## Re: Shortcutting the Lucas-Lehmer Test

Expand Messages
• If we look at the known test: A Mersenne number M(p) is prime if M(p) divides S(p-2) where S(0)=4 and, S(p)=S(p-1)^2 -2(mod 2^p -1) My double case
Message 1 of 3 , Aug 21, 2004
If we look at the known test:

A Mersenne number M(p) is prime if M(p) divides S(p-2) where S(0)=4
and,

S(p)=S(p-1)^2 -2(mod 2^p -1)

My double case test/probable test: (where L(n) are Lucas numbers, via
the golden proportion)(10/19/01) The golden ratio could speed up the
squarings involved. Since there are various algorithmic ways to
generate Lucas numbers, by adding fractals etc. The mod here, would
be subracted at each add, rather than divided out.

L(2^(p-1))mod 2^p -1 = 0 [CASE1]
L(2^(p-1) -1) mod 2^p -1 = 0 [CASE2]
Or L(2^(p-1) -1)mod 2^p +1 /3 = 0[CASE2]

2^2-1=3 divides L(2)=3 [C1]
2^3-1=7 divides, L(4)=7 [C1]
2^5-1=31 divides, L(15)=1364 [C2]
2^7-1=127 divides, L(64)=23725150497407 [C1]
2^13-1=8191 divides, L(4095) [C2]
2^17-1=131071 divides, L(65535) [C2]
2^19-1=524287 divides, L(262144) [C1]

So, also with F(n): (Fibonacci numbers)
F(2^p) mod 2^p -1 = 0[Case 1]
F(2^p -2) mod 2^p -1 = 0[Case 2]
F(2^p -2) mod 2^p +1 /3 = 0[Case 2]

2^2-1=3 divides F(4)=3 [C1]
2^3-1= 7 divides F(8)=21 [C1]
2^5-1=31 divides F(30)=832040, [C2]
2^7-1=127 divides F(128)=251728825683549488150424261 [C1]
2^13-1=8191 divides F(8190) [C2]
2^17-1=131071 divides F(131070) [C2]
2^19-1=524287 divides F(524288) [C1]