decoding RSA. what is the complexity?

II° Lepore primality testing and factorization

with an appropriate n the three equations to decode the 'RSA are: (if not mistaken)

X ^ 2 + n * (X * 6) = number rsa

X ^ 2 + (6n + 2) * X = number rsa

X ^ 2 + (6n + 4) * X = number rsa

X = the first factor (unknown)

the three equations of n are: (if not mistaken)

n = (number rsa - X ^ 2) / (X * 6)

n = (number rsa - X * (X + 2)) / (X * 6)

n = (number rsa - X * (X + 4)) / (X * 6)

oh I forgot applies as primality testing and factorization

and do not forget the 2 and 3

We place two examples of decoding X * Y = number rsa example1 the best case is when the two prime numbers are consecutive, or are closest to each other is faster than the algorithm: 10916407 = number rsa; X = 3301; Y = 3307; the cycle (on) will be from 1 to 1 that is a cycle in fact can occur from the first equation X ^ 2 + n * (X * 6) = number rsa example2 The worst case is when the two prime numbers are the highest or farthest distance between them are the slower the algorithm. 50035 = rsa number; X = 5; Y = 10007 the cycle (on) will be from 1 to 1667 in fact can occur from the first equation X ^ 2 + n * (X * 6) = number rsa

In conclusion we can say that we need a mixed algorithm, that is, test the low numbers with I° Test di fattorizzazione e di primalità di Lepore however integers while the algorithm of the three equations.

The n then will have two functions:

for n = 0; n = k; n + 1

{

rsa / value associated with the nth position * 6;

rsa equations for n;

}

explain better with a numerical example

rsa / 1

equations for n = 0

rsa / 5

equations for n = 1

rsa / 7

equations for n = 2

rsa / 11

equations for n = 3

etc.etc.

Find the meeting point, or k, is the complexity of the algorithm.

I'll post some data

100 for k = 3

1000 k = 7

10000 for k = 20

100000 for k = 62

1000000 for k = 194

what is the complexity?