Using Zeta Zeros to Count Factors (1)

As we have seen there is a close relationship as between the cumulative frequency of factors up to a given number n and the corresponding frequency of zeta zeros to a given number t (where n = t/2π).

As an approximation, the formula n(ln n – 1) can be used to approximate the accumulated total of factors to n, with t/2π(ln t/2π – 1) used to approximate the frequency of zeta zeros to t. And the latter formula for zeta zeros proves an especially good approximation!

So for example the cumulative frequency of factors to n = 100 (excluding 1 as a factor) is 382. And the number of trivial zeros to t = 200π (i.e. where n = t/2π) = 361.

So the percentage accuracy here is 94.5% and this increases towards 100% as n becomes progressively larger)

The question then arises as to whether this relationship can be extended to situations where certain factors are omitted.

For example if we omit all even numbers (where 2 is a factor) can we still establish a connection as between the cumulative number of odd factors to n and trivial zeros to a related number?

And the answer here is yes!

In fact is this case (with once again the trivial case of 1 as a factor excluded from consideration) the odd numbers i.e. 3, 5, 7, 9, 11, … will represent half of the divisors that can be used as factors in generating natural numbers.

So in the former case, where both even and odd divisors are used as factors n = t/2π.

Now where the number of divisors (as odd numbers) that can be used = ½ (of total of divisors), n = t/(2π * ½) = t/π.

In other words if we accumulate the total number of odd factors used in generating all numbers to n, this will be approximated closely by the corresponding number of zeta (non-trivial) zeros to nπ.

In the table below, I show the cumulative number of odd factors of all numbers up to n = 500 (starting with 50 and increasing in uniform blocks of 50).

Then I also show the corresponding number of trivial zeros to 500π (starting with 50π and increasing in uniform blocks of 50π).

And as one can see the approximation steadily improves as we ascend both number scales..

For example up to n = 100, we have 172 cumulative odd factors involved. And the corresponding number of zeta zeros to t = 100π = 145 with a relative accuracy here of 84.3%.

And when n = 500, we have 1262 odd factors and 1132 zeta zeros to 500π. So the relative accuracy has now improved to 89.54%. And this relative accuracy slowly increases towards 100% as n becomes progressively larger.

n	Odd factors to n	Zeta zeros to t (n = t/π)
50	70	55
100	172	145
150	290	249
200	413	361
250	541	479
300	680	602
350	818	729
400	963	860
450	1111	994
500	1262	1130

If we now for example were to exclude all numbers with 3 as a factors this would mean that 2/3 of all possible divisors would be used i.e. 2, 4, 5, 7, 8, 10, 11, …

And we would then in turn consider the cumulative factors of all natural numbers (excluding those with 3 is a factor).

To illustrate I will now accumulate the factors (excluding those with 3 as a factor) up to 20. Remember all numbers must be considered as some though divisible by 3 will contain other factor(s) not divisible by 3. For example, 15 though divisible by 3, contains 5 as a factor!

So 1 as a trivial factor is excluded in this case; 2 contains one factor, i.e. 2 (1);

3 is excluded; 4 contains two factors, i.e. 2 and 4 (2); 5 contains one factor, i.e. 5 (1); 6 is excluded; 7 contains one factor, i.e. 7 (1); 8 contains 3 factors i.e. 2, 4 and 8 (3); 9 is excluded; 10 contains three factors i.e. 2, 5 and 10 (3). 11 contains 1 factor, i.e. 11 (1); 12 is excluded; 13 contains one factor, i.e. 13 (1); 14 contains 3 factors, i.e. 2, 4 and 7 (3); 15 contains one factor i.e. 5 (1); 16 contains 4 factors i.e. 2, 4, 8 and 16 (4); 17 contains one factor i.e. 17; 18 is excluded; 19 contains one factor i.e. 19; 20 contains 5 factors i.e. 2, 4, 5, 10 and 20.

So the cumulative no. of factors to 20

= 1 + 2 + 1 + 1 + 3 + 3 + 1 + 1 + 3 + 1 + 4 + 1 + 1 + 5 = 28.  

As 2/3 of all divisors are involved in this case n = t/(2π * 2/3) = t /(4π/3)

Therefore t = 4nπ/3 = (80 * π)/3 = 84 (rounded to nearest integer).

And the number of zeta zeros to 84 = 22 (which already is a fairly reasonable approximation).

Therefore in general if k = percentage of total divisors involved in accumulating factors of numbers (containing these divisors) up to n, then n = t/(2π * k) where frequency of zeta zeros to t (on imaginary scale) is calculated.

Alternatively t = 2nπk 

Finally to illustrate this further, imagine we exclude from consideration all numbers containing either 2 or 3 as a factor.

Now, 1/2 of all divisors contain 2 as a factor and 1/3 contain 3. However as half of the factors containing 3 as a factor also contain 2, those divisors containing either 2 or 3 as factors =  1/2 + 1/6 = 2/3.

Therefore the remaining divisors i.e. 5, 7, 11, 13, … = 1/3 of total divisors.

So  with k = 1/3, if we accumulate the factors of all numbers (other than those divisible by 2 or 3 up to n) then t = 2nπ/3

The accumulated total of factors of numbers (neither divisible by 2 or 3) up to 100 = 97.

And the number of zeta zeros to 200π/3 = 209.43 = 81. So the percentage accuracy here = 85.5%

Then the accumulated total number of factors (neither divisible by 2 or 3) to 200 = 241.

And the number of zeta zeros to 400π/3 = 418.86 = 213. And the percentage accuracy has now improved to 88.4%.