**Preamble**

The aim here is to discover the frequencies of numbers for the purpose of guiding the selection of a numerical base that best represents the most frequent numbers.

__Proposition 1__:

Suppose that all natural whole numbers are equally probable.

__Caveat__:

It is not actually believed that all whole numbers have an equal probability of being selected.

__Corollary__:

Then, if a number from the set of natural whole numbers be selected at random, each whole number has an equal probability of being selected.

**Expected Frequencies of Primes**

__Query__:

What is the number of times an \(i\)th prime number \(p_i\) is expected to occur in such a randomly selected whole number?

__Response__:

Consider the prime number 2. It occurs at least once in half of all whole numbers. In some of these even numbers, the prime number two as a factor occurs more than once. For example, in the composite number four that is formed by a multiple of prime numbers as two to the power of two, the prime number two as a factor appears twice. In the composite number eight, the prime number two appears three times as a factor.

__Caveat__:

By the number of times a prime factor appears as a factor in a composite number is meant not the number of times that prime number appears added with itself in a sum to make up the composite number, because this number of times is the composite number divided by the prime number, but rather the exponent of the prime number in the prime factorisation of the composite number.

The prime number two appears as a factor at least once in half of all whole numbers, so, given the first Proposition, the frequency of the prime number two appearing as a factor at least once is 1/2.

The prime number two as a factor appears at least twice in a quarter of all numbers. Given Proposition 1, the frequency of the prime number two appearing as a factor at least twice in whole numbers is 1/4.

In general, the frequency of the prime number \(p\) appearing as a factor at least \(n\) times is 1/p^n.

The expected frequency \(F(p)\) of a prime number \(p\) appearing as a factor, taking into account how often that prime number appears

*within*a number, in a randomly selected whole number, given Proposition 1, is the sum $$ S_{\infty} = \sum_{n=1}^{n=\infty}\frac{1}{p^{n}}$$ of the frequencies \(\frac{1}{p^{n}}\) of that prime number \(p\) appearing as a factor at least \(n\) times. This summation is a geometric series with initial term 1/p and common ratio $$\frac{\frac{1}{p^{n+1}} }{\frac{1}{p^{n}}} = \frac{1}{p}$$ between adjacent terms. The sum to infinity \(S_{\infty}\) is therefore $$ \frac{\frac{1}{p}}{1-\frac{1}{p} } = \frac{1}{p-1} = F(p)\text{.}$$

__Caveat__:

Since the expected frequencies were derived assuming Proposition 1, but Proposition 1 is not believed, the expected frequencies so formulated are not believed to be a realistic model.

The expected frequencies \(F(p)\) for the smallest prime numbers can be compared in a table.

__Table 1. Expected Frequencies for Primes__

Prime, \(p\) | Expected Frequency, F(p) = \(\frac{1}{p-1}\) |

2 | 1 |

3 | 1/2 |

5 | 1/4 |

7 | 1/6 |

⑪ | 1/⑩ |

It can be seen from this derivation that the prime number two is twice as frequent as the prime number three, and that the prime number three is twice as frequent as the prime number five. For example, considering the numbers up to eleven, the prime factor two occurs once in the number two, twice in the number four, once in the number six, three times in the number eight, and once in the number ten, or a total of eight times. In the same span of numbers, the prime factor three occurs once in the number three, once in the number six, and twice in the number nine, or a total of four times, which is half as often as for the prime number two. The prime factor five occurs once in the number five and once in the number ten, or a total of twice, which is half the frequency of the prime number three.

For ranking the prime numbers for importance by their frequencies of occurrence, the principles of the preceding argument have been described previously elsewhere.

For the purpose of ascertaining the base most likely to produce rounding in numbers formed from randomly selected prime numbers, yesterday I proceeded with the following reasoning.

**Probabilities of Primes**

__Proposition 2__:

Suppose that the probability \(P(p)\) of a prime number being selected is proportional to its expected frequency \(F(p)\).

__Caveat__:

Since the Expected Frequency as formulated above is not believed because it was derived from Proposition 1 which is not believed, the probabilities of primes are not believed to be proportional to the expected frequencies so formulated, but may nevertheless be proportional to the actual expected frequencies of another formulation.

__Discussion__:

The expected frequency of the prime number two was one. However, the probability of the prime number two appearing in a randomly selected number is less than one, because half of all numbers do not contain the prime number two as a factor. The probability for the prime number two need not be a half, however. All numbers are considered to be broken into their constituent prime numbers, which are then placed into a bag and shuffled. A prime number is then withdrawn at random from this mix. The probability sought is that of the selected prime number being a certain prime number. For example, what is the probability of the withdrawn prime number being the prime number two? Let this probability be called P(2). Then, given the second Proposition, the probabilities of all other prime numbers can be expressed in terms of P(2), as the multiple P(2)F(p) of P(2) and the expected frequency F(p).

**Probabilities of Composites**

__Proposition 3__:

The probability of selecting a composite number is the product of the probabilities that are raised to the exponents of the primes for the prime numbers in the prime factorisation of the composite number.

