1) from dividing n, by each of natural numbers 1, 2,3 ,4 ,......n, and

2) dividing n by each of the primes 2, 3, 5, 7, ....n respectively.

And the formula which entails a simple deduction with respect to two proven results is given as:

1) 1

**–**γ (where γ is the Euler-Mascheroni constant .57722156649...) = .4228 (correct to 4 decimal places), and as:

2) (1

**–**γ)/log n.

To test out this result taking n = 100, I divided 100 by each of the natural numbers from 1 to 100 respectively, focusing on the remainders of the quotients resulting.

The total of remainders = 34.27 (approx) giving an average of .3427.

Now this seems somewhat less than 1

**–**γ (= .4228). However it has to be remembered that because 100 is a very composite number (containing 8 natural number divisors) this entails that no remainders occur in these 8 cases. Therefore we would expect the result to fall somewhat short of the predicted average.

To demonstrate this, I then calculated the corresponding remainders for n = 101 (this time dividing by all integers from 1 to 101 respectively). Because this number is prime, only two divisors i.e. 1 and 101 will have no remainders in this case.

Here the total for remainders = 40.72 (approx) with the average = .4007.

So this result (although n is still very small) approximates much closer to the predicted average!

Then using the same two values for n (i.e. 100 and 101), I divided these numbers by each of the primes (totalling 25 in all).

Here the average for the 25 primes involved = .3713.

However as 2 and 5 are factors of 100, this would suggest that the average for remainders would likely be less than than the predicted average.

When n = 101, now dividing by the 26 primes to 101 (inclusive) the average for the 26 primes involved = .4263 (which is very close indeed to the predicted average).

However in the case of division by primes, when we average with respect to n (and not just the number of primes up to n) the predicted formula for the average remainder is given as

(1

**–**γ)/log n.

And as n becomes progressively larger, this average will fall towards 0.

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