20180226, 17:33  #1 
Feb 2018
2^{2}·3^{2} Posts 
Prime counting function
PRIME NUMBER COUNTING: Eg: Up to 91squared and 121squared on number line (1+n30): 1,31,61,91 etc.
(A) 0.455897334 × (B)276 = 125.8255664157 (C) up to 91squared (D) 0.430387657 × (E)488 = 210.0291769069 (F) up to 121 squared B = (91squared  1)/30. C = Prime numbers up to 91squared on number line (1 + n 30) 1, 31, 61, 91 etc. which = 125(C) E = (121squared  1)/30 F = Prime numbers up to 121 squared on number line (1 + n30) 1, 31, 61 etc. which = 210(F) A = (6 × 10 × 12 × 16 × 18 × 22 × 28 × 30 × 36 × 40 × 42 × 46 × 52 × 58 × 60 × 66 × 70 × 72 × 78 × 82 × 88)divided by (7 × 11 × 13 × 17 × 19 × 23 × 29 × 31 × 37 × 41 × 43 × 47 × 53 × 59 × 61 × 67 × 71 × 73 × 79 × 83 × 89) OR A = ((prime(1)  1) × (prime(2)  1)....× (prime(n)  1)) divided by ((prime(1) × prime(2)......×.prime(n)) Prime 1 is 7. Prime 2 is 11. Prime 3 is 13 etc Prime (n) in this case = 89(prime next down from 91) D = A × ((96 × 100 × 102 × 106 × 108 × 112) divided by (97 × 101 × 103 × 107 × 109 × 113)) In this case prime (n) = 113(prime next down from 121). There are more than 8 times as many primes up to 121 squared in totality because other number lines out of the possible 7+n30, 11+n30, 13+n30, 17+n30, 19+n30, 23+n30 and 29+n30, none except for above 1 + n30 use 112/113 and two don't use 108/109 which increases the number of primes on the number line. 
20180226, 18:21  #2 
Aug 2006
3×1,987 Posts 
Let me see if I understand. You're trying to estimate the number of primes in an interval which is a below (30n + 1)^2, and your estimate is that if the interval has length L, there are about
\[L\prod_{7\le p\le 30n+1}\frac{p1}{p}\] primes. Is this right? 
20180227, 21:37  #3  
Feb 2018
2^{2}×3^{2} Posts 
Quote:
I wish l could say yes or no to what you wrote, but sorry, l do not know mathematical notation so if you will allow l shall write in plain language and maybe you will be able to tell me if that is what l am saying. I am saying that on the number line (1 + n30) 1, 31, 61, 91, 121 etc there are 210 prime numbers up to 121squared. This is calculated by a simple procedure that uses only lowest prime factors to negate primality. Prime(1) is 7. Prime(2) is 11 etc. The equation used is: (Prime(1)minus 1)/Prime(1) × (Prime(2)minus 1)/Prime(2) × (Prime(3)minus 1)/Prime(3).....×(Prime(n)minus 1)/Prime(n) finally × (Prime(n)minus 1)/30 Prime(n) is the highest prime up to, in my example 121, which is 113. The reason that l said there were more primes on the other number lines 7+n30, 11+n30 etc is that being that l am only using lowest prime factors to negate primality; on number line 23+n30 for example, 113 as a lowest prime factor multiplies with 131 which is greater than 121squared, therefore it allows for more primes on that line up to 121squared. If it were that the results l got were mere coincidence, the results would not continue to be correct, as they are. Please check higher numbers for proof. It would be like a broken clock, only correct once. As l said, l don't know mathematical notation, have never in my life studied or spoken maths with anyone. I just like playing with numbers trying to get results. If you can show an error in my reasoning, l have no problem in accepting that. In fact l would be appreciative. But my only interest is in getting results and if the results l get match reality, then l personally count that as success. My ugly writing can be tidied up if correct. Disgarded if not. Again l thank you for at least not dismissing out of hand what l wrote. I am truly interested if you can show an error. Beauty lies in the content, not the style. Please look at my twin prime proof to see if you can find an error there. With great respect, there isn't one. If l have missed any explanation please keep up your noncombative tone. 

20180227, 21:49  #4 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
https://math.illinoisstate.edu/day/c...nnotation.html may help the OP.

20180228, 09:37  #5 
Feb 2018
2^{2}×3^{2} Posts 
Mistake made. Should read ...on number line (13+n30) for example, 113 as a lowest prime factor multiplies with 131 which is greater than 121squared, not (23+n30). AND, the last part of the equation should read...finally × (121squared minus 1)/30, not (prime(n)1)/30. Apologies, l do that too often. I don't seem to be able to edit the post.

20180228, 17:37  #6  
Aug 2006
3×1,987 Posts 
Quote:
Last fiddled with by CRGreathouse on 20180228 at 17:37 

20180302, 12:47  #7  
Feb 2018
2^{2}·3^{2} Posts 
Quote:
210 equals....(primes from 7 up to 121(7  113)each minus 1) multiplied together, divided by (primes 7 up to 121(7  113) multiplied together. Then multiplied by (121squared minus 1)/30. This result if true but irrelevant, and higher numbers that stiil work, need explaining away as to why such huge numbers being divided and multiplied result in numbers that match desired result. 

20180302, 15:52  #8 
Aug 2006
13511_{8} Posts 
Sorry, I can't make any sense of what you've written. Good luck!

20180302, 17:37  #9 
Feb 2018
2^{2}·3^{2} Posts 

20180302, 17:46  #10 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Math has well used definitions, axioms, and notation which is easier to understand for anyone deep into math.
Last fiddled with by science_man_88 on 20180302 at 17:52 
20180302, 18:38  #11 
Aug 2006
3·1,987 Posts 

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