We have looked at the frequency of numbers where one or more prime factors can occur at most 2 and 3 times respectively, showing how all the more refined sub-categories that can then arise, can be calculated.
We will now look briefly in more detail at the case where 1 or more factors can occur at most 4 times before suggesting a general formula with which to calculate all possible cases.
Now once more, the relative frequency of numbers where 1 or more prime factors can occur at most 4 times = 1/ζ(5) – 1/ζ(4) = .040 (approx) i.e. 4.0%.
However, again we can make more refined predictions for the cases where 2 or more, 3 or more or in general terms, n or more primes can occur at most 4 times.
Once again the key to this is the common ratio. So once we have found the common ratio we successively multiply the frequency of the case where 1 or more primes can occur at most 4 times by the common ratio, to obtain all the other results.
Now the common ratio in this case is given as {1/ζ(5) – 1/ζ(4)}/8 = .005 (approx).
Notice that a definite pattern is emerging?
We have already seen how with respect to the (internal) frequency of prime factors associated with each of the major classes, that such a definite pattern emerged.
So for those factors associated with numbers where 1 or more primes occur at most 2 times (Class 2), the overall frequency → 1/2 of Class 1 (where each prime factor occurs just once in a number).
Then, for those factors associated with numbers where 1 or more primes occur at most 3 times (Class 3) the overall frequency → 1/4 of Class 1.
And continuing for those factors associated with numbers where 1 or more primes occur at most 4 times (Class 4) the overall frequency → 1/8 of Class 1.
So the fractions involved are successive terms in the simple geometric series 1/2, 1/4, 1/8, ....
the sum of which = 1.
Thus in this case we simply divide the starting key estimate (for where 1 or more primes can occur at most n times) by the appropriate fraction (as successive terms in this simple geometric series) to calculate the common ratio, which is then applied to all the more restricted sub-classes that we have mentioned.
So in general terms, where we start by considering the relative frequency of numbers where 1 or more primes occur at most n times, the common ratio is given as
{1/ζ(n + 1) – 1/ζ(n)}/2n –
1
Therefore in this case to calculate the frequency where 2 or more primes can occur at most 4 times we obtain {1/ζ(5) – 1/ζ(4)} * {1/ζ(5) – 1/ζ(4)}/8 = .0002 (approx).
So this means that on average only 2 in 10,000 numbers will be comprised of a prime factor combination, where 2 or more factors can occur at most 4 times!
And then to obtain the frequency associated with the numbers where 3 or more primes can occur at most 4 times we multiply in turn .0002 by the common ratio (i.e. .005).
Now once again, armed with these results, we can then calculate the frequency for the occurrence of all exact combinations (made up of 4 prime factors).
Therefore the frequency of occurrence for numbers where exactly 1 prime factor occurs at most 4 times = frequency where 1 or more prime factors occur at most 4 times – frequency where 2 or more prime factors occur at most 4 times
= .04 – .0002 = .0398.
And we can continue on in this fashion to calculate the corresponding frequencies associated with exactly 2, exactly 3,.....exactly n primes occurring at most 4 times.
Now the case where at least 1 prime occurs 4 times in the factor composition of a number, leaves open the possibility of varying combinations of other primes where 2 or 3 occurrences may be involved.
For example continuing in the same stretch of numbers, from which I have been illustrating (8000000000000001 +), the following number occurs,
8000000000000144 = 12758387 * 3221 * 23 * 23 * 23 * 2 * 2 * 2 * 2
Now this represents an example of a number where exactly 1 prime factor occurs at most 4 times, with exactly 1 other prime factor occurring at most 3 times (with the remaining factors occurring but once).
So the issue now is to calculate the relative frequency (with respect to the number system as a whole) of this subset of numbers where exactly 1 prime occurs at most 4 times and exactly 1 other prime occurs 3 times, (with all others occurring once).
Therefore the first requirement is to calculate the relative frequency of numbers where exactly 1 prime factor occurs at most 4 times.
And - as we have seen above the result here = .0398 (approx).
The next step is then to combine this with the relative frequency of numbers where exactly 1 other prime occurs at most 3 times, which as illustrated in yesterday's entry = .0899.
Therefore the relative frequency of that subset of numbers where exactly 1 prime factor occurs at most 4 times and exactly 1 other factor occurs 3 times (with all other factors occurring once),
= .0398 * 0899 = .0036 (approx).
Therefore on average we would expect somewhere between 3 and 4 numbers in every 1000 to belong to this subset of numbers.
We therefore now have the means to calculate for every possible subset of numbers (based on the varying combinations of prime factors) the relative frequency (probability) with which it is associated with respect to the number system as a whole.
And in one way this is quite remarkable and requiring nothing more than the Riemann zeta function (for the positive integers) .
Wednesday, November 30, 2016
Tuesday, November 29, 2016
Extended Riemann Connections (2)
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 - continuing 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.
