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Question

Numbers $n$ of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. For example:

  • $k=1$ then $n=10$.
  • $k=2$ then $n=31$.
  • $k=3$ then $n=73$.
  • $k=4$ then $n=157.$

Conjecture:

the number $(2^k-1)\cdot 10^m+2^{k-1}-1$ where $m$ is the number of decimal digits of $2^{k-1}$ is never prime when it is of the form $7s+6$, that is when it is congruent to $6$ $\pmod 7$. Examples: $n=1023511$ ($k=10$)$\equiv 6 \pmod 7$ and thus it is composite $(1023511=19\times103\times523)$, $n=20471023$ ($k=11$) $\equiv 6 \pmod 7$ and thus it is composite ($20471023=479\times42737)$. With PFGW we arrived to $k=565000$ and all the $n's$ congruent to $6 \pmod 7$ are composite. According to Giovanni Resta's calculations in a post which has been canceled, there should be no probable prime congruent to 6 $\pmod 7$ upto k=800.000. The residue $6$ $\pmod 7$ occurs when either $m=6t+3$ and $k=3l+1$ or $m=6t+4$ and $k=3l+2$ with $k$ and $l$ some non-negative integers, but amazingly when it occurs the number is not prime. Can you find a counter-example or give a proof for the conjecture? Here a link to other interesting questions: Is there a number of the form $f(n)=7k+6=5p$ with prime p? and Why do all residues occur in this similar sequence? For primes of this form see: The on-line Encyclopedia of integer sequences The following vector contains all the exponents k<=366800 leading to a prime

$[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$

Exponent $541456$ leads to another probable prime with residue 5 mod 7 and 325990 digits, but it need not be the next in increasing order.

Remark: we found five-in-a-row probable primes with res 5 mod 7. Probable primes with residue 5 are now twice frequent than expected. Exponents of these primes seem to be NOT random at all. Another thing I noticed, i don't know if it has some importance: the exponents leading to a probable prime $215, 69660, 92020, 541456$ are multiples of $43$. I noticed that $\frac{215}{41}, \frac{69660}{41}, \frac{92020}{41}, \frac{541456}{41}$ all have a periodic decimal expansion equal to $\overline{24390}=29^3+1$. This is equivalent to say that when k is a multiple of 43 and the number $10^{m}(2^{k}−1)+2^{k-1}−1$ is prime, then k is of the form $41s+r$ where r is a number in the set (1,10,16,18,37). Is there some mathematical reason for that?

Enzo Creti
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2 Answers2

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According to your list, a counter-example, if it exists, must have more than $60,000$ digits. So, a counterexample would be a quite gigantic prime.

Unfortunately, a proof of the conjecture will almost certainly be out of reach.

The search for a counter-example can be painful as well, it is well possible that the smallest is already too big for current algorithms for primality testing.

