Tuesday, April 26, 2016

(Distinct) Prime Factors

Just as a connection can be shown as between the harmonic series and the average no. of divisors of n, likewise a connection can be shown as between the prime harmonic series (i.e. the sum of the reciprocals of the primes) and the average no. of (distinct) prime factors of n (especially when n is very large).

The sum of the series of prime reciprocals

= 1/2 + 1/3 + 1/5 + 1/7 +  1/11 + ........

  ~ log log n + B1 , where B1 is the Merten's constant = .2614972...

Now once again we can approach the issue of  the average no. of (distinct) prime factors of n from this perspective.

Now clearly, 2 will be a factor in relation to 1/2, 3 a factor in relation to 1/3, 5 a factor in relation to 1/5, 7 a factor in relation to 1/7, 11 a factor in relation to 1/11 of all numbers and in like manner also for all other primes.

Now of course these prime factors will then often be repeated more than once in a number. For example 4 = 2 * 2. However this not effect the number of distinct factors where each prime factor is counted only once!

Therefore this therefore implies that the average number of (distinct) prime factors is simply given as

1/2 + 1/3 + 1/5 + 1/7 + 1/11 +...... 

However, this method of calculation will lead to factor quotients with remainders occurring in most cases, whereas the no. of factors can only occur in whole (integer) amounts.

Thus for example when n = 20, our suggested method (using prime reciprocals) would suggest that 10 include the factor 2. In this instance, no remainder exists (as 2 is a factor of 20).

However with respect to 3, it would suggest that 6.67 would contain 3 factors (whereas in fact 6 would be the correct answer). In the case of 5 we would get the correct integer result of 4 (as 5 is factor of 20). Then in the case of 7 we would get 2.86  (instead of 2) with 11 we would get 1.82 (instead of 1), with 13, 1.54 (instead of 1) with 17, 1.18 (instead of 1) and finally with 19, 1.05 (instead of 1).

So to this extent the prime reciprocal formula overstates slightly the average no. of (distinct) prime factors.

Once again we would expect with larger n that the remainders would tend to vary in a fairly balanced manner as between 0 and 1 with the average remainder therefore .5 (approx).

And as the total no. of primes to n is simply given as n/log n (approx) this would mean that the  discrepancy with respect to the total no. of factors would be given as n/2(log n) approx.

Therefore the average discrepancy per number would be given as 1/2(log n).

However as n becomes very large, this average discrepancy will keep falling ever closer to zero.

Therefore for very large n, the average no. of distinct prime factors is accurately given by the reciprocal prime formula,

i.e.  log log n + B1.

And as B1 is a small constant, this would imply that the average no. of (distinct) prime factors at n (when n is sufficiently large), is given as log log n. 

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