Factors in the Dirichlet Function with Label 5,4 (2)

Yesterday, we looked at the distribution with respect to the negative sign factors of the Dirichlet function (5,4) i.e.

L(χ,s)

= 1 − 2-s − 3-s + 4-s + 6-s − 7-s − 8-s + 9-s + 11-s − 12-s − 13-s + 14-s + 16-s −

We will now look at the corresponding positive sign factors

So again if on division by 5 a natural number has a remainder of either 1 or 4, then it is included.

n	Rem 1	Rem 4	Total	Zeta Zeros	% acc.
50	21	29	50	40	80.0
100	55	72	127	108	85.0
150	95	105	200	186	93.0
200	139	157	296	271	91.6
250	184	208	392	361	92.1
300	231	262	493	454	92.1
350	276	313	589	552	93.7
400	324	371	695	652	93.8
450	374	431	805	755	93.8
500	425	490	915	860	94.0

Again column 1 includes accumulated factors (of a specified type) up to a given number. Column 2 then relates to all the factors or divisors of each number (excluding the factor 1) which leave a remainder of 1 on division by 5.

So again to illustrate 12 has the factors 2, 3, 4, 6 and 12. Of these only 6 leaves a remainder of 1 (on division by 5). Therefore 12 contributes 1 to the accumulation of factors in Column 2.

Then 4 leaves a remainder of 4 (on division by 5). So 12 equally contributes 1 to the accumulation of factors in Column 3.

Column 4 then gives the total accumulation of factors that leave either a remainder of 1 or 4 (on division by 5).

Column 5 then relates these factor totals (in column 4) to the zeros of the Riemann Zeta function.

Now once again to estimate the total of all factors up to n we measure the frequency of Riemann zeros to t (on an imaginary scale) where n = t/2π.

However we are concerned specifically here − not with all factors − but rather those that leave a remainder of either 1 or 4 (on division by 5).

As these account for 2/5 (40%) of all factors to estimate the total number of such factors to n, we estimate the total number of all factors (without restriction) to 2n/5.

So for example to estimate the total number of factors that leave a remainder of 1 or 4 (on division by 5) to 500, we estimate the total number of all factors (without restriction) to 200.  

Now there are 898 such factors (without restriction) to 200 and this equates very well with the corresponding total of factors (that leave a remainder of 1 or 4 on division by 5) to 500 = 915. In fact this represents 98.1% accuracy.

However because there is a corresponding relationship as between the sum of all factors to n and the corresponding frequency of zeta zeros to t (where n = t/2π), we can use this latter figure to likewise estimate the sum of factors.

Now strictly, we use the formula t/2π(log t/2π − 1) to estimate the frequency of zeta zeros. However this is amazingly accurate (even in absolute terms) generally being just 1 less than the correct answer.   

However in this case, as we are not estimating the total of all factors to n, but rather the total to 2n/5, 2n/5 = t/5π.

So for example our last entry of zeta zeros i.e. 860 (in  column 5) which estimates total number of all factors to 200, which in turn estimates the frequency of those factors, leaving a remainder of 1 or 4 on division by 5, to 500, is the calculation of zeta zeros to 200 * 2π,

And as we can see from column 6, this already gives 94% accuracy in terms of the true number, which will steadily increase as the value of n (and t) gets progressively larger.

We also can point to a similar pattern in terms of the factor totals associated with the respective remainders.

In this case the smaller total (associated with factors giving a remainder of 1 on division by 5) measures the frequency of the combined factors giving a remainder of either 1 or 4 up to n/2, whereas the larger total (associated with factors giving a remainder of 4 on division by 5) measures the frequency of the combined factors from n/2 to n.

So for example the total of all factors that leave a remainder of 1 (on division by 5) up to 500, as we can see from the table = 425.

This in turn should give a good estimate of the total number of all factors (leaving a remainder of either 1 or 4) to 250.

And as we can see from the table this total to 250 = 392. So our estimate though far from perfect is about 92% accurate.

However the important point is that this relative accuracy will steadily improve towards 100% as the value on n progressively increases.

Alternatively the total of all factors that leave a remainder of 4 (on division by 5) up to 500 = 490.

