Factorization and primality test O([log_9(N)]^3)

[Alberico Lepore]

I thought

6^X

where X is

N^(1/X)=1

[Alberico Lepore]

I am not practical in mathematics.
It should be about 6 ^ [log_2 (log_10 (N))]

[Dr Sardonicus]

Let's see here: Subject says O([log_9(N)]^3)

Post #5 says

sorry it's 3 ^ [log_9 (N)] = sqrt (N).
but I will try again

Post #23 says

I thought

6^X

where X is

N^(1/X)=1

I am not practical in mathematics.

Apparently not. if N > 1, and N^(1/X) = 1, then X = ln(N)/(2*Pi*I*k) for a non-zero integer k, where Pi is the circle number and I^2 = -1. So exp(ln(6)*X) is a non-real complex number of absolute value 1.

Post #24 says

It should be about 6 ^ [log_2 (log_10 (N))]

So now you're claiming, if I did the algebra correctly,

O(log(N)^(log(6)/log(2))).

Since log(6)/log(2) < 3, that's smaller than your original claim.

I can hardly wait to see what's next

[Alberico Lepore]

Originally Posted by Dr Sardonicus

Let's see here: Subject says O([log_9(N)]^3)

Post #5 says

Post #23 says

Apparently not. if N > 1, and N^(1/X) = 1, then X = ln(N)/(2*Pi*I*k) for a non-zero integer k, where Pi is the circle number and I^2 = -1. So exp(ln(6)*X) is a non-real complex number of absolute value 1.

Post #24 says

So now you're claiming, if I did the algebra correctly,

O(log(N)^(log(6)/log(2))).

Since log(6)/log(2) < 3, that's smaller than your original claim.

I can hardly wait to see what's next

thank you

[CRGreathouse]

Originally Posted by Dr Sardonicus

So now you're claiming, if I did the algebra correctly,

O(log(N)^(log(6)/log(2))).

Since log(6)/log(2) < 3, that's smaller than your original claim.

I get the same.