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I'm reviewing the cyclotomic form $f_b(n)= {b^n-1 \over b-1}$ for various properties to extend an older treatize of mine on that form.

With respect to primality there is the Lucas-Lehmer-test for primeness of $f_2(p)$ where of course $p$ itself must be a prime.

I was now looking, whether I can say some things for primes of the form $f_3(p) = {3^p-1 \over 2}$ ,for instance $f_3(3)=13, f_3(7)=1093, f_3(13)=797161, ...$ (more terms see bottom). For this I was looking for a comparable test, similar to the scheme in the Lucas-Lehmer test.

There is a short remark at Weisstein's mathworld involving the concept of Lucas-sequences for a generalized primality test (eq (2) to (4)), of which then the Lucas-Lehmer-test is only a special case, but I could not decode the formulae & recipes into some algorithm.

So Q: Is there a primality test for numbers of the form $f_3(p) = {3^p-1 \over 2}$ similar to the scheme in the Lucas-Lehmer test?

More terms for $(3^{a(n)}-1)/2 \in \Bbb P$ 
 a(n)=[3, 7, 13, 71, 103, 541, 1091, 1367, 1627, 4177,
 9011, 9551, 36913, 43063, 49681, 57917, 483611, 877843 ]

source: OEIS:A028491

Gottfried Helms
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  • I would like to know the same thing. It is very surprising that nobody seems interested. Having such a test would make it much more efficient to find large primes (provided each iteration is not much more expensive), as the number of iterations could be relatively tiny with large prime bases. – Alan Gee Aug 24 '17 at 19:49
  • If you want to know the current status of the facorization of numbers of the form $3^n-1$ (and many similar forms), go to https://homes.cerias.purdue.edu/~ssw/cun/index.html – Gerry Myerson Aug 26 '17 at 09:47
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    The Lucas-Lehmer test works for testing if $P$ is prime when $P+1$ has a known factorization. Therefore it works for $2, 3^p-1$ but not for $\frac{3^p -1}{2}$ @GerryMyerson – reuns Aug 26 '17 at 18:12
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    I know that the Lucas-Lehmer test doesn't work for numbers of this form, but an adaptation of it to work for such numbers would be so useful that it should have been done by now - if feasible. – Alan Gee Aug 26 '17 at 21:44
  • That comment of @AlanGee meets perfectly the intention of my question. At the moment and up to beginning of september I'm on holyday and cannot make detailed contribution, but I've time to read here frequently and to think forward unsystematically... "under the sun" (quoting a bit of Qohelet ;-) ) – Gottfried Helms Aug 27 '17 at 08:09
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    I presume that you know this paper, any case I add it here, but it isn't directly related with your question. Denoting your sequence as $G_p$ it is easy to deduce a relationship involving your $G_p$ and the $N_p$ of the paper, feel free to ask me about it but I don't know if such relationship that I evoke should be useful. The paper that I say is Pedro Berrizbeitia, Florian Luca, and Ray Melham, On a Compositeness Test for $(2^p+1)/3$, Journal of Integer Sequences, Volume 13, Issue 1, Article 10.1.7 (2010). –  Sep 15 '17 at 18:28
  • Ah - thanks to @user243301. No I didn't know that article, but it is easy reading and also has links to more PRP tests. I'll see where this leads me to ... – Gottfried Helms Sep 15 '17 at 18:39
  • The Lucas-Lehmer test works too for testing if Fermat numbers are prime. There are 3 proofs on the web, including mine. My proof shows that this LLT test is equivalent to the Pépin's test. – Tony Reix Jul 19 '22 at 08:52
  • The LLT for Mersenne and Fermat numbers is based on the big tree of the DiGraph under x^2-2 modulo N : start from a universal seed and reach 0. It may be possible to use instead a cycle of the DiGraph. There are conjectures for using a cycle of the DiGraph for proving that a Mersenne, Fermat, or Wagstaff number is prime. The necessity part has been provided for these 3 kinds of numbers. For Wagastaff numbers ((2^q+1)/3), for which there are no big tree, a big PRP has been found using the Vbra-Reix method: (2^4031399+1)/3 . – Tony Reix Jul 19 '22 at 09:03

6 Answers6

4

I refer you to Brillhart et al., Factorizations of $b^n\pm1$, published by the American Mathematical Society. The section, Introduction to the Main Tables, subsection Developments Contributing to the Present Tables, subsubsection Developments in Primality Testing, will tell you what methods there are for factoring such numbers.