And the total of factors (leaving a remainder of either 1 or 4) from 250 to 500 = 915 − 392 = 523. Again this compares reasonably well with 490 (93.7% accurate) and this relative accuracy will increase towards 100% as the value of n progressively increases.

Factors in the Dirichlet Function with Label 5,4 (1)

We are looking now at the Dirichlet series,

L(χ,s)

= 1 − 2-s − 3-s + 4-s + 6-s − 7-s − 8-s + 9-s + 11-s − 12-s − 13-s + 14-s + 16-s −  where χ is the Dirichlet character with label 5, 4.

So here when we divide each natural number by 5, when the remainder is either 2 or 3, we give it a negative sign; when the remainder is 1 or 4 we give it a positive sign and when the reminder is 0, we omit the number.

To calculate the zeros of this Dirichlet (5,4) function to t (on the imaginary scale) we calculate the corresponding zeros of the Riemann zeta function to 5t before then dividing the total number of zeros k by 5.

n	Rem. 2	Rem. 3	Total	Freq. of zeros	% acc.
50	45	34	79	72	91.1
100	102	83	185	172	93.0
150	161	135	296	282	95.3
200	228	191	419	399	95.2
250	296	246	542	521	96.1
300	367	309	676	648	95.9
350	439	370	809	777	96.0
400	513	435	948	910	96.0
450	588	500	1088	1044	96.0
500	663	565	1228	1182	96.3

In the above table, the 1^st column shows the accumulated number of factors (of a given type) up to n.

Then the 2^nd column shows the accumulated no. of factors that leave a remainder of 2 when divided by 5, while the 3^rd column shows the accumulated no. of factors that leave a remainder of 3 when divided by 5.

So for example 12 is made up of 5 divisors (i.e. factors) i.e. 2, 3, 4, 6 and 12. Because 1 is a factor of all numbers, we exclude it in this case.

So two of these i.e. 2 and 12 leave a remainder of 2 when divided by 5 and one i.e. 3 leaves a remainder of 3 when divided by 5. So 12 therefore contributes 2 factors to the Rem. 2 total and 1 factor to the Rem. 3 total.

The 4^th column then represents the total accumulation of factors (where the factors of a number leave a remainder of either 2 or 3 when divided by 5).

And remember that in the Dirichlet series (5,4) all natural numbers that leave a remainder of either 2 or 3 (when divided by 5) are given a negative sign.

The 5^th column then shows the frequency of zeros for this particular Dirichlet function with label (5, 4) up to a given no. which is calculated as follows.

Now if we look at numbers that leave a remainder of 2 or 3, when divided by 5, these constitute 2/5 (or 40%) of all natural numbers

Therefore to get an estimate of the accumulated factors of numbers up to n (that leave a remainder of 2 or 3 when divided by 5, we estimate the corresponding total accumulation of all factors up to 2n/5.

So for example   to estimate the accumulated factors of all numbers up to 100 (that leave a remainder of 2 or 3 on division by 5) we estimate the corresponding total of all factors up to 40.

There are 118 such factors up to (and including) 40.

The total of factors (that leave a remainder of 2 or 3 on division by 5) up to 100 is from Col. 3 of the table above 185.

Now in all these cases, where numbers carry positive or negative signs, the numbers associated with each sign will follow one of two distributions where the distribution will either be the standard Riemann distribution or alternatively the Dirichlet distribution associated with the specific function (5,4).

There are 118 such factors up to (and including) 40.

However the total of factors (that leave a remainder of 2 or 3 on division by 5) up to 100 is from Col. 3 of the table above 185.

So the discrepancy here is very marked.

Therefore we can safely assume that the numbers with negative signs i.e. that leave a remainder of 2 or 3 when divided by 5, are associated with the corresponding Dirichlet distribution (5, 4).

And as the conductor here is 5, to estimate the accumulated no. of factors (that leave a remainder of 2 or 3 on division by 5) to 100 we estimate the accumulated total of all factors to 200 i.e. 40 * 5) and then divide this total by 5.

So the total number of factors to 200 = 898, and 898/5 = 180 (to nearest integer)

And this compares very favourably with the actual no. of such factors to 100 = 185.

However we can use the frequency of Riemann zeta zeros directly to make a surprisingly accurate estimate of such factors.

