Yes, it is.

[QUOTE=Andi47;292112]How did George know that only the last iteration was zeroed out, and not one of the previous ones? Would zeroing out an iteration halfway through (and thus having zeros all the further iterations)[/quote]No, a zero does not lead to a zero in the next iteration. See below.

[quote]produce a non-zero error code?)[/QUOTE]Suppose an iteration k, not one of the last three iterations, produces (erroneously) a zero value for S[sub]k[/sub].

Since the LL iteration calculation is S[sub]k[/sub][sup]2[/sup] -2 => S[sub]k+1[/sub], if S[sub]k[/sub] = 0 then S[sub]k+1[/sub] = 0[sup]2[/sup] - 2 = -2.

The next iteration after that will produce S[sub]k+2[/sub] = (-2)[sup]2[/sup] - 2 = 2.

The iteration after that will produce S[sub]k+3[/sub] = (2)[sup]2[/sup] - 2 = 2, so all remaining LL iterations will just keep repeating the 2 value, and that will be the value of the final residue.

(The above assumes that the calculations of S[sub]k+1[/sub], S[sub]k+2[/sub], S[sub]k+3[/sub] ... proceed without error.)

This fact (repeated 2 values) by itself will not cause an error code to be generated.

[QUOTE=cheesehead;292168]No, a zero does not lead to a zero in the next iteration. See below.
<snip>[/QUOTE]

Thanks. Seems my knowledge of the LL-test is quite rusted... :redface:

[QUOTE=Batalov;292160]While you would be at it, you may want to add the output of a few last RES64 values to the results.txt. The usual question people ask is if the penultimate iteration value was + or -[TEX]\sqrt 2 (mod M) [\eq \pm 2^{(p+1)/2} ][/TEX] -- there will be an answer for free.

Ze math:
[SPOILER]The next to last iteration can be only one of two values, the square of each equals 2. It is easy to see that they are sqrt2=2[SUP](p+1)/2[/SUP] or Mp-sqrt2. Hint: square them and subtract 2; the result will be 0. Their RES64 will be 0000000000000000 or FFFFFFFFFFFFFFFF. The second next to last could be one of four values, etc.[/SPOILER][/QUOTE]

I've been told the problem with an idea like this ( I thought I expressed a similar idea in the theory on Mersenne primes thread) is will it take longer to do the checks than to do the next iteration

[QUOTE=Batalov;292160] The second next to last could be one of four values, etc[/QUOTE]
That only for prime mersennes. There is no sqrt(sqrt(2)+2) for non-prime mersenne, the value is QNR.

M( 5xxxxxxx )C, 0xdb8f585db0b58a3e, n = 3145728, CUDALucas v1.64
It was a match. Good news for CUDALucas v1.64 and the card.

Blast.

Just to let you all know that there's a new competitor in the house. The leading edge of Generalized Fermat Prime Search at BOINC PrimeGrid is now in world record territory. More info in here: [URL]http://www.primegrid.com/forum_forum.php?id=75[/URL]

Here is the exponent in question: [url]http://mersenne.org/report_exponent/?exp_lo=50823037&exp_hi=10000&B1=Get+status[/url]

It looks like the bad result has been removed.

[QUOTE=ixfd64;292209]

It looks like the bad result has been removed.[/QUOTE]
It had been gone for a while, since the test was turned in I think. It might even be that PrimeNet automatically deletes the result as soon as it's reported.

[QUOTE=LaurV;292198]That only for prime mersennes. There is no sqrt(sqrt(2)+2) for non-prime mersenne, the value is QNR.[/QUOTE]
But was I talking about other Mersennes? For them, penultimate iteration value [tex]\neq \pm \sqrt 2[/tex].