Gerry Myerson
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    The numbers we are referring to are of the form 111 base(p) where the number of 1s is also a prime. These numbers are all either prime or Fermat Liars for witness p just as standard Mersenne numbers are either prime or Fermat Liars for witness 2. b^n +- 1 is of little use for bases other than 2. – Alan Gee Aug 26 '17 at 21:51
  • The beauty of Lucas-Lehmer is that the number of iterations is dependent on the number of powers of 2 in the power to which 2 is raised not on the number being tested. Imagine if this type of test worked with prime number bases in the trillions rather than 2. The test iterations would be reduced drastically. – Alan Gee Aug 26 '17 at 22:19
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    Brillhart et al are well aware that $3^n-1$ is even, hence composite, for $n\ge2$. They are interested in obtaining the complete factorization, if possible, for numbers of the form $b^n\pm1$, $b\le12$ (so we are not talking about bases in the trillions), and once you've factored the 2 out of $3^p-1$ you want to test for primality before attempting to factor. So if you haven't looked at the book, I'd suggest doing that before dismissing it out of hand. And if it turns out there's nothing there like a Lucas-Lehmer test, then that would strongly suggest no such thing exists. – Gerry Myerson Aug 27 '17 at 00:18
  • Thank you for the hint; I've downloads of the factorization-tables from several years ago. However it was that I wanted to understand the logic of the Lucas-Lehmer-test in such a way that I could possibly create a similar process with the $(3^n-1)/2$ numbers. The actual factors/the factoring methods are not the focus of this current question. – Gottfried Helms Aug 27 '17 at 08:05
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    I've found the article; possibly the information on pg lxii , par (7) "Lucas theorem" is the one which is relevant. I hope I'll get a practical exercise working next week, when I'm home. Thanks for the hint so far! – Gottfried Helms Aug 27 '17 at 09:01
  • @GerryMyerson - It was the mention of factorization that put me off going into the details, although you are right that I shouldn't have dismissed it so readily. The reason I think there should be a way of adapting Lucas Lehmer is because I believe it woks for Mersenne numbers due to the fact that N + 1 has only a single prime factor, whereas for other bases (to which the question refers) we can say that KN + 1 has only a single prime factor. So on the surface there is a lot in common. – Alan Gee Aug 28 '17 at 14:05
  • Lucas primality test works when we know how to factor $p \pm 1$, more generally $p^n-1$ (the structure of $\mathbf{F}_{p^n}^\times$). Thus it doesn't work there. @GottfriedHelms – reuns Sep 09 '17 at 14:02
  • @AlanGee The LLT works quite the same way for $2 a^b -1$ (if you know the factorization of $a$). The difficulty is the very low density of such primes for $b$ large. – reuns Sep 10 '17 at 15:53
2

There is the basic idea behind the Lucas-Lehmer primality test. For $n= 2^p-1$ we can choose $d= 3$ (quadratic reciprocity) and $\alpha= 2+\sqrt{3}$

If $n$ is prime and $d$ is not a square $\bmod n$ then $$\mathbb{Z}/n\mathbb{Z}[\sqrt{d}] = \{ a+b \sqrt{d}, (a,b) \in \mathbb{Z}/n\mathbb{Z}\}$$ is a field with $n^2$ elements, and its multiplicative group is cyclic with $n^2-1$ elements. If we find an element $\alpha$ of multiplicative order $n+1$ (ie. $\alpha^{n+1} \equiv 1 \bmod n, \alpha^{(n+1)/p} \not\equiv 1 \bmod n $ for every prime divisors $p | n+1$) then $n$ is prime.