In general terms to estimate the no. of all factors to n, we estimate the corresponding frequency of all zeros to t (where n = t/2π).

However in this case to estimate the frequency of Dirichlet zeros (5,4) to t, we estimate the frequency of Riemann zeros to 5t and then divide total by 5.

So if k = frequency of all zeros to 5t, then k/5 measures frequency of zeros to t

And once again to measure the frequency of factors that leave a remainder of 2 or 3 on division by 5 up to n, we measure the frequency of all factors to 2n/5.

And as n = t/2π, 2n/5 = t/5π.

Therefore to estimate for example accumulated total of factors (that leave a remainder of 2 or on division by 5) up to 100 we measure the Dirichlet zeros to 40 * 2π, which in turn means estimating the frequency of Riemann zeros (k) to 200 * 2π and then dividing by 5.

So the number of  Riemann zeros (k) to 200 * 2π  = 860 (to nearest integer).

And k/5 = no. of Dirichlet zeros to 40 * 2π  = 172.

And this is already a very good estimate (93% accurate) of the actual total of such factors i.e. 185.

Therefore because the frequency of zeros of each Dirichlet function is directly related to the corresponding frequency of zeros for the Riemann zeta function, we can always relate the accumulated total of factors for any Dirichlet function to the corresponding frequency of Riemann zeros, thereby using these latter well-known zeros as a goof estimate for the corresponding total of factors.

As we can see from the tables when we estimate the total no. of factors (leaving a remainder of 2 or on division by 5) up to 500, the corresponding estimate of Dirichlet (5,4) zeros to 2t/5 which in turn is related to the frequency of Riemann zeros (k) to 2t as k/5 already gives an estimate that is more than 96% accurate.

There is another fascinating relationship that is worth relating

In the table above, I have given separate totals factors that leave a remainder of 2 from those that leave a remainder of 3 (when divided by 5).

These two in turn follow a fascinating pattern.

As there are just two options here by which the natural numbers can be given a negative sign in the Dirichlet function (5,4) i.e. leaving a remainder of 2 or 3 on division by 5, this means that if we measure the total frequency of both types to n, then one the lower total will represent the frequency of factors to n/2 and the other the corresponding frequency from n/2 to n.

So in the above case the accumulated factors that leave a remainder of 3 (on division by 5) is lower than the corresponding accumulation of factors that leave a reminder of 3

Therefore the accumulated amount for the former with a remainder of 3, gives a good estimate for the total number of factors (with a remainder of either 2 or 3) up to n/2, with the accumulated amount with a remainder of 2 giving a good estimate of the total from /2 to n.

For example the total no. of factors to 500 (with a remainder of 2 or 3 on division by 5) = 1228.

Now the accumulated no. of factors (with a remainder of 3) to 500 = 565. And this corresponds well to the total no. of factors (with a remainder of either 2 or 3 to 250) = 542.

And the accumulated no. of factors (with a remainder of 2) to 500 = 663.

The total no. of factors from 250 to 500 = 1228 − 542 = 686. Again, this compares well with the reminder 2 figure, i.e. 663 (over 96% accurate).

Dirichlet Divisor Problem

Riemann in his original article on the zeta function in 1859, was enabled through appreciation of the important nature of the zeta zeros to provide a formula – given the incorporation of the influence of a sufficient number of zeros – could exactly predict the magnitude of primes up to a given number.

However the other side of the coin as it were is the corresponding accumulated total of all the divisors (or factors) of the natural numbers up to a given magnitude.

Just as we can give the simple formula n/ln n that will approximate the primes up to a given number n, we can give the corresponding simple formula n(log n) to calculate the accumulated number of factors.

So – counting all divisors of a number (including 1 and the number in question) – the total number of factors to 100 = 482. And the corresponding answer given by our simple formula is 461 (when rounded to the nearest integer) which is already surprisingly accurate (over 95%).