__Discussion__:

By one of the Fundamental Principles of Counting, the probability of selecting a combination of prime numbers, such as a prime number \(p_a\)

*and*a prime number \(p_b\), where \(a\) and \(b\) may be the same or different,

*and forming a composite number out of them*, shall be the multiple P(\(p_{a}\))P(\(p_{b}\)) of the probabilities P(\(p_{a}\)) and P(\(p_{b}\)) of selecting the prime numbers individually.

Since P(3) = 2P(5), multiplying both sides of the equation by [P(2)]^2 gives

[P(2)]^2 P(3) = 2[P(2)]^2 P(5),

thus:

P(①⓪) = 2P(2) P(⑩)

and the ratio of the probabilities for the numbers twelve and ten is:

P(①⓪)/P(⑩) = 2P(2).

If the numbers twelve and ten were equally probable, then P(①⓪)/P(⑩) = 1 and 2P(2) = 1, so P(2) = 1/2.

Suppose the probability P(2) to be a half. Then, probabilities of numbers being selected from a pool of prime numbers could be tabulated as follows.

__Table 2. Probabilities for Whole Numbers__

Whole Numbers, n | Probability, P(n) |

2 | 1/2 |

3, 4 | 1/4 |

5, 6, 8 | 1/8 |

7 | 1/①⓪ |

9, ⑩, ①⓪, ①④ | 1/①④ |

⑪ | 1/①⑧ |

Regardless of the value of P(2), the sum of the probabilities of the prime numbers would be $$P(2)\sum_{i=1}^{i=\infty} \frac{1}{p_i - 1}\text{.}$$ However, if the summation does not converge then this scheme of probabilities cannot be the correct one. The series diverges because the sum of the reciprocals of the primes diverges. This means that as the sizes of the prime numbers being considered in the sum increase, the value of P(2) must decrease towards zero in order for the sum of probabilities to be one, with the implication that the probability of the prime number two being selected decreases because of the increasingly large number of other prime numbers expected to be found in the large composite numbers.

Hence, and because the larger numbers approaching infinity are less likely to be encountered than the smaller numbers, it is necessary to adjust the probabilities of numbers being selected at random dependent on their sizes, contrary to Proposition 1.

It is expected that the effect of such an essential adjustment would be to make the smaller prime numbers, such as two and three, more probable and the larger prime numbers, such as five and seven, less probable.

It may be that the adjustment can be done in such a way that maintains the equally of the probabilities of the numbers three and four as seen in Table 2, if P(3) = [P(2)]^2. On the other hand, the prime number five can be expected to move down the rows of the table such that it would become less probable than six or eight. Consequently, the composite number ten would move down to being less probable than the numbers nine, twelve, and ①④. Similarly, the prime number seven would be expected to become less probable, such that it might end up below the numbers nine, twelve, ①④, or being even less probable than ten.

Consider a ranking of the numbers that has been proposed previously elsewhere which places the numbers in an order that fulfils these expectations.

**Unimportance Rank of Numbers**

Let the unimportance, say U(n), of a number \(n\) be calculated by subtracting one from each prime factor in the prime factorisation of a number and adding up these subtractions. This produces the following table ranking the first few numbers.

__Table 3. Unimportances of Numbers__

Whole Numbers, \(n\) | Unimportance Score |

2 | 1 |

3, 4 | 2 |

6, 8 | 3 |

5, 9, ①⓪, ①④ | 4 |

⑩ | 5 |

7, ①③, ①⑧ | 6 |

**Probabilities from Unimportances**

Since the unimportance score of a composite number is calculated by the addition of the unimportance scores of its prime factors, but for probabilities a multiplication or product rather than addition of scores is required, to convert the unimportance to a probability, exponentiate the unimportances by raising the unimportance score as the exponent of the base two. Suppose that the resulting scores are proportional to the probabilities for the numbers. Summing the probability scores for all the prime numbers should be equal to one, so to achieve this the sum can be normalised by multiplying it by a constant \(k\) such that this sum of all probabilities is one. This constant then distributes into the sum, multiplying by each term to produce the probabilities of the primes.

__Table 4. Prime Probabilities from their Unimportances__

Prime, \(p\) | Probability, \(P(p) = \frac{k}{2^{p-1}}\) |

2 | k/2 |

3 | k/4 |

5 | k/①④ |

7 | k/⑤④ |

$$ \sum P(p_{i}) = k\sum_{i=1}^{i=\infty} \frac{1}{2^{p_{i}-1}} = 1$$

This series converges by the ratio test. The sum is found to converge at ≈0.829365019, about ⓪⁏⑨⑪⑤①⑧⑥. The reciprocal ≈1.205741715 of this is \(k\). Then the probability P(2) for the prime number two being selected is k/2 ≈ 0.602870857. From this probability, the probabilities for the other prime numbers and composite numbers can be calculated and tabulated.

__Discussion__:

From these results of probabilities from unimportances, it seems that the probability for the prime number five is far too low. This implies that the unimportance scores and rank that others started advocating produce unrealistic probabilities. It seems that a more realistic scheme would be intermediate between those of Tables 2 and 3. Thus, it is expected that the prime number five would appear less probable than the numbers six and eight, but more probable than the numbers nine, twelve, and ①④.

**Reference**

https://www.tapatalk.com/groups/dozensonline/viewtopic.php?p=40010332#p40010332

https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes

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