mersenneforum.org > Math Repeating residues in LL tests of composite Mersenne numbers
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 2020-08-03, 17:30 #12 JeppeSN     "Jeppe" Jan 2016 Denmark 2·34 Posts It is perhaps also interesting to note the lengths of the "pre-periods", or offsets. That is the number of terms in the LL sequence preceding the first occurrence of the period. With Batalov's data as input, that is easy (PARI/GP): Code: findOffset(p,t)=s=Mod(4,2^p-1);S=s;for(i=1,t,S=S^2-2);i=0;while(s!=S,s=s^2-2;S=S^2-2;i++);i Pass the Mersenne exponent as p, and the period length (from Batalov's post; of course Batalov's technique for finding the period length is also easy to write in PARI/GP) as t. In the function, the S sequence is always t positions ahead of the s sequence, so we just check to see how long it takes until they agree. Result: Code: findOffset(11,60) 1 findOffset(23,32340) 1 findOffset(29,252) 1 findOffset(37,516924) 5 findOffset(41,822960) 1 findOffset(43,420) 3 findOffset(47,20338900) 4 findOffset(53,1309620) 2 The last one, findOffset(59,345603421440), would take a long time with PARI/GP with this approach. Not sure if the above values have any interesting interpretation. We see that for the cases I was able to resolve, the constant 120000 in Batalov's C program was on the safe side. Of course, Viliam Furik's approach (searching for "14") works exactly in the cases where findOffset gives 1. /JeppeSN
2020-08-03, 18:13   #13
JeppeSN

"Jeppe"
Jan 2016
Denmark

2×34 Posts

Quote:
 Originally Posted by Batalov Periods may (?) be different if we use the other popular seed values S0 = 10, or S0 = 2/3 (mod Mp).
That is true: This PARI/GP function imitates your program, with an optional seed argument:
Code:
findPeriod(p,seed=4)=s=Mod(seed,2^p-1);for(i=1,120000,s=s^2-2);S=s;i=0;until(s==S,S=S^2-2;i++);i
Results:
Code:
findPeriod(11)
60

findPeriod(11,10)
10

findPeriod(11,2/3)
5
So they do differ. The one for 2/3 is not even a multiple of 11 - 1 = 10. You may think PARI handles 2/3 wrong, but it does not: Evaluating findPeriod(11,683) also gives 5.

Explicitly, the LL sequence for 2^11 - 1, starting from seed 2/3 == 683, is:

683 -> 1818 -> 1264 -> 1034 -> 620 -> 1609 -> 1471 -> 160 -> 1034 -> etc.

Similarly, for 2^29 - 1, and seed 2/3, the period length is 154, which is 14 mod 28.

