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Abstract

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.
Conclusions
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)) = -
(99364166484840112256708922645041637995448020289168347414383980959309
77604225271478451567997266651943887682613902625240630756400260258807
4672130259396141609459384320000000000000000000000000
(14124738050219081504823475627787268312082340753306253593396010464249
47134801096089880492729526564485794206039961909522697735028587280446
3187506556774268539415656610053022758274688636348233364040819
x^(2/199) +
74395815037759038466048139180476073256924007194049059793540321633040
15375586945896427977202396635263570951463768739415871568529686983470
388556886477432807087259005918095557574075248654438621000088))/
(76790525741798814147739722024656541832379779544942834289320631222830
12863136141226080950493607708198950026290145142696888920938866374089
05005446892190734624961059132500532096789390465051335714291661355241
05744097027866266597980001 x^(19901/199))
The exponent in blue is the number given.
We choose the number in yellow
The factorization is
2^3×3×19^2×23×41×59×61×73×83×151×181×193×211×227×229×233×239×263×271
×281×293×311×347×359×389×419×421×439×457×607×617×643×647×661×709×743
×787×797×821×853×929×977×1123×1277×1327×1433×1439×1493×1559×1811×183
1×2239×2389×2521×2687×2753×3433×3583×3821×3881×4229×4909×5573×6269×9
851×11941×13931×16319×17911
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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