Now Dirichlet proposed a slight modification on this formula as n(log n + 2γ – 1) + Δ(n), where γ (the Euler-Mascheroni constant) = .5772156649…  and Δ(n) represents a remaining variable which can have a positive or negative value

Using this modified formula, the total number of factors to 100 = 476 (to the nearest integer). Now the reason why we still get a slight underestimate here of 6 is mainly due to the fact that the cut-off point we are using for n, i.e. 100 is in itself a highly composite number with 9 divisors. For example if we had taken the estimate with respect to 101 (which is prime) the total number of factors would increase by 2 to 478 while the estimated formula would give 476 (when rounded) that reduces the underestimate to 2.

However, because of the random nature in which the factor composition of numbers varies, there will still always be a certain absolute discrepancy (positive or negative) using the Dirichlet formula in the estimate of the total number of factors.

Now various attempts have been made to put an upper bound on the value of this last variable term. Dirichlet himself suggested n1/2. Later estimates have reduced the value down close to .3, though it is widely believed that .25 would effectively be the lowest power that can be assigned. However this would only strictly apply as n approaches ∞. This would entail for example that when n = 100, the difference between our estimate and the actual number of divisors would be less than 10 for Dirichlet’s power of ½. In fact it is 6! However if the power is taken as .25 this would means the limit on our bound would be just over 3, whereas in fact the actual discrepancy is 6. So it would not work in this case where n is a relatively small finite value.

In one way however it is surprising that an exact formula cannot be provided – at least in principle to measure the sum of divisors up to a given number – when a formula exists i.e. the Van Mangoldt formula, to do this for the primes. 

There are however two distinct distributions with respect to the primes.

The commonly understood distribution relates to the frequency of individual primes up to a given natural number magnitude. So for example we have 25 such primes up to 100 and the primes in this regard are randomly distributed.

However the alternative distribution relates to the collective distribution of prime factors within a given natural number. In this sense 100 for example has just 2 (distinct) prime factors i.e. 2 and 5. 

The question then arises as to the nature of the distribution of all the divisors of a natural number. Again the presumption would be that this distribution is likewise random in nature but its precise relationship to the corresponding distribution of prime factors may be much more difficult to define.

Going back however to Dirichlet’s formula for the (accumulated) sum of divisors, I have generally used the slightly modified version where one divisor for each number is treated as redundant.

This in fact is what happens for example in the case of perfect numbers. 

So 6 for example has 4 divisors i.e. 1, 2, 3 and 6. However the last of these is excluded giving a measurement therefore of what are sometimes referred to as the proper factors of the number. (There is however considerable confusion evident in the definition of proper factors with some definitions excluding both 1 and the number in question). And 1 + 2 + 3 = 6. Therefore 6 is a perfect number.

So the formula for the accumulated sum of proper factors can be given as 

n(log n + 2γ – 2) + Δ(n).

Therefore for estimates we can leave out the variable term and use 

n(log n + 2γ – 2) = n(log n – .8456...)

This is then quite close to the simpler approximation n(ln n – 1).

n	Actual No. of Factors (not including 1 as factor)	Estimated number (using formula n(log n + 2γ – 2) with frequency of zeros to t , where n = t/2π (in brackets)
– 100	382	376       (361)
– 200	898	891       (860)
– 300	1467	1457      (1411)
– 400	2064	2058      (1997)
– 500	2684	2684      (2607)
– 600	3338	3331      (3238)
– 700	3994	3994      (3886)
– 800	4676	4671      (4548)
– 900	5370	5361      (5222)
– 1000	6061	6062      (5908)

Therefore using the former more precise measurement we have 376 accumulated (proper) factors up to 100, whereas the latter estimate gives 361. The actual number of factors is 382. Then in relative terms as n becomes progressively larger, the latter simpler estimate will approach closer and closer to 100% accuracy.

The advantage of all of this is that the corresponding formula for frequency of zeta zeros up to t is t/2π(ln t/2π – 1) where n = t/2π. And as this formula is remarkably accurate in terms of predicting the actual frequency of zeta zeros to t, we can then use it in turn to calculate the (accumulated) frequency of proper factors to n.

For example as shown in the above table. the actual number of divisors up to n =1000, is 6061. The corresponding estimate of the frequency of the Riemann zeta zeros to t, where n = t/2π, is 5908.

And this is already 97.46% accurate.

And as we shall see in future contribution all factor distributions for Dirichlet L-functions can in turn be linked to the distribution of the Riemann zeta zeros.