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 - continuing 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.
Monday, November 28, 2016
Extended Riemann Connections (1)
As we have already seen, the frequency of - what I term - the Type 2 Class - where 1 or more prime factors of a number occur at most 2 times (with therefore all other factors occurring just once) is given as 1/{ζ(3) – 1/ζ(2)} = .224 (approx) i.e. 22.4%.
Once again as I illustrated in the last entry for the numbers between 8000000000000001 - 16,
8000000000000001 = 4447 * 1423 * 409 * 163 * 43 * 7 * 7 * 3 * 3 ,
8000000000000004 = 37884167 * 138563 * 127 * 3 * 2 * 2 ,
8000000000000006 = 601775236949 * 23 * 17 * 17 * 2,
8000000000000010 = 88888888888889 * 5 * 3 * 3 * 2 and
8000000000000012 = 105263157894737 * 19 * 2 * 2,
belong to this class (i.e. Class 2).
In the first case, we have two factors that occur at most twice; in each of the other cases just one factor that occurs at most twice!
And - though with much less probability - it is possible that 2 or more, 3 or more, 4 or more and in general terms, n or more prime factors could occur at most twice.
Therefore a more refined estimation should allow one to calculate the exact frequency of occurrence (with respect to the number system as a whole) of all these sub-classes of Class 2.
So once again, 1/ζ(3) – 1/ζ(2) = .224 (approx) represents the frequency of occurrence for all cases where 1 or more prime factors occur at most 2 times.
To calculate the more restricted frequency of occurrence for all cases where now 2 or more prime factors occur at most 2 times we obtain {ζ(3) – 1/ζ(2)}2/2 = .02508... i.e. .025 (approx).
So we would expect on average just 2.5% of all numbers to belong to this subclass (of Class 2)
Then to calculate the frequency of occurrence now for all cases where 3 or more prime factors occur at most 2 times we obtain {ζ(3) – 1/ζ(2)}3/4 = .0028 (approx).
Therefore, we would expect just .28% (i.e. slightly less than 3 numbers in a 1000) to belong to this class.
Then the frequency of occurrence where 4 or more prime factors occur at most 2 times = {ζ(3) – 1/ζ(2)}4/8 = .0003 (approx) i.e. just 3 in 10,000 numbers.
In more general terms to calculate the frequency of occurrence, when n or more factors occur at most 2 times we obtain {ζ(3) – 1/ζ(2)}n/2n – 1
Alternatively, we could start from the first result i.e. the frequency of occurrence where 1 or more prime factors occur at most 2 times and continually multiply by a constant factor of {ζ(3) – 1/ζ(2)}/2 to find the more restricted - and progressively less frequent - occurrence of all the other subclasses.
Armed with the above information we are then in a position to find the frequency of occurrence where exactly 1, exactly 2, exactly 3, and - in more general terms - exactly n prime factors occur at most twice. And these now comprise further subclasses of the subclasses already defined!
So for example the frequency of occurrence of all those numbers where 1 factor occurs at most exactly twice represents a further subclass of the - already defined - subclass, where 1 or more factors occurs at most twice!
Thus to find the frequency of occurrence of this new subclass we simply subtract the frequency of occurrence where 2 or more primes occur at most twice from the corresponding frequency of occurrence where 1 or more primes occur at most twice,
i.e. 1/ζ(3) – 1/ζ(2) – {ζ(3) – 1/ζ(2)}2/2 = .224 – .025 = .199 (approx) = 19.9%.
Therefore we can conclude that with respect to the number system as a whole, that very close to 1 in 5 members will belong to that subclass of numbers, where 1 prime factor occurs at most twice (with all other factors occurring just once).
Then, to find the frequency of occurrence of all those numbers, where 2 factors occur at most exactly twice, we subtract the frequency of occurrence, where 3 or more primes occur at most twice from the corresponding frequency of occurrence, where 2 or more occur at most twice,
i.e. {ζ(3) – 1/ζ(2)}2/2 – {ζ(3) – 1/ζ(2)}3/4 = .025 – .0028 = .022 (approx) = 2.2%.
In more general terms to calculate the frequency of occurrence, where n factors occur at most exactly twice, we subtract the frequency of occurrence, where (n + 1) or more primes occur at most twice from the corresponding frequency of occurrence where n or more occur at most twice,
i.e. {ζ(3) – 1/ζ(2)}n/2n – 1 – {ζ(3) – 1/ζ(2)}(n + 1)/2n
It should of course be clear that the a geometric type progression connect the various terms where 1, 2, 3, ... n primes occur exactly twice. So here again, the 1st term, corresponding to the frequency of occurrence where exactly 1 prime factor occurs at most twice =
1/ζ(3) – 1/ζ(2) – {ζ(3) – 1/ζ(2)}2/2 = .224 – .025 = .199 (approx)
The corresponding terms, corresponding to the frequency of occurrence of exactly 1, 2, 3, ... n prime factors occurring at most twice require the successive multiplication of .199 by a common ratio =
{1/ζ(3) – 1/ζ(2)}/2 = .112 (approx).
