Some Number Magic (2)

In yesterday’s blog entry, I sought to show how the sums of reciprocals of the unique number sequences associated with (x – 1)ⁿ = 0 (where n is an integer > 1) - what I refer to more simply as the alternative Zeta 2 function - can be shown to have fascinating connections with the corresponding Zeta 1 (Riemann) function.

In particular, I showed how to generate the respective terms of the Zeta 1 function i.e. ζ₁(s), for s = 2 from the alternative Zeta 2 function.

However, it is time now to attempt to generalise this procedure for to all values of ζ₁(s) where s is an integer ≥ 1.

Indeed the procedure appears trivial where s = 1.

So ζ₁(1) = 1 + 1/2 + 1/3 + 1/4 + ….

And this equates directly with the sum of reciprocals of the unique digit sequence for (x – 1)² = 1 + 1/2 + 1/3 + 1/4 + ….

Thus for simplicity we will refer to this latter expression of the same harmonic series as Alt ζ₂(2) i.e. where n in the polynomial expression (x – 1)ⁿ, from which the reciprocals of the unique number sequence involved is derived = 2.

However there is an alternative way of relating the two zeta expressions, which will be very useful for deriving the terms of ζ₁(s), where s > 1.

ζ₁(1), as the sum of  the terms of the harmonic series, diverges to infinity.

However we can express each term of this expression as half the sum of all Alt ζ₂(s), where s > 3.

So Alt ζ₂(3) = 2/1;  Alt ζ₂(4) = 3/2; Alt ζ₂(5) = 4/3; ; Alt ζ₂(6) = 5/4 and so on.

Therefore ζ₁(1) = 1/2{∑Alt ζ₂(s)} for s ≥ 3.    

So Alt ζ₂(2) = 1/2{∑Alt ζ₂(s)} for s ≥ 3.

Then as we saw yesterday to obtain the appropriate “conversion” for the terms of ζ₁(2),

we then use the respective terms of Alt ζ₂(3) multiplied by 1/2. Now 1/2 in this context can be more precisely expressed as 1/2!

So 1/2!{Alt ζ₂(3)} = 1/2(1 + 1/3 + 1/6 + 1/10 + 1/15 + …)

= 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + …

We then multiply each successive term by the corresponding sum of infinite series values for Alt ζ₂(3), Alt ζ₂(4), Alt ζ₂(5), Alt ζ₂(6), …

So   1/2{Alt ζ₂(3)} =   1/2 * 2/1 = 1       i.e. (1/1²)

       1/6{Alt ζ₂(4)} =   1/6 * 3/2 = 1/4    i.e. (1/2²)

     1/12{Alt ζ₂(5)} = 1/12 * 4/3 = 1/9    i.e. (1/3²)

     1/20{Alt ζ₂(6}   = 1/20 * 5/4 = 1/16  i.e. (1/4²)

            …                       …                  ….     

And we can continue on in this general manner to generate the further terms in the infinite series for ζ₁(2).

Then to generate the respective terms of ζ₁(3), we move on to consideration of the respective terms of Alt ζ₂(4) i.e. 1 + 1/4 + 1/10 + 1/20 + 1/35 + …

However in this case each individual term is now multiplied by 1/3! = 1/6.

So we now have 1/6 + 1/24 + 1/60 + 1/120 + 1/210 + …

Each of these individual terms is now successively multiplied by Alt ζ₂(4), Alt ζ₂(5),

Alt ζ₂(6), Alt ζ₂(7) … (Note that Alt ζ₂(3) is now omitted).

Now the value of Alt ζ₂(4)= 3/2. So we multiply the 1st term by this value; however we also multiply by (3 – 2)/2 = 1/2.

So in general terms we multiply each successive term by,

 (n – 1)/(n – 2)} * (n – 3)/(n – 2)

Therefore using the 1st term,        1/6 * 3/2 * 1/2     = 1/8 (i.e. 1/2³).

Then with the 2^nd term we have    1/24 * 4/3 * 2/3 =  1/27 (i.e. 1/3³).

With the 3^rd term, we have          1/60 * 5/4 * 3/4   = 1/64  (i.e. 1/4³).

Then with the 4^th, we have        1/120 * 6/5 * 4/5 =  1/125 (i.e. 1/5³).

                                                             …                          … 

Though the derivation might appear arbitrary, once the correct procedure is used, it works universally to generate all further terms.

Also this procedure preserves unique hidden number patterns.

For instance the 1^st case 1/6 * 3/2 * 1/2 = 1/2³ .

