sum of  terms  = ∑ k/1! with (infinite) sum of reciprocals = ∑1!/k.    
                           k=1                                                                 n=1 

For (x – 1)3, the unique number sequence is 1, 3, 6, 10, … with the kth term = k(k + 1)/2!
                                                                                                         
And sum of terms  = ∑k(k + 1)/2! with sum of reciprocals = ∑ 2!/{k(k + 1)}.
                                  k=1                                                                     k=1 

Then, giving one more example to illustrate a consistent pattern, for (x – 1)4, the unique number sequence is 1, 4, 10, 20, …, with kth term = k(k + 1)(k + 2)/3!   
                                                       
So the (infinite) sum of reciprocals = ∑ 3!/{k(k + 1)(k + 2)}.
                                                                       k=1

Therefore for the general case (x – 1)n = 0, the (infinite) sum of reciprocals of the associated unique number sequences of this simple polynomial equation i.e. Alt ζ2(n), is given as
∑ (k – 1)!/{k(k + 1)(k + 2) … (k + n – 2)}.
k=1

Note, however that the Alt ζ2(n) function remains undefined for n = 1!


A fascinating geometrical way of looking at the unique number sequences given by
(x – 1)n = 0, is through number patterns that are associated with the simplest perfectly symmetrical objects that can be constructed in the various dimensions.

For example the simplest symmetrical polygon figure (with least number of sides) that can be constructed in 2-dimensional space is a 3-sided equilateral triangle.

The diagram in Mathworld then illustrates how the triangular numbers - which correspond to the unique number sequences associated with (x – 1)3 = 0 - then arise.

So we start with 1 dot and with the construction of an equilateral triangle we then place a dot at each vertex = 3. Then by placing a dot midway along each line, the total no. of points = 6.

And then with 4 dots equally spaced out along each line (with another placed within the triangle), when the total no. = 10 (overall) we can preserve the perfect symmetry of equal distance between adjacent dots. 

Then in the final example we place 5 dots equally spaced on each side of the triangle with 3 more internally = 15 (overall). 

What is also interesting here is that we can join up the dots in each triangle to form a number of smaller triangles that represents the previous number in the series.

Thus in the simplest case in the 3 dot case we can join the points to form 1 triangle (which represents the previous number in the sequence). Then in the case of the 6 dot triangle we can join the dots to form 3 upright triangles (with same orientation to the 1 larger triangle contained). Then in the case of the 10 dot triangle we can form 6 upright triangles and in the 15 dot triangle 10 upright triangles of same orientation to larger triangle and so on.


Then the simplest 3-dimensional figure that is perfectly symmetrical with respect to its angular rotations is the tetrahedron which has the equilateral triangle as its base.

Though harder to represent (in 2-dimensional) terms, it is easy enough to see how the number of symmetrically arranged points (associated with this 4-sided figure follow the unique number sequence associated with (x – 1)4 = 0, i.e. 1, 4, 10, 20, …

So once again we start with 1 dot represented by the red dot at the top of the diagram (from Mathworld). Then to picture the simplest symmetrical tetrahedral figure we combine the 3 blue dots (representing the base of a tetrahedron with the red dot above) to obtain 4.

Then at the next level where each line of the base of the tetrahedron is divided in 2 we have now 6 green dots which combine the 3 blue and 1 red above to complete the new tetrahedron with 10 dots. Then with the next larger tetrahedron, we have 10 brown points at the base combined with the 6, 3 and 1 respectively above to complete the tetrahedron with 20 points.

So we can easily see here how this sequence of dots, representing the fully symmetrical nature of the 3-dimensional tetrahedron (with the least number of sides i.e. 4, possible in 3-dimensional space) is the unique digit sequence for (x – 1)4 = 0, i.e. 1, 4, 10, 20, 35, … 


However what is perhaps remarkable is that these insights can then be extended for the simplest fully symmetrical objects in n-dimensional space.

Thus - though impossible to properly visualise in 3 or less dimensions - we can equally envisage for example a 4-dimensional polytope, as a 5-sided equivalent in 4 space dimensions to the 4-sided tetrahedron in 3 space. This is known as a 5-cell or tehrahedral pyramid. (See Wikipedia entry).
Thus once again this contains the least number of sides i.e. 5, that a fully symmetrical object can possess in 4 space dimensions.  

So what we can now say therefore is that the unique number sequence associated with
(x – 1)5 = 0, i.e. 1, 5, 15, 35, 70, …  now describes the appropriate number of equally spaced points with respect to this object (as the number of points on each side progressively increases). 


Thus in general terms, the unique sequence of numbers associated with the polynomial equation
(x – 1)n = 0, encodes the manner, in which the equally spaced points of the simplex n-sided fully symmetrical geometrical objects, occurs in (n – 1) space dimensions.