0の比率

　素数の非ランダムな振る舞いについての解説　　いつか、日本語のも書き込みます
I computed two sequences S_01 and S_02 of which law is the following I also computed the number of 0
S_01 :
If Prime(m) = 1 Mod 3 then a(m) = Prime(m) + Prime(m-1) Mod 4
If Prime(m) = 2 Mod 3 then a(m) = Prime(m) + Prime(m+1) Mod 4


0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 2, 2, 2, 2

Number of 0 , 0<m<10^3, 10^3<m<10^3+10^3, 10^4<m<10^4+10^3, 10^5<m<10^5+10^3, 10^6<m<10^6+10^3, 10^7<m<10^7+10^3, 10^8<m<10^8+10^3

721, 665, 661, 601, 621, 571, 565

S_02 :
If Prime(m) = 1 Mod 3 then b(m) = Prime(m) + Prime(m+1) Mod 4
If Prime(m) = 2 Mod 3 then b(m) = Prime(m) + Prime(m-1) Mod 4

0, 2, 2, 2, 2, 2, 2, 0, 0, 2, 2, 2, 2, 0, 0, 0, 2, 2, 0, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 2, 2, 2, 2, 0, 0, 0, 2, 2, 0, 0, 0, 2, 2, 0, 2, 2, 0

513, 497, 531, 496, 515, 513, 524


S_03 :
c(m) = Prime(m) + Prime(m+1) Mod 4 : A103271

617, 576, 596, 548, 568, 542, 545

If Prime number behave randomly then the inequality of 0, 2 on S_01 never appears It should be like S_02 But in fact ,it appears I compared them and observed as follows

The inequality of third one is the standard, because it depends on the probability of (Prime(m), Prime(m+1)) Mod 4 that is (1, 1), (1, 3), (3, 3), (3, 1)
Read the comment on A103271

The inequality of the first one has two parts, the standard and the other one, so it becomes much larger than the standard , on contrary , in the case of the second one , both inequalities offset each other . It behaves randomly as if Prime were random. It is interesting and strange.

These three ratios have some fluctuation. It is also interesting


I have understood my observation is incorrect I explain the phenomenon of S_01
The S_01 is possible to be described as follows
a(m) = f(c(m))
In general, it is possible to consider inequality of the following sequence
b(m) = h(c(m)) where h is any mapping from c(m) which is A103271 to sequence
The ratio of inequality of b(m) depends on mapping h, sometimes it is more than the standard' one, sometimes, it is less than it

Problem_0 : Why does the mapping f make c(m) to have larger inequality ?

It is easy to explain what the f is
Here is the matrix of probabilities of (Prime(m), Prime(m+1)) Mod 12 m<800
M_1
1 5 7 11
1 12,83,52,41
5 35,15,98,56
7 58,48,16,83
11 83,57,39,22
Here is the matrix of probabilities of (Prime(m), Prime(m-1)) Mod 12 m<800
M_2
1 5 7 11
1 12,35,58,83
5 83,15,48,57
7 52,98,16,39
11 40,56,83,22

The probability of 0 of a(m) is mean of probabilities of four cases that (5, 7) + (5, 11), (11, 1) + (11, 5) on M_1 and (1, 7) + (1, 11), (7, 1) + (7, 5) on M_2
They are (98+56)/204. (83+57)/201, (58+83)/188, (52+98)/205 = 75%. 70%, 73%. 75%
Hence mean is 73%

Mapping h of b(m) is the reverse of mapping f, so probability of 0 of b(m) is mean of (52+41)/188, (58+48)/205, (48+57)/203, (40+5 )/20

Hence Problem_0 is the same as the following

Problem_1 : Why does the law of a(m) choose the largest and the second probabilities on M_1 and M_2 ?
Its reverse one
Problem_2 : Why does the law of b(m) choose the second and the third large probabilities on M__1 and M_2 ?
I don't understand the reason

I am searching interesting mapping which has, for instance, three times larger inequality than the standard
It is difficult to find mapping like that, because the number of mapping from sequence to sequence is Omega_2 I wonder if AI might be able to find them