## Thursday, May 15, 2014

### Interesting Observations

I have commented on previously the important complementary link as between the simple formula for estimating the frequency of primes up to n, i.e. n/(log n – 1) and the corresponding simple formula for estimating the frequency of the factors of composite numbers up to n i.e. n(log n – 1).

The product of the two formulae is therefore n2.

This means that if we multiply the frequency of primes by the corresponding frequency of factors of composite numbers (up to n) that the result will closely approximate n2.

Alternatively, if we know the frequency of primes to a given number, then we can use this result to estimate the corresponding frequency of the sum of factors (of the composites). Alternatively, we can use the frequency of the sum of factors to estimate the corresponding frequency of the primes.

For example, we know that there are 25 primes up to 100. So n= 10,000 with the corresponding frequency of factors of composite numbers estimated at 10,000/25 = 400.

In fact as we have seen the actual number of such factors (up to 100) = 357.

So this approach already gives a reasonably accurate estimate (which improves in relative terms for higher n ultimately approaching 100% accuracy).

Alternatively we can use knowledge of the frequency of the sum of factors to estimate the corresponding frequency of primes.

So the estimate of frequency of primes (up to 100) = 10,000/357 = 28 (which is just 3 more than the correct result).

An even more striking expression of this result can be given with respect to the average frequency of primes and factors of composites up to a given number.

So quite simply, when we multiply the average frequency of primes (up to a given number) by the corresponding average frequency of factors of the composites (to the same number), the result will closely approximate 1 (with the accuracy increasing towards 100% in relative terms at higher n).

The average frequency of primes up to 100 = 25/100 = .25 and the average frequency of factors = 357/100 = 3.57.

The product of these two numbers = .8925 (which is already reasonably close to 1).

Now at 200, the average frequency of primes = 46/200 = .23 and the average frequency of factors = 852/200 = 4.26. So the product of both = .23 * 4.26 = .9798.

So we can see, that the relative accuracy has greatly increased in moving from 100 to 200 and already is very close to 1!

There is another very interesting point that can be made here!

As we have seen, up to n = 100, the average frequency of primes is 1 in 4. This means that the average unbroken sequence of composite numbers (between primes) = 3.

So the average number of factors  is closely related to this average unbroken sequence of composite numbers. In fact we just add 1 to the unbroken sequence.

We know for example that there are 46 primes up to 200. So the average is 1 in every 4.35 numbers (approx).

This means that the average unbroken sequence of composite numbers (between each prime) = 3.35.

So the estimate for average number of  factors per number (for n = 200) = 3.35 + 1 = 4.35.

And as we have seen the actual number of factors (up to 200) = 852.

This gives an average for each number of 852/200 = 4.26 (which compares very well with our estimate of 4.35).

I will make just one more observation at this point.

As we have seen the (accumulated) sum of the factors of the composite nos. up to n, on the real scale, is very closely related to the corresponding frequency of the Riemann (Zeta 1) zeros up to t, on the imaginary scale (where n = t/2π).

Therefore we can equally say that the product of the frequency of primes (up to n) multiplied by the corresponding frequency of zeros (up to t) approximates very closely to n2.

Thus, once more, we can use the frequency of primes to estimate the corresponding frequency of (non-trivial) zeros or alternatively use the frequency of the zeros to estimate the frequency of the primes.

Again, as we have seen the frequency of primes to 200 =  46.

Therefore we can estimate the corresponding frequency of zeros to 1256.63... (i.e. 200 * 2π),

as 2002/46 = 40,000/46 = 870 (to nearest integer).

As the the actual number of zeros  to 1256.63 = 861, we can see already the relative closeness of the two results!

Alternatively by making use of the actual existence of 861 zeros to 1256.63..., we can estimate the frequency of primes to 200 as 40,000/861 = 46 (to nearest integer) which in fact is the exact result in this case!