So, alternatively, for example the estimated frequency of occurrence of numbers (with respect to the number system as a whole) where exactly 2 prime factors occur at most twice = .199 * .112 = .022 (approx).
Then the estimated frequency of occurrence of numbers where exactly 3 prime factors occur at most twice = .022 * .112 = .0025 (approx).
And we can continue on in this manner to generate the corresponding frequency of occurrence where exactly 4, 5,....n prime factors occur at most twice.
Once again as I illustrated in the last entry for the numbers between 8000000000000001 - 16,
8000000000000001 = 4447 * 1423 * 409 * 163 * 43 * 7 * 7 * 3 * 3 ,
8000000000000004 = 37884167 * 138563 * 127 * 3 * 2 * 2 ,
8000000000000006 = 601775236949 * 23 * 17 * 17 * 2,
8000000000000010 = 88888888888889 * 5 * 3 * 3 * 2 and
8000000000000012 = 105263157894737 * 19 * 2 * 2,
belong to this class (i.e. Class 2).
In the first case, we have two factors that occur at most twice; in each of the other cases just one factor that occurs at most twice!
And - though with much less probability - it is possible that 2 or more, 3 or more, 4 or more and in general terms, n or more prime factors could occur at most twice.
Therefore a more refined estimation should allow one to calculate the exact frequency of occurrence (with respect to the number system as a whole) of all these sub-classes of Class 2.
So once again, 1/ζ(3) – 1/ζ(2) = .224 (approx) represents the frequency of occurrence for all cases where 1 or more prime factors occur at most 2 times.
To calculate the more restricted frequency of occurrence for all cases where now 2 or more prime factors occur at most 2 times we obtain {ζ(3) – 1/ζ(2)}2/2 = .02508... i.e. .025 (approx).
So we would expect on average just 2.5% of all numbers to belong to this subclass (of Class 2)
Then to calculate the frequency of occurrence now for all cases where 3 or more prime factors occur at most 2 times we obtain {ζ(3) – 1/ζ(2)}3/4 = .0028 (approx).
Therefore, we would expect just .28% (i.e. slightly less than 3 numbers in a 1000) to belong to this class.
Then the frequency of occurrence where 4 or more prime factors occur at most 2 times = {ζ(3) – 1/ζ(2)}4/8 = .0003 (approx) i.e. just 3 in 10,000 numbers.
In more general terms to calculate the frequency of occurrence, when n or more factors occur at most 2 times we obtain {ζ(3) – 1/ζ(2)}n/2n – 1
Alternatively, we could start from the first result i.e. the frequency of occurrence where 1 or more prime factors occur at most 2 times and continually multiply by a constant factor of {ζ(3) – 1/ζ(2)}/2 to find the more restricted - and progressively less frequent - occurrence of all the other subclasses.
Armed with the above information we are then in a position to find the frequency of occurrence where exactly 1, exactly 2, exactly 3, and - in more general terms - exactly n prime factors occur at most twice. And these now comprise further subclasses of the subclasses already defined!
So for example the frequency of occurrence of all those numbers where 1 factor occurs at most exactly twice represents a further subclass of the - already defined - subclass, where 1 or more factors occurs at most twice!
Thus to find the frequency of occurrence of this new subclass we simply subtract the frequency of occurrence where 2 or more primes occur at most twice from the corresponding frequency of occurrence where 1 or more primes occur at most twice,
i.e. 1/ζ(3) – 1/ζ(2) – {ζ(3) – 1/ζ(2)}2/2 = .224 – .025 = .199 (approx) = 19.9%.
Therefore we can conclude that with respect to the number system as a whole, that very close to 1 in 5 members will belong to that subclass of numbers, where 1 prime factor occurs at most twice (with all other factors occurring just once).
Then, to find the frequency of occurrence of all those numbers, where 2 factors occur at most exactly twice, we subtract the frequency of occurrence, where 3 or more primes occur at most twice from the corresponding frequency of occurrence, where 2 or more occur at most twice,
i.e. {ζ(3) – 1/ζ(2)}2/2 – {ζ(3) – 1/ζ(2)}3/4 = .025 – .0028 = .022 (approx) = 2.2%.
In more general terms to calculate the frequency of occurrence, where n factors occur at most exactly twice, we subtract the frequency of occurrence, where (n + 1) or more primes occur at most twice from the corresponding frequency of occurrence where n or more occur at most twice,
i.e. {ζ(3) – 1/ζ(2)}n/2n – 1 – {ζ(3) – 1/ζ(2)}(n + 1)/2n
It should of course be clear that the a geometric type progression connect the various terms where 1, 2, 3, ... n primes occur exactly twice. So here again, the 1st term, corresponding to the frequency of occurrence where exactly 1 prime factor occurs at most twice =
The corresponding terms, corresponding to the frequency of occurrence of exactly 1, 2, 3, ... n prime factors occurring at most twice require the successive multiplication of .199 by a common ratio =
{1/ζ(3) – 1/ζ(2)}/2 = .112 (approx).