editor XD
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Peter
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  • @EnzoCreti How did you get this list ? It shows that a counterexample must have more than $60\ 000$ digits. I think that the numbers of this form (no matter of the congruences) can have least prime factor very high. The probability that the least prime factor is very high is much higher than one might expect. A large "random" number has no prime factor less than $10^{20}$ with probability about $1.2$%. I do not think there is a special reason. Moreover, my list of candidates has become obsolete. – Peter Feb 05 '18 at 11:41
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    @EnzoCreti No, this is only known for arithmetic progressions of the form $an+b$ with coprime $a$ and $b$ (Dirichlet). The numbers here will behave like "random" numbers , hence we won't be able to rule out a large prime. – Peter Feb 05 '18 at 12:33
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    @Peter Every number is of the form $10^{m}(2^k-1)+2^{k-1}-1$, where $m$ is the number of decimal digits of $2^{k-1}$. As $\log_{2}10$ is irrational, the residues of $m$ and $k$ modulo 6, and hence the residues of $10^{m}$ and $2^{k-1}$ modulo 7, are independent. You can find out the distribution simply by tabulating the possible cases. – Taneli Huuskonen Feb 06 '18 at 16:03
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    My observation just explains why different residues modulo 7 occur at different frequencies. – Taneli Huuskonen Feb 06 '18 at 19:34
  • @Taneli Huuskonen And can you extimate in what percentage residue 6 mod 7 should occur? – Enzo Creti Feb 06 '18 at 20:36
  • Yes. The residue 3 occurs 4/9 of the time; the residues 2, 4, 5, and 6 occur 1/9 of the time each; and the residues 0 and 1 occur 1/18 of the time each. The residue 6 occurs exactly when either $m=6n+3$ and $k=3l+1$, or $m=6n+4$ and $k=3l+2$, for some nonnegative integers $n$ and $l$. – Taneli Huuskonen Feb 06 '18 at 21:35
  • Let us continue this discussion in chat. – Enzo Creti Feb 07 '18 at 12:06
  • @EnzoCreti Not really realistic. Sometimes, we only have a very small prime factor and a prime cofactor and sometimes the smallest prime factor has more than $15$ digits. This does not indicate hidden factors. So, it seems to be a pure coincidence that $6$ does not appear early. If factordb would allow concatenating numbers, that would be fine! But it seems that factordb does not support this yet. I have passed $120\ 000$ now without success. – Peter Feb 11 '18 at 13:40
  • Let us continue this discussion in chat. – Enzo Creti Feb 11 '18 at 17:59
  • Primes are of limited importance. They are used for cryptography, but there are better methods and the safety of the RSA is debatable. Such primes as the one you search are purely for curiousity. Nevertheless, many such primes were calculated. It is a bit sad that this kind of primes is not considered to be interesting by most of the people here. Of course, special primes, as the Mersenne primes, are more important because they answer important mathematical questions. – Peter Feb 14 '18 at 09:54
  • Let us continue this discussion in chat. – Enzo Creti Feb 14 '18 at 12:38
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    https://math.stackexchange.com/questions/2658464/is-frac-2716-915-680-41713-a-wagstaff-prime#comment5492093_2658464 – Peter Feb 20 '18 at 18:51
  • @Peter Yes I have noticed. – Enzo Creti Feb 21 '18 at 09:50
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    For $k=32$, we have the factorization $$131\cdot 4463\cdot 21601\cdot 44623 \cdot 76213$$ This is a counterexample with residue $6$ – Peter Feb 21 '18 at 10:57
  • For $k=586916$, we get a number of the form $7k+6$ divisible by $83339$ – Peter Feb 21 '18 at 11:13
  • Let us continue this discussion in chat. – Enzo Creti Feb 21 '18 at 11:29
  • New project : https://math.stackexchange.com/questions/2666033/is-there-a-number-of-the-form-fn-7k6-5p-with-prime-p – Peter Feb 25 '18 at 14:09
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    @EnzoCreti Concerning the structure of the factors of those numbers, I search a prime factor of $f(530)$ and in a new project I search $f(n)$ of the form $7k+6$ such that $\frac{f(n)}{5}$ is prime which turns out to be tough as well. – Peter Feb 25 '18 at 14:52
  • @Peter ok I wait news from you. – Enzo Creti Feb 25 '18 at 15:17
  • @EnzoCreti To show that the smallest prime factor can be very large. – Peter Feb 25 '18 at 15:53
  • @Peter have you find a factor for f(530)? – Enzo Creti Feb 26 '18 at 18:24
  • @EnzoCreti No, seems that the smallest prime factor has more than $25$ , probably even more than $30$ digits. Of course, trial division does not work anymore to find such factors. The elliptic curve method is the key. With factordb , the concatenation for numbers works, if the numbers are concrete or depend linear on $n$ , for example $c(123,456)$ gives $123456$ and $c(n,n+2)$ correctly concatenates $n$ and $n+2$. Strangely, $c(M(n+1),M(n))$ does NOT cancatenate $2^{n+1}-1$ and $2^n-1$ correctly, otherwise this would be the perfect way to search further primes. – Peter Feb 26 '18 at 21:17
  • For clarification, $M(n)=2^n-1$ , the $n-th$ Mersenne number is a valid syntax in factordb. I also tried "c(2^(n+1)-1,2^n-1)" , but without success either. – Peter Feb 26 '18 at 21:21
  • https://math.stackexchange.com/questions/2668139/concatenation-with-factordb – Peter Feb 26 '18 at 21:55
  • My post have 870 views but only 13 votes...so I think that there is skepticism about that conjecture or nobody believes it has some importance... – Enzo Creti Feb 27 '18 at 19:08
  • @Peter between 10^5 and 2*10^5 there are about 2000 candidates (or even less)...if the probability of a random number to be prime is 1 out of 4200, perhaps it will be still difficult to find a prime? – Enzo Creti Mar 01 '18 at 10:46
