20061219, 16:58  #1 
Dec 2006
3 Posts 
New LLT formula
i use this formula to check primality of Mersenne numbers.
S(0)=2 S(n)=2*S(n1)  1 and generalized form is : S(0)=2^(2r) S(n)=(2^r)*S(n1)  2^(1r) 
20061219, 21:12  #2  
Feb 2005
2^{2}×5×13 Posts 
Quote:
How do you test primality of M_p using these numbers S(n)? Last fiddled with by maxal on 20061219 at 21:13 

20061220, 10:09  #3 
Dec 2006
3 Posts 
opps sorry i forgot square of S(n1)
S(0)=2 S(n)=2*S(n1)^2  1; and generalized form is : S(0)=2^(2r) S(n)=(2^r)*S(n1)^2  2^(1r) forexample: S(0)=2 S(1)=2*2^21=7 =0 Mod(2^31) s(2)=2*7^21=97 s(3)=2*97^21=18817 =0 mod(2^51) and so on... OR s(n)=4*s(n1)^21/2 s(0)=1 s(1)=4*1^21/2=7/2 =0 mod(7) s(2)=4*(7/2)^21/2=97/2 s(3)=4*(97/2)^21/2=18817/2=0 mod(31) .... pls any comment to proof or explain 
20061220, 14:40  #4 
"Bob Silverman"
Nov 2003
North of Boston
2^{3}×937 Posts 
Working mod M_p throughout:
This is just LL slightly disguised. Multiply by 2 giving: 2S(n) = 4S(n1)^2  2 = [2 S(n1)]^2  2 Compare the arithmetic for (say) M_5. In Your sequence we get: 2,7,4,1,0, while the ordinary LL gives 4,14,8,2,0. All you have done is divided the original sequence by 2. (or multiplied by 2^1 mod M_p) Working (say) mod 127, LL gives 4,14,67,42,111,0 while you sequence gives 2,7,97, ... etc. Note that 97 = 67/2 mod 127 etc. 
20061220, 20:13  #5 
Aug 2003
Snicker, AL
1700_{8} Posts 
Translated into common terms, it means your method is not an improvement of the existing method. Its just a variant.
Fusion 
20061221, 01:10  #6 
"Bob Silverman"
Nov 2003
North of Boston
2^{3}×937 Posts 

20070305, 17:36  #7 
Feb 2007
2^{4}·3^{3} Posts 
(*stumble in, some months late*)
This sequence is also called A2812 on OEIS. It gives S(k) not only through the already mentioned relations, but also through the following great formula : ((c) by M_F_H !) For k>3, S(k) = 2 + 2^(2k) x 3 x product( A2812(i), i = 1..k3 )^2 Unfortunately, this did not help me for my ultimate goal : calculate S(p1) mod 2^p1 as an explicit function of p and S(p2) mod 2^(p1)1 ... Last fiddled with by m_f_h on 20070305 at 17:47 Reason: removed title which consisted in illdefined clipboard contents (copypaste did not work directly into editing area...) 
20070305, 17:41  #8 
∂^{2}ω=0
Sep 2002
República de California
2·3^{2}·653 Posts 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
How do I run this formula in PFGW?  Stargate38  Software  1  20140819 15:23 
Interesting formulæ  Uncwilly  Lounge  29  20130505 16:12 
Fibonacci Formula  MattcAnderson  Math  7  20130114 23:29 
P1 formula (help wanted)  Prime95  Data  18  20120212 18:07 
Combination Formula?  drake2  Math  10  20061208 22:43 