So, alternatively, for example the estimated frequency of occurrence of numbers (with respect to the number system as a whole) where exactly 2 prime factors occur at most twice = .199 * .112 = .022 (approx).
Then the estimated frequency of occurrence of numbers where exactly 3 prime factors occur at most twice = .022 * .112 = .0025 (approx).
And we can continue on in this manner to generate the corresponding frequency of occurrence where exactly 4, 5,....n prime factors occur at most twice.
Friday, November 11, 2016
Riemann Zeta Function: Important Number Relationships (12)
Just to illustrate a little better the terminology, I have been using in recent blog entries with respect to the distribution of the prime factors of the natural numbers, I include below the factor breakdown for the 16 consecutive numbers between 800000000000001 and 800000000000016.
8000000000000001 = 4447 * 1423 * 409 * 163 * 43 * 7 * 7 * 3 * 3
8000000000000002 = 23121387283237 * 173 * 2
8000000000000003 = 159897739 * 112939 * 443
8000000000000004 = 37884167 * 138563 * 127 * 3 * 2 * 2
8000000000000005 = 3470715835141 * 461 * 5
8000000000000006 = 601775236949 * 23 * 17 * 17 * 2
8000000000000007 = 33471823001 * 79669 * 3
8000000000000008 = 9091 * 2161 * 241 * 211 * 13 * 11 * 7 * 2 * 2 * 2
8000000000000009 = 556261 * 42683 * 107 * 67 * 47
8000000000000010 = 88888888888889 * 5 * 3 * 3 * 2
8000000000000011 = 8000000000000011 * 1
8000000000000012 = 105263157894737 * 19 * 2 * 2
8000000000000013 = 99826551367 * 26713 * 3
8000000000000014 = 9364339 * 468883 * 911 * 2
8000000000000015 = 20476333 * 11162713 * 7 * 5
8000000000000016 = 166666666666667 * 3 * 2 * 2 * 2 * 2
Now the factors of the first number belong to that class where 1 or more primes occur at most 2 times. As we can see both 7 and 3 occur here twice (with all other factors occurring just once).
As we have seen for the number system as a whole, the proportion of all numbers belonging to each of the designated factor classes is governed by the Riemann zeta function (for the positive integers).
The proportion belonging to this class i.e. where again 1 or more prime factors occur at most 2 times = 1/ζ(3) – 1/ζ(2) = .224 (approx) i.e. 22.4%. For convenience we will refer to this as Class 2.
The factors of the second number belong to the most frequent class of numbers, where each prime occurs on just 1 occasion (i.e. where no prime occurs more than once).
The proportion belonging to this class = 1/ζ(2) – 1/ζ(1) = .608 (approx) i.e. 60.8%. For convenience we will refer to this as Class 1.
The prime factors of the third number again belongs to to Class 1 (where each prime occurs just once).
The prime factors of the 4th number then belong to Class 2. (On this occasion only 1 prime occurs at most twice!)
The prime factors of the 5th number again belong to Class 1; the factors of the 6th belong to Class 2 and the factors of the 7th number to Class 1.
The factors of the 8th number belong however to a new class i.e. Class 3, where 1 or more factors occur at most 3 times. On this occasion only one factor i.e. 2 occurs 3 times.
The proportion of all numbers belonging to Class 3 = 1/ζ(4) – 1/ζ(3) = .092 (approx) i.e. 9.2% of the accumulated total of factors for all numbers.
Then the factors of the next 7 numbers belong to Classes 1, 2, 1, 2, 1, 1 and 1 respectively.
Finally, the factors of the 16th number belongs to a new class i.e. Class 4, where 1 or more primes occur at most 4 times. (Here, 2 occurs 4 times!)
The proportion of all numbers belonging to Class 4 = 1/ζ(5) – 1/ζ(4) = .040 (approx) i.e. 4.0%.
Of course the factors of other numbers may belong with less frequency to other classes (that do not arise in this example).
In general, Class n can be defined as where 1 or more prime factors of a number occur at most n times.
And the probability of such occurrence = 1/{ζ(n + 1) – 1/ζ(n)}.
What is illustrated above refers to - what I term - the external nature of the number system, Here the proportion of all Classes - as defined by the Riemann zeta function (for the positive integers) - relates to the overall frequency with which the natural numbers (as defined by prime factors of a certain type above) occur.