(since otherwise with $q$ the least prime divisor of $n$ then $\mathbb{Z}/q\mathbb{Z}[\sqrt{d}]^\times$ is a group with less than $q^2-1 \le n-1$ elements, so the order of $\alpha\bmod q$ can't be $n+1$)

Thus all we need is to know the prime divisors of $n+1$ and compute $\alpha^{(n+1)/p} \bmod n$ for many $\alpha$. The same idea works in $\mathbb{Z}/n\mathbb{Z}$ if we know the prime divisors of $n-1$, and in $\mathbb{Z}/n\mathbb{Z}[x]/(f(x))$ for some irreducible polynomial $f$ of degree $k$ if we know a large part of the factorization of $n^k-1$.

$$\boxed{\ \ \text{Thus it doesn't work for }\ \frac{3^a-1}{2} \quad(\text{but it does for }2\cdot 3^a-1)\ \ }$$

Community
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reuns
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This is not an answer but Lucas-Lehmer test for $(3^p-1)/2$ is maybe possible if you change the starting value and the sequence. I will show you the PRP test I found :

Let the sequence $S_i = 28 \cdot S_{i-1}^3 + 294 \cdot S_{i-1}^2 + 1029 \cdot S_{i-1} + 1197$ with $S_0 = 1197$

Then $(3^p-1)/2$ is prime if $S_{p-2} \equiv 0 (mod (3^p-1)/2)$

I have checked until 10000 with PARI GP and it seems it works for higher exponent

kijinSeija
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    Ah- thank you very much for this late contribution. Unfortunately my mind is far away of those problems currently, and I can perhaps come back to this only next days. – Gottfried Helms Apr 30 '23 at 10:29
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Having offered and awarded a bounty on this question for effort more than results, after considerable analysis and effort I have come up with a solution - probably.

Given a number of this form $\sum\limits_{k=0}^{P-1}{b^k}$
Where b is an odd Prime , P is Prime

the following primality test can be used to either confirm primality or highly probable primality (I have seen plenty of evidence to suggest the high probability but none to disprove the confirmation). Assume all variables are BigInteger (or equivalent)

initialize with res = $2^b$, test = $\frac{b^P-1}{b-1}$ 

Repeat P-1 times
{
    res = (res ^ b) Mod(test);
}
if after repeating (res = 2^b)
then test is Prime

So for the case where b = 3, and to answer the question:

Initialize with

res = 8 and test = $\frac{3^P-1}{2}$

Repeat P-1 times
{
    res = (res ^ 3) MOD(test);
}
if after repeating (res = 8)
then test is Prime

I will explain how I came up with this and why I am not at all confident that it proves primality as the Lucas-Lehmer test has been shown to.

My initial aim was to produce an efficient Fermat Primality test for 2, as I had noticed that nearly all (if not all?) numbers of this form with b=3, that passed that test were primes.

So I set out to find an efficient way to calculate $2^N mod(N)$ where N is the sum of powers of 3. Noticing that $2^{3^{n+1}} = {({2^{3^n}}})^3$ where n>0, and being familiar with the repeated squaring method, I recognized this as a candidate for repeated cubing.

However, when I tested this out I was surprised to find that the final phase of this method (i.e. multiplying all remainders mod(N)) was not necessary as the $P^{th}$ element was always 8. The reason for this is unclear to me, but maybe someone can answer this. I was also surprised to find that I could not get the test to fail when tested with a number of different prime bases and for thousands of prime exponents with base 3.

So the questions I would have about this would be:

