20070515, 00:34  #12  
∂^{2}ω=0
Sep 2002
República de California
2^{6}·181 Posts 
Quote:


20070515, 00:58  #13  
Feb 2007
2^{4}×3^{3} Posts 
Quote:
(there are no "editing privileges"  submitting the form sends an email to N.Sloane who will incorporate the specially formatted data within some days, and mention the contributor & date for any new data, comment etc.). Quote:


20070516, 00:23  #14 
Feb 2007
660_{8} Posts 
Can't we exploit the fact that D3=X(X^23)
successively gives all elements of the tree at given height, and commutes with LL=X^22 ? 
20081004, 12:17  #15 
Einyen
Dec 2003
Denmark
BC1_{16} Posts 
M37156667 interim Wd1 residue 833185DE0326B429 at iteration 37156663
M37156667 interim Wd1 residue 21F10ABD65A37DF4 at iteration 37156664 M37156667 interim Wd1 residue 36E916675C4F0334 at iteration 37156665 M37156667 interim Wd1 residue 0000000000000000 at iteration 37156666 M37156667 interim Wd1 residue 0000000000000000 at iteration 37156667 M37156667 is prime! Wd1: EA38283B,00000000 M43112609 interim Wd4 residue F042423AF370FBF0 at iteration 43112605 M43112609 interim Wd4 residue C395044781A6672D at iteration 43112606 M43112609 interim Wd4 residue 138F82F361A9E5B9 at iteration 43112607 M43112609 interim Wd4 residue 0000000000000000 at iteration 43112608 M43112609 interim Wd4 residue 0000000000000000 at iteration 43112609 M43112609 is prime! Wd4: 5B377150,00000000 So: M45: p=37156667 s[p3]== 2^((p+1)/2) (mod Mp). M46: p=43112609 s[p3]== 2^((p+1)/2) (mod Mp). Now we have 24+ and 21. penultimate.txt Last fiddled with by ATH on 20081004 at 12:17 
20081005, 03:45  #16 
Dec 2003
Hopefully Near M48
1758_{10} Posts 
Isn't the nexttolast residue of a Mersenne prime never supposed to be 0? Because then the final residue would be 0^2 2 = 2. Or am I forgetting something?

20081005, 04:19  #17  
May 2008
3×5×73 Posts 
Quote:
If the nexttolast residue is positive, it is of the form: 2^((p+1)/2) In binary, a power of two has only a single '1' bit and the rest are zeroes. When p is bigger than or equal to 127, then the '1' bit in the p1 residue will be outside the 64bit residue, making the 64bit residue have allzeroes, but the full residue is not zero. Look at the p1 residue of M61: 0000000080000000 or 2^31. 

20081006, 02:06  #18 
Einyen
Dec 2003
Denmark
3009_{10} Posts 
Yeah its either 2^{(p+1)/2} or +2^{(p+1)/2}. If its + in binary it will be 10000...millions of zeros..00000 and the 64bit residue will be 0x0000000000000000. If its  the residue will be 2^^{p}1  2^{(p+1)/2}, which in binary is 1111...millions of 1's...111101111...millions of 1's...11111, so the 64 bit residue will in hex be 0xFFFFFFFFFFFFFFFF.
Last fiddled with by ATH on 20081006 at 02:08 
20090622, 12:38  #19 
Einyen
Dec 2003
Denmark
3×17×59 Posts 
M42643801 interim Wd4 residue 56D4857CEECAE826 at iteration 42643797
M42643801 interim Wd4 residue 1FE903F0A3186A6F at iteration 42643798 M42643801 interim Wd4 residue DAACA15663F030C8 at iteration 42643799 M42643801 interim Wd4 residue FFFFFFFFFFFFFFFF at iteration 42643800 M42643801 interim Wd4 residue 0000000000000000 at iteration 42643801 M42643801 is prime! Wd4: B7E878A9,00000000 M46: p=42643801 s[p3]==  2^((p+1)/2) (mod Mp). Now there is 24+ and 22 : penultimate.txt resultspenultimate.txt Last fiddled with by ATH on 20090622 at 12:38 
20090622, 15:57  #20 
∂^{2}ω=0
Sep 2002
República de California
2^{6}×181 Posts 
Thanks for the table addition, ATH.
Now, anyone want to trot out a curiousthoughlikelyspurious statistical correlation, perhaps with the phase of the moon or one of the major stock market indices at time of discovery? 
20090622, 16:07  #21 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5794_{10} Posts 

20090625, 10:07  #22  
May 2009
Loughborough, UK
2^{2}·11 Posts 
Quote:
Maybe it is best just to submit the entries up to the completely checked range and add the rest as a comment to Sloanes. Otherwise A123271 will have to be corrected if another prime is discovered out of sequence. 

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