However there is a corresponding internal nature to the number system which - properly understood - is dynamically complementary with the external aspect. This relates to the corresponding proportion with respect to the number system as a whole with which the average frequency of the combined number of factors (associated with each Class Type) occurs.
So for example, we would intuitively expect on average to find more prime factors belonging to those classes i.e. 2 or higher (where one or more factors occur more than once), than Class 1 (where no such repetition of factors occurs).
The important implication here is that neither external nor internal aspects are independent of each other but rather synchronistically arise in a holistic manner that - ultimately - is of a purely relative nature.
So the internal aspect of measurement relates to calculation of the average frequency (with respect to the overall average frequency of prime factors for the number system as a whole).
Again, for convenience, we will refer to this category of numbers (which of course represents all the natural numbers) as Class 0.
Therefore, the task in terms of the internal distribution of the prime factors of numbers is to assess the average proportion of prime factors in each class (i.e. Class 1, Class 2, Class 3,......Class n) with respect to the average proportion of prime factors for the number system as a whole (i.e. Class 0).
Now, given acceptance with respect to the conjecture of the last entry that the total number of prime factors belonging to Class 1 represents 50% of the factors of Class 0 (i.e. the cumulative sum of factors of all numbers), the total number of prime factors belonging to Class 2 represents 25% of all factors, the total number of prime factors belonging to Class 3 represents 12.5% of all factors and so on, we thereby have a ready means of calculating this internal distribution of prime factors.
For example though Class 1 accounts for 1/ζ(2) – 1/ζ(1) = .608 (approx) of the frequency of all numbers with no repeating prime factors, it accounts, as conjectured yesterday, for only 50% of the cumulative sum of all such factors (with respect to the number system as a whole).
Therefore to obtain the average frequency of such factors (with respect to the average frequency of factors for the number system as a whole) we divide .5 by 1/ζ(2) – 1/ζ(1) to obtain 1/2{ζ(2) – 1/ζ(1)} = .822 (approx).
Then to obtain the average frequency of prime factors of Class 2 - where 1 or more primes occur at most twice - we divide .25 by 1/ζ(3) – 1/ζ(2) = .224 (approx) to obtain 1/4{ζ(3) – 1/ζ(2)} = 1.116 (approx). So the average frequency of prime factors for this class is somewhat greater than for the number system as a whole.
Likewise to obtain the average frequency of prime factors of Class 3 - where 1 or more primes occur at most 3 times - we divide .125 by 1/ζ(4) – 1/ζ(3) = .092 (approx) to obtain 1/8{ζ(4) – 1/ζ(3)} = 1.358 (approx). So the average frequency of prime factors for this class as indeed all subsequent classes has further increased, and therefore again greater than for the number system as a whole.
In fact we can provide a simple general means of calculating all these proportions.
So to obtain the average frequency of prime factors of Class n - where 1 or more primes occur at most n times - we divide 1/2n by 1/{ζ(n + 1) – 1/ζ(n)} to obtain
1/2n{ζ(n + 1) – 1/ζ(n)}.
8000000000000001 = 4447 * 1423 * 409 * 163 * 43 * 7 * 7 * 3 * 3
8000000000000002 = 23121387283237 * 173 * 2
8000000000000003 = 159897739 * 112939 * 443
8000000000000004 = 37884167 * 138563 * 127 * 3 * 2 * 2
8000000000000005 = 3470715835141 * 461 * 5
8000000000000006 = 601775236949 * 23 * 17 * 17 * 2
8000000000000007 = 33471823001 * 79669 * 3
8000000000000008 = 9091 * 2161 * 241 * 211 * 13 * 11 * 7 * 2 * 2 * 2
8000000000000009 = 556261 * 42683 * 107 * 67 * 47
8000000000000010 = 88888888888889 * 5 * 3 * 3 * 2
8000000000000011 = 8000000000000011 * 1
8000000000000012 = 105263157894737 * 19 * 2 * 2
8000000000000013 = 99826551367 * 26713 * 3
8000000000000014 = 9364339 * 468883 * 911 * 2
8000000000000015 = 20476333 * 11162713 * 7 * 5
8000000000000016 = 166666666666667 * 3 * 2 * 2 * 2 * 2
Now the factors of the first number belong to that class where 1 or more primes occur at most 2 times. As we can see both 7 and 3 occur here twice (with all other factors occurring just once).
As we have seen for the number system as a whole, the proportion of all numbers belonging to each of the designated factor classes is governed by the Riemann zeta function (for the positive integers).
The proportion belonging to this class i.e. where again 1 or more prime factors occur at most 2 times = 1/ζ(3) – 1/ζ(2) = .224 (approx) i.e. 22.4%. For convenience we will refer to this as Class 2.
The factors of the second number belong to the most frequent class of numbers, where each prime occurs on just 1 occasion (i.e. where no prime occurs more than once).