/JeppeSN

 2020-08-03, 18:15 #14 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 217368 Posts You have a solid footing to retrace Tony's DiGraphs adventure, now! It is a bit of a déjà vu, but a good warm up to visit his old threads.
 2020-10-01, 22:15 #15 jshort   "James Short" Mar 2019 Canada 1710 Posts Let $S_{1} = 4 = (2 + \sqrt{3}) + (2 - \sqrt{3})$ and $S_{n+1} = S_{n}^{2} - 2$. We'll first prove via induction that $S_{n} = (2 + \sqrt{3})^{2^{n-1}} + (2 - \sqrt{3})^{2^{n-1}}$. Clearly $S_{1} = (2 + \sqrt{3})^{2^{1-1}} + (2 - \sqrt{3})^{2^{1-1} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$. Assume that it holds for all $k < m$. Now $S_{m} = S_{m-1}^{2} - 2 = ((2 + \sqrt{3})^{2^{m-2}} + (2 - \sqrt{3})^{2^{m-2}})^{2} - 2 = (2 + \sqrt{3})^{2^{m-1}} + (2 - \sqrt{3})^{2^{m-1}} + 2(2 + \sqrt{3})^{2^{m-2}} \cdot (2 - \sqrt{3})^{2^{m-2}} - 2 = (2 + \sqrt{3})^{2^{m-1}} + (2 - \sqrt{3})^{2^{m-1}} + 2 - 2 = (2 + \sqrt{3})^{2^{m-1}} + (2 - \sqrt{3})^{2^{m-1}}$ Note that in the above, I used the fact that $(2 + \sqrt{3}) \cdot (2 - \sqrt{3}) = 1$. Their inverses. Since their inverses of each other, they also have the same order over $M_{p}$. Suppose the order of $2 + \sqrt{3}$ is $n = 2^{i} \cdot m$. If $i=0$, the order is odd. In any case, I believe $n$ will have to be a factor of $\prod_{i} (p_{i}^{2} - 1)$ where $p_{i}$ are the prime factors of $M_{P}$.** First assume that $n$ is odd. Let $r$ be the smallest integer such that $2^{r} \equiv 1 mod(n)$ In this case, we have that $(2 + \sqrt{3})^{2^{r} - 1} \equiv 1$ or $(2 + \sqrt{3})^{2^{r}} \equiv 2 + \sqrt{3}$. The same can be repeated with $2 - \sqrt{3}$. The result is that $S_{r+1} = (2 + \sqrt{3})^{2^{r}} + (2 - \sqrt{3})^{2^{r}} \equiv (2 + \sqrt{3}) + (2 - \sqrt{3}) \equiv 4 \equiv S_{1}$. However what if $n$ is even? In this case, $i > 0$ where $n = 2^{i} \cdot m$. We can repeat the procedure above, however we have to start at $S_{i+1}$ and instead of asking ourselves what the order of $2$ is, we're asking what the order of $2^{i}$ is over $m$. ............................. **My guess that the order $n$ needs to be a factor of $\prod_{i} (p_{i}^{2} - 1)$ is based on the assumption that the number of invertible elements $a + b \sqrt{3}$ over $Z_{p}$ is always going to be $p^{2} - 1$, however my memory of group theory is fuzzy, so someone might want to check this! Last fiddled with by jshort on 2020-10-01 at 22:42
2020-10-20, 17:23   #16
Viliam Furik

Jul 2018
Martin, Slovakia

22·5·13 Posts

Quote:
 Originally Posted by Batalov Periods may (?) be different if we use the other popular seed values S0 = 10, or S0 = 2/3 (mod Mp).
Period values for starting value 10:
M11 -> 10 (1 * 10)
M23 -> 32340 (1470 * 22)
M29 -> 252 (9 * 28)
M37 -> 516924 (14359 * 36)
M41 -> 822960 (20574 * 40)
M43 -> 420 (10 * 42)

They are all the same (at least for these exponents) as with S(0) = 4, except for the M11.

I have also looked at periods in PRP testing:
M11 -> 10 (1 * 10)
M23 -> 528 (24 * 22)
M29 -> 252 (9 * 28)
M37 -> 516924 (14359 * 36)
M41 -> x > 3077786
M43 -> 9492 (226 * 42)

Here they are more different. But still sometimes the same as for LL, and also preserve the k*(p-1) property.

Last fiddled with by Viliam Furik on 2020-10-20 at 17:25

 2020-10-20, 23:22 #17 Viliam Furik   Jul 2018 Martin, Slovakia 1000001002 Posts I have done the PRP part because I realized that if we know the period of the PRP test of a composite exponent, we can run P-1 in a different way (which may or may not be faster, probably not), by simply computing the product of primes * 2p, finding its binary representation, and multiplying residues corresponding to powers of 1s in the representation, modulo the period, and doing the GCD. Of course, that only pays out if the period is smaller than the log2 of the (2p * product of primes smaller than smallest B1 needed ). It depends on the smoothness of the factors of Mp. EXAMPLE: p = 11 B1=1 E(B1) --> product of all primes less than B1 x = E(1) * 2 * p = 2210 = 101102 3x = 32^4 * 32^2 * 32^1 (I compute the As all powers here are smaller than the period length, the method is of no use in this case. If there was a power of 3 higher than the period length + the length of the aperiodic part, then we can find a corresponding power of 3 that is less by some multiple of the period length. The obvious upper bound for the period is the number of quadratic residues, phi(Mp)/2k, where k is the number of factors of Mp, and phi(n) is the Euler totient function. ................................... Because I realize that this method might never be possible to use, due to needed bound being much lower than necessary, I have been thinking about it for about 6 hours now. If you think it's unusable, you are most probably right.

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