Sunday, 2 January 2011

Factor 2,5

Formula for calculating the result of the sieve of Eratosthenes
Provides an opportunity to work for (min) intervals. This is a great advantage compared to the known formulas, determine the number of primes in the intervals (0, x).
For example:
The formula for calculating the number of primes in the interval  

has the form
 
For values of (m)
The formula is
 
Value in the denominator
 
taken only over the integers, by Antje

Below I give a table of calculations based on this formula for (k) = 2,5. Please note the change in error. Agree to obtain such a result, all (min) intervals,



Old methods can not.

Coefficient (k). How to tie a factor and the value of P_n. Here's a question that needs solving? Tabulated calculations show (k) = 2,5 satisfies all the values that I checked. And for the other values remains an open question.
The initial value (k) = 3, I found out from a simple calculation (k) for
P_n = 2    (k)=2 (m) = 2, ie, no interval. P_n=m
You can take the initial value (k) and more than three, but this leads to an increase in the interval, my goal, to investigate the result of (min) interval.




0 n=1 k=3






0 n=2 k=3







0,1666666666666667 n=3 k=2.5







-0,625 n=4 k=2.5







-0,125







0,5







-0,5







0,5







0,5







0,0833333333333333







-0,0833333333333333







-0,5833333333333333







-0,9166666666666667







0,3571428571428571







0,7857142857142857







-0,0714285714285714







0,0714285714285714







0,7857142857142857







-0,0714285714285714







-0,6428571428571429







0,0714285714285714







0,6875







1,9375







1,8125







0,3125







0,5625







0,1875







-0,5625







0,0625







1,3125







0,6875







-0,0625







-0,1875







-0,5625







1,388888888888889







1,944444444444444







1,611111111111111







2,277777777777778







2,388888888888889







1,055555555555556







-0,2777777777777778







0,2777777777777778







0,0555555555555556







-0,3888888888888889







-0,2777777777777778







1,277777777777778







1,611111111111111







-1,055555555555556







-0,9444444444444444







-1,388888888888889







-0,2777777777777778







-1,611111111111111







-2,055555555555556







-2,277777777777778







2,25







0,75







0,25







1,25







1,25







2,25







2,75







1,25







0,75







1,75







2,25







1,25







-0,25







0,25







-0,25







0,25







0,25







-0,25







-0,25







-0,75







-0,25







-1,25







-1,75







-2,75







-1,75







-3,75







-4,25







-2,75







-3,25







-1,75







-2,25







-1,25







-0,75







-2,75







1,772727272727273







3,227272727272727







3,136363636363636







2,863636363636364







1,681818181818182







1,590909090909091







2,409090909090909







3,318181818181818







1,681818181818173







-1,590909090909091







-0,1363636363636364







0,9545454545454545







2,318181818181818







1,590909090909054







0,9545454545454545







0,3181818181818182







0,7727272727272727







-1,863636363636364

Continuing the theme (2) K = 2,5
Criterion for evaluating the formula for calculating the number of primes in the interval, there is a margin of error. If you take multiple values P_n, (there are several intervals) then the criterion for evaluating the formula can take the (E + -) = number of changes of signs of error.
For example: Factor (k) = 2,3 (E + -) = 68
(k) = 2,35 (E + -) = 129
(k) = 2,4 (E + -) = 151
(k) = 2,45 (E + -) = 77
(k) = 2,5 (E + -) = 64
Of course, it's all wet, but I think the more the quantity (E + -) = the more complete formula. What do you think?
Grind a little search on small values, and you can try larger values (P_n). Certainly within the capabilities of the program, the possibility of computing.
 
Sergey Sitnikov 

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