The proportion belonging to this class = 1/ζ(2) – 1/ζ(1) = .608 (approx) i.e. 60.8%. For convenience we will refer to this as Class 1.
The prime factors of the third number again belongs to to Class 1 (where each prime occurs just once).
The prime factors of the 4th number then belong to Class 2. (On this occasion only 1 prime occurs at most twice!)
The prime factors of the 5th number again belong to Class 1; the factors of the 6th belong to Class 2 and the factors of the 7th number to Class 1.
The factors of the 8th number belong however to a new class i.e. Class 3, where 1 or more factors occur at most 3 times. On this occasion only one factor i.e. 2 occurs 3 times.
The proportion of all numbers belonging to Class 3 = 1/ζ(4) – 1/ζ(3) = .092 (approx) i.e. 9.2% of the accumulated total of factors for all numbers.
Then the factors of the next 7 numbers belong to Classes 1, 2, 1, 2, 1, 1 and 1 respectively.
Finally, the factors of the 16th number belongs to a new class i.e. Class 4, where 1 or more primes occur at most 4 times. (Here, 2 occurs 4 times!)
The proportion of all numbers belonging to Class 4 = 1/ζ(5) – 1/ζ(4) = .040 (approx) i.e. 4.0%.
Of course the factors of other numbers may belong with less frequency to other classes (that do not arise in this example).
In general, Class n can be defined as where 1 or more prime factors of a number occur at most n times.
And the probability of such occurrence = 1/{ζ(n + 1) – 1/ζ(n)}.
What is illustrated above refers to - what I term - the external nature of the number system, Here the proportion of all Classes - as defined by the Riemann zeta function (for the positive integers) - relates to the overall frequency with which the natural numbers (as defined by prime factors of a certain type above) occur.
However there is a corresponding internal nature to the number system which - properly understood - is dynamically complementary with the external aspect. This relates to the corresponding proportion with respect to the number system as a whole with which the average frequency of the combined number of factors (associated with each Class Type) occurs.
So for example, we would intuitively expect on average to find more prime factors belonging to those classes i.e. 2 or higher (where one or more factors occur more than once), than Class 1 (where no such repetition of factors occurs).
The important implication here is that neither external nor internal aspects are independent of each other but rather synchronistically arise in a holistic manner that - ultimately - is of a purely relative nature.
So the internal aspect of measurement relates to calculation of the average frequency (with respect to the overall average frequency of prime factors for the number system as a whole).
Again, for convenience, we will refer to this category of numbers (which of course represents all the natural numbers) as Class 0.
Therefore, the task in terms of the internal distribution of the prime factors of numbers is to assess the average proportion of prime factors in each class (i.e. Class 1, Class 2, Class 3,......Class n) with respect to the average proportion of prime factors for the number system as a whole (i.e. Class 0).
Now, given acceptance with respect to the conjecture of the last entry that the total number of prime factors belonging to Class 1 represents 50% of the factors of Class 0 (i.e. the cumulative sum of factors of all numbers), the total number of prime factors belonging to Class 2 represents 25% of all factors, the total number of prime factors belonging to Class 3 represents 12.5% of all factors and so on, we thereby have a ready means of calculating this internal distribution of prime factors.
For example though Class 1 accounts for 1/ζ(2) – 1/ζ(1) = .608 (approx) of the frequency of all numbers with no repeating prime factors, it accounts, as conjectured yesterday, for only 50% of the cumulative sum of all such factors (with respect to the number system as a whole).
Therefore to obtain the average frequency of such factors (with respect to the average frequency of factors for the number system as a whole) we divide .5 by 1/ζ(2) – 1/ζ(1) to obtain 1/2{ζ(2) – 1/ζ(1)} = .822 (approx).
Then to obtain the average frequency of prime factors of Class 2 - where 1 or more primes occur at most twice - we divide .25 by 1/ζ(3) – 1/ζ(2) = .224 (approx) to obtain 1/4{ζ(3) – 1/ζ(2)} = 1.116 (approx). So the average frequency of prime factors for this class is somewhat greater than for the number system as a whole.
Likewise to obtain the average frequency of prime factors of Class 3 - where 1 or more primes occur at most 3 times - we divide .125 by 1/ζ(4) – 1/ζ(3) = .092 (approx) to obtain 1/8{ζ(4) – 1/ζ(3)} = 1.358 (approx). So the average frequency of prime factors for this class as indeed all subsequent classes has further increased, and therefore again greater than for the number system as a whole.
In fact we can provide a simple general means of calculating all these proportions.
So to obtain the average frequency of prime factors of Class n - where 1 or more primes occur at most n times - we divide 1/2n by 1/{ζ(n + 1) – 1/ζ(n)} to obtain
1/2n{ζ(n + 1) – 1/ζ(n)}.
Wednesday, November 9, 2016
Riemann Zeta Function: Important Number Relationships (11)
To illustrate the subject matter of the previous blog entry "Riemann Zeta Function: Important Number Relations (10)", I am now providing some preliminary sample evidence that I conducted in assessing the ratios mentioned yesterday.