  • @EnzoCreti The logarithm is roughly $n\cdot \ln(4)$ , hence the probability that a candidate is prime is about $1:\frac{e^{-\gamma}}{\ln(10^6)}\cdot \ln(4)\cdot n\approx 1:0.0563\cdot n$ – Peter Mar 01 '18 at 12:28
  • A large number has no prime factor less than $x$ with probability about $\frac{e^{-\gamma}}{\ln(x)}$, if $x$ is large. The approximation is already good for $x=10^6$ – Peter Mar 01 '18 at 12:36
  • Didn't continue, probably no factor below $10^{30}$ – Peter Mar 02 '18 at 13:54
  • https://math.stackexchange.com/questions/2674050/why-do-all-residues-occur-in-this-similar-sequence – Peter Mar 02 '18 at 18:54
  • Most of the conjectures on primes hold good only in a certain interval. – Vishaal Selvaraj Mar 05 '18 at 06:54
  • I know that, I myself am astonished with Legendre's conjecture that there exists a prime between every square number and it is true and has been put in landau's problems. – Vishaal Selvaraj Mar 05 '18 at 07:02
  • @Peter Could we start towards k=300.000? – Enzo Creti Mar 05 '18 at 08:35
  • Let us continue this discussion in chat. – Enzo Creti Mar 05 '18 at 18:01
  • Enzo creti, your discovery is remarkable. I don't believe I am saying this, but I think you should patent your discoveries before uploading them. – Vishaal Selvaraj Mar 07 '18 at 07:13
  • I have asked my friends to siphon some computer power from cyber cafes to help you calculate. – Vishaal Selvaraj Mar 07 '18 at 07:14
  • If possible could you take me in your chat room? – Vishaal Selvaraj Mar 07 '18 at 07:18
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    @VishaalSelvaraj A nice observation, but I don't think that is considered to be of great importance. Otherwise we would already have found helpers. r.e.s was a helper, but quit at $10^5$, now we are only two despite of 1k+ views. And without some luck or more helpers, we won't be able to decide the conjecture. – Peter Mar 07 '18 at 11:45
  • @Peter this morning I arrived at 248.000, the search has slowed down. I was hoping that on Saturday morning we would have arrived at k=300.000 but now I think that this is not the case... – Enzo Creti Mar 08 '18 at 07:20
  • @Vishaal Selvaraj you told me about a library...you said that it is necessary high power of the computer. How much do you believe? – Enzo Creti Mar 08 '18 at 14:29
  • 1.8 GHz give or take, but 4 GB RAM IS recommended – Vishaal Selvaraj Mar 08 '18 at 14:59
  • Actually the library uses custom data types with user defined precision.. – Vishaal Selvaraj Mar 08 '18 at 15:00
  • @EnzoCreti It is not a matter of space, it is a matter of computational power and of parallelization. – Peter Mar 08 '18 at 15:00
  • @Vishaal Selvaraj Ok thank you very much, I have half a mind to buy a new computer (mine at home is broken I'm using the computer office). So I could help also Peter... – Enzo Creti Mar 08 '18 at 15:03
  • @Peter f(n)/5 conjecture is another good question... – Enzo Creti Mar 08 '18 at 15:05
  • No , it is a matter of memory space too since, the data types are 100~200 bytes in size and running operations on them requires more ram. – Vishaal Selvaraj Mar 08 '18 at 15:05
  • @EnzoCreti I know no primes of the forms given in this question : https://math.stackexchange.com/questions/2374537/are-there-prime-numbers-cn-or-rcn?rq=1 – Peter Mar 08 '18 at 20:26
  • @EnzoCreti Great! – Peter Mar 08 '18 at 22:32
  • @EnzoCreti You passed the $150k$-digit-mark. The largest known prime is only roughly $150$ times larger than the numbers you now verify. – Peter Mar 09 '18 at 08:02
  • Let us continue this discussion in chat. – Enzo Creti Mar 09 '18 at 08:35
  • PARI/GP is slow with modulo-calculations involving very large numbers. With a better software we could sieve out more factors with pollard-rho, p-1-method, etc. So, perhaps you are looking for a powerful program that can do this. Or, alternatively, you are content with the fact that counterexamples will be very rare, but likely exist. What I finally can make : Estimate the number of primes in very large ranges based on the weak candidates (no prime factor less than $100$) – Peter Mar 14 '18 at 08:24
  • Surprisingly, the expected number of primes in the ranges $[10^k,10^{k+1}]$ for $k=3,4,5,6$ is about $5.4$. Whether this pattern continues ? – Peter Mar 14 '18 at 08:50
  • In fact , the pattern should continue. Since $\ln(f(n))\approx \ln(4^n)$ for large $k$, the expected number of primes in the range $[10^k,10^{k+1}]$ is approximately $5$, if we only consider that $2$ and $3$ can never be a prime factor. The special structure of the numbers seems to slightly increase this number to about $5.4$. So, we probably have infinite many counter-examples. – Peter Mar 14 '18 at 09:34
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    You should at least present all primes upto $n=10^5$ – Peter Mar 14 '18 at 10:01
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    Let us continue this discussion in chat. – Peter Mar 14 '18 at 10:07
  • @EnzoCreti When do you have your internet at home ? – Peter Mar 15 '18 at 12:22
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    @Peter today at 16.00 – Enzo Creti Mar 15 '18 at 12:43
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    @Peter the sum of the reciprocals of the ec-primes is about $0.052408...$ Curiously $0.05240...$ is about equal to $\frac{360}{\sqrt{360*2^{17}}}$ – Enzo Creti Aug 27 '18 at 14:27
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@peter told me about this and i found it very interesting.
I made this tool, so anybody can help computing. Simply download and run to participate. Just refresh the page to update stats.

With 30-40 person, getting to $n=10^7$ should not be to long.

François Huppé
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