Yesterday we looked at the extended case where 1 or more primes can occur at most twice in the unique factor composition of a number, showing how - using again the Riemann zeta function (for positive integers) - further sub-classifications for the frequency of occurrence of such numbers can be made.

So we demonstrated from one perspective how further numerical estimates can be made for the frequency of occurrence of numbers with 2 or more, 3 or more, 4 or more and in general terms n or more prime factors.

And then we showed, how using these estimates, we can then successively calculate the frequency of occurrence where exactly 1, exactly 2, exactly 3 and in general terms exactly n prime factors occur at most twice.

Not surprisingly these results can be extended universally for the factor compositions of numbers where primes can occur at most 3, 4, and - again in more general terms - n times.

We will look now at the case - which we have already partially addressed - where 1 or more prime factors of a number can occur at most 3 times (which I refer to as Class 3).

As we have seen, the relative frequency or probability occurrence of such a number in Class 3 (with respect to the number system as a whole) = 1/ζ(4) – 1/ζ(3) = .092 (approx) i.e. 9.2%.

However once again we can refine our results further, so as to provide an estimate for the frequency - or alternatively the probability - of a number where 2 or more, 3 or more, 4 or more and in general terms n or more prime factors can occur at most 3 times.

Now in the previous case where factors occurat most twice we, converted from 1 or more case to the more restricted case where 2 or more factors occur at most twice, by multiplying 1/ζ(3) – 1/ζ(2) i.e. .224 (approx) by {1/ζ(3) – 1/ζ(2}/2 i.e. .112 (approx) to obtain .025 (approx) i.e. 2.5%.

Thus the frequency of occurrence i.e. the probability of a number, where 2 or more prime factors occur at most twice = .025.

Then to further calculate the frequency of occurrence (the probability) of a number where 3 or more prime factors occur at most twice, we multiply the previous result by {1/ζ(3) – 1/ζ(2}/2 i.e. .112 (approx) to obtain .00028 (approx).

However in the present case, where initially 1 or more prime factors of a number can occur at most 3 times, we now multiply by {1/ζ(4) – 1/ζ(3)}/4 = .023 (approx) to convert from 1 or more to the more restricted 2 or more case.

Therefore the relative frequency or probability of a number, where in its composition 2 or more prime factors occur at most 3 times = 1/ζ(4) – 1/ζ(3) * {1/ζ(4) – 1/ζ(3)}/4 = .0021 (approx).

In other words, we would expect roughly 2 in a 1000 numbers to be comprised of unique factor combinations where 2 or more primes occur at most 3 times.

And then for further results - where the probability becomes increasingly small - we keep multiplying by the common ratio.

So for example, the relative frequency (the probability) of a number where 3 or more primes occur at most 3 times = .0021 * .023 = .000048 (approx) i.e. roughly 1 in 20,000!

As before we can then use these results to calculate the relative frequency (probability) of a number where exactly 1, 2, 3...n primes occur at most 3 times.

So for example, the relative frequency of a number where exactly 1 prime factor occurs at most 3 times = the frequency that 1 or more prime factors occur minus the frequency that 2 or more prime factors occur at most 3 times

= .092 – .0021 = .0899 (approx).

In other words, on average we would expect close to 9 in a 100 numbers to have a factor composition, where exactly 1 prime occurs at most 3 times.

In a previous illustration where we looked at the factor composition of the 16 numbers from 8000000000000001 to 8000000000000016 the number

8000000000000008 = 9091 * 2161 * 241 * 211 * 13 * 11 * 7 * 2 * 2 * 2,

provides an example of this case (where exactly 1 prime occurs at most 3 times).

Then to further calculate for example, the much smaller (relative) frequency (probability) of a number where exactly 2 primes occur at most 3 times we calculate the frequency of 2 or more minus 3 or more factors occurring at most 3 times = .0021 – .000048 = .00205 (approx)

However a further complication arises with this case, where prime factors occur at most 3 times, in that the remaining factors are therefore not necessarily made up of single (non-repeating) primes.

Where for example, exactly 1 prime factor occurs at most 3 times, 1 or more other factors may occur twice along with any remaining single prime factors.

So the question then arises as how to calculate the proportion attributable to prime factors that may occur twice.

Basically, this can be obtained quite simply.

For example one may wish to calculate the relative frequency (in relation to all numbers) of that subclass of numbers, where 1 prime factor occurs exactly 3 times, 1 other prime occurs exactly 2 times (with all the rest necessarily comprising factors that occur just once).

For example - cintinuing in the same stretch of numbers from which we previously illustrated -

8000000000000050 = 58771 * 997 * 419 * 19 * 7 * 7 * 7 * 5 * 5 * 2,

serves as an example of such a prime.

The first thing to observe here is that this represents a subclass of Class C, where now just 1 prime occurs exactly 3 times.

And as already stated above, the relative frequency (probability) of occurrence of this subclass = .0899 (approx).

Now, the relative frequency, where 1 other prime occurs at most 2 times (as explained in yesterday's blog entry) = .199.

Therefore to obtain the combined relative frequency or probability, where in a number, 1 prime factor occurs exactly 3 times and another prime factor occurs exactly 2 times, we multiply the two results

= .0899 * .199 = .0179.

In other words, slightly less than 2 in 100 numbers would belong to this prime factor classification.

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