In this sample, I surveyed the 500 numbers of the 14 digit numbers in sequence from 15000000006001 - 150000000006500 with respect to frequency and also with respect to the combined number of factors contained.
Thus for example with reference to the first 100 of these numbers, 61 represents the frequency of those numbers which contain at most 1 prime factor (i.e. no repeating primes) and 215 then represents the cumulative number of prime factors contained with respect to these 61 numbers!
So in the case of each prime factor repeating category, I then worked out both the total frequency for the 500 numbers and the corresponding total cumulative frequency of the prime factors involved.
Prop. 1 .4635 .2576 .1273 .0623
I then worked out the average no. of factors involved in each case, which then provides estimates for the ratios involved (as mentioned in the last entry).
For example, the ratio here of the average factor frequency of those numbers with at most 1 prime factor to those where at most 1 or more factors occur 2 times = 3.485/5.435 = .641.
The suggested estimate directly based on the Riemann zeta function (for positive integers) of 1/ζ(2) = .6079. So the estimate is not that accurate. However our sample is relatively small with the average number of factors per number involved still at a very low rate!
In this sample, I surveyed the 500 numbers of the 14 digit numbers in sequence from 15000000006001 - 150000000006500 with respect to frequency and also with respect to the combined number of factors contained.
Thus for example with reference to the first 100 of these numbers, 61 represents the frequency of those numbers which contain at most 1 prime factor (i.e. no repeating primes) and 215 then represents the cumulative number of prime factors contained with respect to these 61 numbers!
So in the case of each prime factor repeating category, I then worked out both the total frequency for the 500 numbers and the corresponding total cumulative frequency of the prime factors involved.
All Nos.
Fr. Factors |
At Most 1
Fr. Factors
|
At
Most 2
Fr. Factors
|
At Most 3
Fr. Factors
|
At
Most 4
Fr. Factors
|
100 456
|
60 215
|
22 119
|
8 55
|
4 27
|
100 451
|
61 207
|
22 119
|
9 55
|
3 20
|
100 453
|
60 204
|
23 128
|
8 53
|
4 34
|
100 464
|
63 227
|
17 89
|
10 69
|
4 30
|
100 454
|
59 203
|
24 132
|
9 58
|
4 30
|
500 2278
|
303 1056
|
108 587
|
44 290
|
19 142
|
Prop. 1 .4635 .2576 .1273 .0623
Av. 4.556 3.485 5.435 6.591 7.474
For example, the ratio here of the average factor frequency of those numbers with at most 1 prime factor to those where at most 1 or more factors occur 2 times = 3.485/5.435 = .641.
The suggested estimate directly based on the Riemann zeta function (for positive integers) of 1/ζ(2) = .6079. So the estimate is not that accurate. However our sample is relatively small with the average number of factors per number involved still at a very low rate!
The next ratio estimate of the corresponding average factor frequency of those numbers where at most 1 or more factors occur 2 times to those where at most 1 or more prime factors occur 3 times = 5.435/6.591 = .825.
In this case, the estimate is very close to the true postulated value i.e. 1/ζ(3) = .832, though not too much significance should be read into this fact.
Then the corresponding estimate of the ratio of average factor frequency of those numbers where at most 1 or more prime factors occur 4 times = 6.591/7.474 = .882. This compares with the suggested result of 1ζ(4) = .924. Again given the limited data on which the sample estimate is based, this result gives support to the view that the underlying pattern here is indeed described by the Riemann zeta function for 1/ζ(2), 1/ζ(3), 1/ζ(4) and so on!
Then the corresponding estimate of the ratio of average factor frequency of those numbers where at most 1 or more prime factors occur 4 times = 6.591/7.474 = .882. This compares with the suggested result of 1ζ(4) = .924. Again given the limited data on which the sample estimate is based, this result gives support to the view that the underlying pattern here is indeed described by the Riemann zeta function for 1/ζ(2), 1/ζ(3), 1/ζ(4) and so on!
However as I made further estimates, I became aware of an important trend that eventually led me to abandon my initial estimates.
Basically, what I found is that these estimates of ratios tended to vary depending on how high up the number scale numbers (with their constituent prime factors) are taken. For example, the important first ratio (measuring the average factor frequency of those numbers where at most factors occur once (i.e. where all are non-repeating) to those with at most 1 or more factors occur 2 times tended to increase for numbers higher up the number scale.
While still confident that a relationship entailing the Riemann zeta function (for positive integer values) still governed the internal distribution of the primes, this led me to look in another direction (which I return to in the next entry).
Basically, what I found is that these estimates of ratios tended to vary depending on how high up the number scale numbers (with their constituent prime factors) are taken. For example, the important first ratio (measuring the average factor frequency of those numbers where at most factors occur once (i.e. where all are non-repeating) to those with at most 1 or more factors occur 2 times tended to increase for numbers higher up the number scale.