Now if we replace 3/2 with 3 * 2 we obtain 1/6 * (3 * 2) * 1/2 = 1/2.

Alternatively if we replace 1/2 by 1 * 2, we get 1/6 * 3/2 * 2 = 1/2 

And this pattern will universally hold.

For example in the last case we have 1/120 * 6/5 * 4/5 =  1/5³.

If we replace 6/5 by 6 * 5 we get 1/120 * (6 * 5) * 4/5 = 1/5

Then if we replace 4/5 by 4 * 5 we get 1/120 * 6/5 * 4 * 5 = 1/5.

So in this way we generate all the terms of the original harmonic series (except 1).

Admittedly it does get more cumbersome attempting to generate terms for ζ₁(s), when s > 2.

However a structured pattern for doing this exists in every case.

For example to create the individual terms of ζ₁(4) - less the 1^st two terms - we move to consideration of the corresponding terms of Alt ζ₂(5) i.e.

1 + 1/5 + 1/15 + 1/35 + 1/70 + …

We now multiply each of these terms by 1/4! = 1/24 to obtain

1/24 + 1/120 + 1/360 + 1/840  + 1/1680 + …

Alt ζ₂(5) = 4/3.

We now use a more complex conversion factor based on this value which in general terms is given as,

 (n – 1)/(n – 2) * (n – 3)/(n – 2) * (n – 4)/(n – 2)

So for Alt ζ₂(5), this gives 4/3 * 2/3 * 1/3

So we using the 1st term, 1/24 we obtain 1/24 * 4/3 * 2/3 * 1/3 = 1/81 (i.e. 1/3⁴)

Once again there are lovely hidden number patterns preserved in this formulation.

So replacing 4/3 by 4 * 3 we get 1/24 * (4 * 3) * 2/3 * 1/3 = 1/9 (i.e. 1/3²).

Then replacing 2/3 by 2 * 3 we get 1/24 * 4/3 * (2 * 3) * 1/3 = 1/9 (i.e. 1/3²).

Finally replacing 1/3 by 1 * 3 we get 1/24 * 4/3 * 2/3 * (1 * 3) = 1/9 (i.e. 1/3²).

And a similar pattern presents itself with all subsequent “conversions” so that we have the emergence of ζ₁(2), less the 1^st two terms.

Just to illustrate further, using the 2^nd term 1/120,

we obtain 1/120 * 5/4 * 3/4 * 2/4 = 1/256 (i.e. 1/4⁴).

Using the 3^rd term 1/360, we obtain,

1/360 * 6/5 * 4/5 * 3/5 = 1/625 (i.e. 1/5⁴).

Then using the 4^th term 1/840, we obtain

1/840 * 7/6 * 5/6 * 4/6 = 1/1296 (1/6⁴).

And again we can continue in this manner to generate the further terms for

ζ₁(4).

I will just finish by setting up the “conversion” to generate the computable term

i.e. the 4^th term of ζ₁(5).

Here we use the next series for Alt ζ₂(6) i.e.

1 + 1/6 + 1/21 + 1/56 + 1/126 + …

Each of these individual terms is now multiplied by 1/5! = 1/120 to obtain

1/120 + 1/720 + 1/2520 + 1/6720 + 1/15120 + …

The general conversion factor now contains 4 product terms

i.e. (n – 1)/(n – 2) * (n – 3)/(n – 2) * (n – 4)/(n – 2) * (n – 5)/(n – 2)

So when n = 6, this conversion factor = 5/4 * 3/4 * 2/4 * 1/4.

Therefore using the 1^st term, we get,

1/120 * 5/4 * 3/4 * 2/4 * 1/4 = 1/1024 = (i.e. 1/4⁵).

Again we can illustrate even more impressively this time, the number magic preserved within this formulation.

So replacing 5/4 with 5 * 4, we obtain,

1/120 * (5 * 4) * 3/4 * 2/4 * 1 /4 = 1/64 (i.e. 1/4³).

Then replacing 3/4 with 3 * 4 we obtain

1/120 * 5/4 * (3 * 4) * 2/4 * 1/4  = 1/64 (i.e. 1/4³).

Replacing 2/4 with 2 * 4 we obtain

1/120 * 5/4 * 3/4 * (2 * 4) * 1/4  = 1/64 (i.e. 1/4³).

Finally replacing 1/4 with 1 * 4 we obtain,

1/120 * 5/4 * 3/4 * 2/4 * (1 * 4) = 1/64 (i.e. 1/4³).