Cumulative sum of Goldbach Partitions

[Paul Mackenzie]

I noted the following concerning the cumulative sum of Goldbach partitions


C[2N] = sum[ G(2N) ;from 6 to 2N] is greater than pi[2N]*(pi[2N] -1)/2

where
2N is an even number 2N=6,,,,,
C[2N] is the cumulative sum of the goldbach partitions of the even numbers 6,...2N
G(2N) is the number of Goldbach partitions for the even number 2N and
pi[2N] is the number of primes less than 2N.

I checked the above on my computer for values of 2N upto 100,000

What would a proof of this be? I got as far as formulating every possible combinations
of primes (p,q) where p and q are both less than 2N, This lead me to a rough guess of
C[2N] = pi^2[2N] /2 but why is it greater?.

Paul

[Paul Mackenzie]

I have been able to prove a weak version of the above.

Let the cumulative sum C[2N] = Sum([G[2N]; 2N= 6 to 2N)
where G[2N] is the number of goldbach partitions for the even number 2N.

Consider every combination of odd primes less than N,
note that the sum of these combination pairs is less than 2N
and note that this set of combination pairs forms a subset
of all combination pairs of odd primes that form the goldbach
partitions constituting the cumulative sum thus
we can state

pi[N]*pi[N] < C[2N]

where pi[N] is the number of primes less than N

though I still am unable to prove

pi[2N]*(pi[2N]-1)/2 < C[2N]

the latter is closer to the function C[2N]

I found another function which appears to be
even more closer to C[2N] for large 2N,
though it does not hold for some small 2N


pi[2N]*pi[N] < C[2N]

Any suggestions?

[RamaWolf]

To get familiar with this topic, I started to establish the function
Goldbach2 to calculate the number of partitons of 2n; the result
r2(2n) is the number of prime pairs (p,q) with p not greater q;
up to 40 the list looks like this:

r(2n)|2n|Sum(r(2n))
-------------------
1|4|1
1|6|2
1|8|3
2|10|5
1|12|6
2|14|8
2|16|10
2|18|12
2|20|14

Perhaps, you did not use r2(n), but R(2n), which is simply 2*r(2n) if
n is not prime, and 2*r(2n) - 1 else.

Regards

[Paul Mackenzie]

To better understand the terminology I have included a word document with a graph with values 2N= 6 to 72,000. It appears the term R[2N] you refered to and G[2N] are interchangeable.

Regards

Paul

[RamaWolf]

Below is a table with the following entries:


R2(2*n) is definied as in my previous post as the
number of binary Goldbach partitions
(eg R2(10) = 3; the partiitions are {(3+7),(5+5),(7+3)}

C(2*n) := Sum(R2(2*n)), the sum is taken from 4 to 2*n
Pi(k) are the number of primes not exceeding k
H1 := Pi(n) * Pi(n)
H2 := Pi(n) * Pi(2*n)
H3 := Pi(2*n)*(Pi(2*n) - 1) / 2

2*n C(2*n) Pi(n) Pi(2*n) H1 H2 H3
-------------------------------------------------------------
4 1 1 2 1 2 1
6 2 2 3 4 6 3
8 4 2 4 4 8 6
10 7 3 4 9 12 6
20 24 4 8 16 32 28
30 50 6 10 36 60 45
40 78 8 12 64 96 66
50 117 9 15 81 135 105
60 158 10 17 100 170 136
70 199 11 19 121 209 171
80 252 12 22 144 264 231
90 312 14 24 196 336 276
100 361 15 25 225 375 300
200 1135 25 46 625 1150 1035
300 2233 35 62 1225 2170 1891
400 3540 46 78 2116 3588 3003
500 5139 53 95 2809 5035 4465
600 6946 62 109 3844 6758 5886
700 9012 70 125 4900 8750 7750
800 11274 78 139 6084 10842 9591
900 13759 87 154 7569 13398 11781
1000 16349 95 168 9025 15960 14028
2000 52934 168 303 28224 50904 45753
3000 105975 239 430 57121 102770 92235
4000 173401 303 550 91809 166650 150975
5000 254879 367 669 134689 245523 223446
6000 349452 430 783 184900 336690 306153
7000 456847 489 900 239121 440100 404550
8000 576324 550 1007 302500 553850 506521
9000 707138 610 1117 372100 681370 623286
10000 850833 669 1229 447561 822201 754606

H1, H2 and H3 serve as lower estimates for C2(2*n)
and H2 seems to serve best

Regards

[Paul Mackenzie]

Hi

I think I have found an upper bound for C[2N] [though it does not hold for some small 2N]. Again an intuitive guess.

C[2N] < 4*Pi[2N]*Pi[N] - 2*Pi[N]*Pi[N] - P[2N]*Pi[2N]

so we have

Pi[N]*Pi[2N] < C[2N] < 4*Pi[2N]*Pi[N] - 2*Pi[N]*Pi[N] - P[2N]*Pi[2N]

I checked this for odd primes upto 2N= 100,000

Regards