While still confident that a relationship entailing the Riemann zeta function (for positive integer values) still governed the internal distribution of the primes, this led me to look in another direction (which I return to in the next entry).
This direction came from the additional consideration of the the total number of accumulated prime factors associated with each of our categories in the number system.
This entails combining distributions with respect to both the external and internal aspects of the number system.
For example though 61% (approx) of all numbers are externally composed of prime factor combinations (where no prime occurs more than once) the internal average frequency of overall factor occurrence for such numbers is less than any of the other categories.
Thus combining both external and internal considerations - in the line marked as prop. (i.e. proportion) - the proportion of all accumulated prime factors (2278 in our example) in relation to the total = 1.
The with respect to the accumulation of prime factors (for those numbers where each factor occurs at most 1 time (= 1056), the proportion - in our example - is slightly less than 1/2 of total (i.e. .4635).
The with respect to the accumulation of prime factors (for those numbers where each factor occurs at most 1 time (= 1056), the proportion - in our example - is slightly less than 1/2 of total (i.e. .4635).
Then with relation to those numbers where at most 1 or more prime factors occur 2 times (= 587), the proportion with respect to the total accumulated prime factors is very close to 1/4 (i.e. .2576).
Next, in relation to those numbers where at most 1 or more prime factors occur 3 times (= 290), the proportion is very close to 1/8 (i.e. .1273).
Next, in relation to those numbers where at most 1 or more prime factors occur 3 times (= 290), the proportion is very close to 1/8 (i.e. .1273).
Then finally (in our example) in relation to those numbers where at most 1 or more prime factors occur 4 times (= 142), the proportion is very close to 1/16 (i.e. .0623).
Now the only estimate in or sample that is not very close to the postulated proportion is that for the accumulated factors of those numbers where no prime occurs more than once.
Once again our estimated value here is .4635 whereas the postulated value is .5.
However there is reason to believe that this proportion steadily rises towards .5 as we move up the number scale.
For example when I earlier measured this proportion with respect to the 500 numbers (200,001 - 200,500), the estimate for this proportion (i.e. ratio of accumulated number of factors of those numbers where no prime factor occurs more than 1, with respect to accumulation of prime factors for all numbers in this range) the estimated value was less than .45.
Then from various other estimates, I found that this proportion tended to steadily rise as we move higher up the number scale.
For example, I measured this proportion for the 1000 16 digit numbers from 800000000000001 - 800000000001000 and obtained the value .4688.
Now this might appear but a small increase (with respect to the 14 digit number estimate of .4635), but one must remember that the average number of prime factors per number changes at a very slow rate. Therefore to obtain numbers with an appreciably higher number of prime factors we would need to move significantly higher up the number scale and unfortunately way beyond the range for which prime factors are readily obtainable!
Then
bearing in mind the nature of the Riemann zeta function (for positive
integers > 1) and the need to ensure that the overall sum of
percentages (for all prime factor combinations) = 100%, this would imply
that these successive proportions would closely follow the series 1 + 1/2 + 1/4 + 1/8 + 1/16 .....
Now the only estimate in or sample that is not very close to the postulated proportion is that for the accumulated factors of those numbers where no prime occurs more than once.
Once again our estimated value here is .4635 whereas the postulated value is .5.
However there is reason to believe that this proportion steadily rises towards .5 as we move up the number scale.
For example when I earlier measured this proportion with respect to the 500 numbers (200,001 - 200,500), the estimate for this proportion (i.e. ratio of accumulated number of factors of those numbers where no prime factor occurs more than 1, with respect to accumulation of prime factors for all numbers in this range) the estimated value was less than .45.
Then from various other estimates, I found that this proportion tended to steadily rise as we move higher up the number scale.
For example, I measured this proportion for the 1000 16 digit numbers from 800000000000001 - 800000000001000 and obtained the value .4688.
Now this might appear but a small increase (with respect to the 14 digit number estimate of .4635), but one must remember that the average number of prime factors per number changes at a very slow rate. Therefore to obtain numbers with an appreciably higher number of prime factors we would need to move significantly higher up the number scale and unfortunately way beyond the range for which prime factors are readily obtainable!
However in view of the fact that the total proportion is necessarily equal to 1 and likewise in view of the fact that subsequent proportions (for numbers where 1 or prime factors occur more than once) conform so closely to the series 1/4 + 1/8/ + 1/16 + ....., then it is eminently reasonable to assume that the proportion corresponding to the first term ultimately → 1/2 for numbers sufficiently high up the number scale.
So, in general, the proportion of the total accumulated frequency of all factors for those numbers where at most 1 or more prime factors occur n times → 1/2n (with the accuracy of this steadily improving as the number increases).
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