  1. Is this just a shortcut way of doing a Fermat Primality test for witness 2?
  2. If it is, is there something special about numbers of this form with odd prime bases that means they are never Fermat Liars for witness 2?
  3. Is it just that there are very few such liars around and I, and probably others, have not yet exposed one?
Alan Gee
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  • You meant $\frac{3^p-1}{2}$. And no the order of $2$ modulo $\frac{3^p-1}{2}$ doesn't tell that it is prime (see the LLT) – reuns Sep 09 '17 at 14:51
  • @reuns - Thanks, I accidentally put P-1 instead of b-1 in the denominator and repeated the mistake. But I couldn't see that disproof in the reference you gave. Maybe I missed something. – Alan Gee Sep 09 '17 at 15:25
  • $2^{11}-1$ is composite. – reuns Sep 09 '17 at 15:26
  • but I stated that b must be odd prime. – Alan Gee Sep 09 '17 at 15:28
  • otherwise I realise that all Mersenne numbers with prime exponents would have to be prime, which they are clearly not. – Alan Gee Sep 09 '17 at 15:30
  • Very nice! I did it in Pari/GP with generalization of the base $3$ to base $5$ and $7$ and the testvalues being repunits in that bases (also called "q-analogue powers") and it worked fine for some small primes in the exponents. I don't know yet why this works, but it looks promising, so (+1) and if I tested some more it also gets the 'accept'-mark – Gottfried Helms Sep 09 '17 at 16:11
  • I compared it with Miller Rabin for base 3 up to the largest one I know of (((3 ^ 4177)-1)/2 which is the tenth) and it agreed in all cases. – Alan Gee Sep 09 '17 at 16:22
  • And I can explain how I formulated it if you are interested. – Alan Gee Sep 09 '17 at 16:25
  • Yes, anything which can teach me a lesson in this would be much appreciated! The Lucas-Lehmer-Test hase always been something lovely "magic" for me. I've even tried to come nearer to ot myself and did a small treatize but became distracted by a very nice observation, see here (http://go.helms-net.de/math/expdioph/lucasLehmer.pdf), and did then not try to continue the initial path... – Gottfried Helms Sep 09 '17 at 18:12
  • See that script and a list at my own answer. – Gottfried Helms Sep 09 '17 at 18:41
  • For those of us who don't speak computer, can you translate the test into English? Is it this: working modulo $(b^P-1)/(b-1)$, compute $2^b,2^{b^2},2^{b^3},\dots,2^{b^{P-1}}$; if the last residue is $2^b$, then the modulus is (probably) prime? – Gerry Myerson Sep 09 '17 at 22:25
  • @GerryMyerson: yes, that seems to be the correct transformation of the algorithm. (But better Alan Gee steps in) – Gottfried Helms Sep 10 '17 at 00:43
  • @GerryMyerson: However, seeing that in this simple expression (instead of including some operation like $x_{k+1}=x_k^p-c$ as it is in the LLT with its iteration $x_{k+a}=x_k^2-2)$ makes me now a bit suspicious about the basic reliability of that test, but let's see ... – Gottfried Helms Sep 10 '17 at 01:07
  • Did you study the maths of the LLT ?.. @GottfriedHelms – reuns Sep 10 '17 at 15:58
  • @reuns - I did my best but I am not as fluent with the numbertheory as it were needed. I've read into descriptions of the procedure and the proof at various places and a couple of years casually and have done exercises but I never got that full understanding of the basic principle why this method can work at all. – Gottfried Helms Sep 10 '17 at 16:04
  • assigned the bounty because of the effort done giving me another push to re-consider the question deeper. "Accept"-mark removed because after user @reuns remarks it came out that the method is not yet proved and very likely not really deterministic. – Gottfried Helms Sep 15 '17 at 17:43
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    @GottfriedHelms - Thanks for the bounty, but now I will have to come up with a proof to claim back my tick. At least the question got the exposure it deserved in the end. – Alan Gee Sep 15 '17 at 19:05
1

This is not an answer but an extended "comment" (example) to @AlanGee's answer

Pari script for testing Lucas-Lehmer-analoguous tests (LLaT) on repunits/q-analogues with $q$ the odd primes up to $q=19$ and $[m]_q$ for prime $m$ up to $399$.
We find the small quirk that the LLaT's are not safe when $m<q$ .
In the script base is $q$ which defines the "generalizations" of the mersernne-numbers, p is the exponent in the generalized "mersenne-numbers" ${\text{base}^{p}-1 \over \text{base}-1}$ or in the notation of $q$-analogues $[p]_\text{base}$

{forprime(base=3,19,
     print("base ",base);print("----------------");
     forprime(p=3,399, 
       test=(base^p-1)/(base-1); \\ repunit of prime length p to be tested

          t0=isprime(test);      \\ t0 indicates primeness by standard primetest            

       cmp=res=2^base; \\ now testing with LLaT
          for(k=1,p-1,res=res^base % test); 
          t1= (res == cmp);  \\ t1 indicates primeness by LLaT

       if((t0==0) && (t1==0),next()); \\ if composite by both tests,
                                      \\ forget it and do next test

       print(dfmt(p,6,0), ": ",t0," ",t1," ",    \\ otherwise print
              if(t0,"is prime","")," :  ",
              if(test<1e16,test,1.0*test) );
      );
      print("========================");
     ); }

