I also demonstrated how the simple formula n(n + 1)(log n – 1)/2 can be used to estimate both sums.
Having completed this entry, I then considered another aggregate sum entailing the Riemann zeros.
Here we multiply each prime in ascending sequence by the matching entry from the list of (nontrivial) zeros (likewise arranged in ascending sequence).
So the 1st prime number is multiplied by the 1st Riemann zero, the 2nd prime by the 2nd Riemann zero, and so on.
Up to n in this case relates to the value of the primes in question.
So to illustrate this new aggregate up to 10, we find the sum of (2 * 14.13) + (3 * 21.02) + (5 * 25.01) + (7 * 30.42) with zeros expressed correct to 2 decimal places = 429.31.
Once again, because the nontrivial zeros are estimated with respect to the imaginary scale, t (where n = t/2π), to convert to n, we divide 429.31 by 2π = 68.33 (or 68 rounded to nearest integer).
In the table below I show these accumulated totals. (in col 2).
Up to n

Agg. of primes * by
zeros = (2)

Accumulated sum of
zeros = (3)

(2)/(2/π) = (4)

(4)/(3) as %

10

68

91

107

117.58

20

228

493

359

72.82

30

634

1234

996

80.71

40

1228

2677

1928

72.02

50

2518

4221

3956

93.72

60

3736

6370

5869

92.14

70

5244

8767

8236

93.94

80

8079

12200

12691

104.02

90

10437

15858

16394

103.38

100

11808

20133

18548

92.13

110

15060

24958

23656

94.78

Then in col. 3 I show again the total accumulated sum of zeros (appropriately rescaled to n) as dealt with in yesterday's entry.
In fact, there is an unexpected link as between the two aggregates (in cols 2 and 3 respectively).
If we divide the total in col 2 by 2/π (or alternatively multiply by π /2), we obtain a new set of figures (in col. 4) which bears close comparison with the previous aggregates measure for Riemann zeros (in yesterday's entry).
The link between the two sets of figures (2/π) is not accidental. We have already seen how this shows up as very important measurement with respect to the Zeta 2 zeros (with the average reduced value for both the cos and sin parts of all the roots of 1 converging to 2/π).
In the final column (col 5), I show the relationship as % as between the measurements in cols. 3 and 4 respectively.
The formula for the new estimate in col. 2 (i.e. aggregate of each prime multiplied by corresponding nontrivial zero), is given as {n(n + 1)(log n – 1)}/π.
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