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.