Document:

 base 3
     p:t0 t1          : test (or: [p]_base) 
                remark: t0==1 means exact primetest gives "prime", t1==1 LLaT gives "prime"
--------------------------------------------
      3: 1 1 is prime :  13
      7: 1 1 is prime :  1093
     13: 1 1 is prime :  797161
     71: 1 1 is prime :  3.75473325749 E33
    103: 1 1 is prime :  6.95759652988 E48
    - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
    541: 1 1 is prime :  6.63084395472 E257  \\appended on second run. 
   1091: 1 1 is prime :  1.73084789203 E520  \\exact primetest-data t0 taken
   1367: 1 1 is prime :  8.38928997716 E651  \\from [OEIS:A028491][1]
   1627: 1 1 is prime :  9.44607595177 E775   
   4177: 1 1 is prime :  4.30973898516 E1992
   9011: 1 1 is prime :  1.09293990543 E4299
   9551: 1 1 is prime :  4.83140930986 E4556
=============================================
base 5
----------------
      3: 1 0 is prime :  31
      7: 1 1 is prime :  19531
     11: 1 1 is prime :  12207031
     13: 1 1 is prime :  305175781
     47: 1 1 is prime :  1.77635683940 E32
    127: 1 1 is prime :  1.46936793853 E88
    149: 1 1 is prime :  3.50324616081 E103
    181: 1 1 is prime :  8.15663058500 E125
    - - - - - - - - - - - - - - - - - - - - - - - 
    619: 1 1 is prime :  1.14913933997 E432  \\appended on second run.
    929: 1 1 is prime :  5.50901603963 E648  \\exact primetest-data t0 taken
   3407: 1 1 is prime :  6.14815462633 E2380 \\from [OEIS:A004061][2]
  10949: 1 1 is prime :  2.63340395660 E7652
========================
base 7
----------------
      5: 1 1 is prime :  2801
     13: 1 1 is prime :  16148168401
    131: 1 1 is prime :  8.50534611648 E109
    149: 1 1 is prime :  1.38502212710 E125
    - - - - - - - - - - - - - - - - - - - - - - - 
   1699: 1 1 is prime :  1.10514365306 E1435  \\appended on second run.
  14221: 1 1 is prime :  2.29654940319 E12017 \\exact primetest-data t0 taken
                                              \\from [OEIS:A004063][3]
========================
base 11
----------------
     17: 1 1 is prime :  5.05447028499 E16
     19: 1 1 is prime :  6.11590904484 E18
     73: 1 1 is prime :  1.05115319950 E75
    139: 1 1 is prime :  5.67000232522 E143
========================
base 13
----------------
      5: 1 1 is prime :  30941
      7: 1 1 is prime :  5229043
    137: 1 1 is prime :  3.39670648186 E151
    283: 1 1 is prime :  1.46820756272 E314
========================
base 17
----------------
      3: 1 0 is prime :  307
      5: 1 0 is prime :  88741
      7: 1 1 is prime :  25646167
     11: 1 1 is prime :  2141993519227
     47: 1 1 is prime :  4.23622795799 E56
     71: 1 1 is prime :  1.43798195172 E86
========================
base 19
----------------
     19: 1 1 is prime :  1.09912203092 E23
     31: 1 1 is prime :  2.43270318891 E38
     47: 1 1 is prime :  7.01692346601 E58
     59: 1 1 is prime :  1.55306613933 E74
     61: 1 1 is prime :  5.60656876297 E76
    107: 1 1 is prime :  3.72702921316 E135
    337: 1 1 is prime :  4.83828019876 E429
=========================

for base 3 see OEIS:A028491
for base 5 see OEIS:A004061
for base 7 see OEIS:A004063
Note: for high indexes the primetest in the OEIS-lists was often only probabilistic! See details at the OEIS-pages!

Gottfried Helms
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    I have not yet had a chance to understand this, but I did see that there were some very large primes and probable primes amongst them. It made me realise even more that my own test is only of any real use if it can be shown to be a proof of primality. – Alan Gee Sep 10 '17 at 21:50
1

The Lucas-Lehmer Test works for Mersenne numbers and for Fermat numbers. There are 4 proofs, at least, including mine. About your numbers, it will be very difficult to build a proof.

Tony Reix
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