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Formula that returns prime numbers of the form (m)*(n)+1 how many prime numbers between 1 and 1000000 I used a polynomyal that it´s roots are the golden ratio squared and the golden ratio conjugate. When I aply fractional exponents to the x the polinomyal it´s roots returns lucas and fibonacci numbers exactly. And when I derivate this function I obtain prime numbers included in one number of the structure of the derivative, but there are more that prime numbers come in relation with the order of the derivative multiplied for the fractional exponent. Is like there a relation between fibonacci and lucas numbers and prime number where the original polynomial derivated adapted his form to returns special prime numbers.
25th of September of 2020
We can find any prime of the form ((m)*(n))+1 just under of a number given.
Because that number given depends of the order of the derivative * The
coefficient (m)+1
Better with an example
If we want to search the prime of the form (199*n)+1 just under than 19,900
we use the derivative 100 in our formula.
100th derivative (-2/x^(1/199) + (4 x^(1/199))/1
d^100/dx^100(-2/x^(1/199) + 4 x^(1/199)) = -
x^(2/199) +
05744097027866266597980001 x^(19901/199))
The exponent in blue is the number given.
We choose the number in yellow
The factorization is
We can see that the bigger of it´s primes is 17911 and 17910/199=90 so (n)=90
and (199*90)+1=17911 and this is the biggest prime number of the form
199*(n)+1 just under 19901
To obtain this prime we have to divide the number in yellow from (i=19,900 to
i=1) The first integer is the prime we are looking for. In our case that integer
will be when we divide the number in yellow by 17910
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