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Here :

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}$.

the numbers $$f(n):=(2^n-1)\cdot 10^d+2^{n-1}-1$$ are introduced , where $d$ denotes the number of digits of $2^{n-1}-1$ in the decimal expansion. So, we simply concatenate two neighboured Mersenne-numbers, for example $f(10)=1023511$. I know no prime $f(n)$ of the form $7k+6$.

I also did not find a number $f(n)$ yet which is of the form $7k+6=5p$ with a prime $p$, so a number $f(n)$ with $f(n)\equiv 6\mod 7$ and $\frac{f(n)}{5}$ is a prime number.

With PFGW, I passed $n=122\ 000$ without finding such a number. Does such a number exist ?

Peter
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  • I believe that maybe there was a typo in your code, the first counterexample is $5\cdot 11=7\cdot 7+6$. I've deleted the comment below, this was a bad counterexaple. –  Feb 25 '18 at 14:41
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    @user243301 This is not the concatenation of two neigboured Mersenne-numbers as described in the question. Look at the linkes question to see which kind of numbers are considered here. – Peter Feb 25 '18 at 14:43
  • @Peter I wonder if these are all coincidences...??? – Enzo Creti Feb 27 '18 at 06:34
  • @EnzoCreti Perhaps in this special case, it can be proven that there must be another divisor. At least , $f(190)$ has the prime factors $19$ and a relatively large prime cofactor. Maybe I find further examples. – Peter Feb 27 '18 at 10:07
  • @Peter I'm going to take a walk around Lugano's lake, perhaps fresh air of Switzerland, home of Euler and Bernoulli, will give me some hint...I think that a proof would be quite difficult to find... – Enzo Creti Feb 27 '18 at 10:17
  • $\frac{f(2062)}{29}$ is prime and has $1241$ digits – Peter Feb 27 '18 at 10:20
  • Let us continue this discussion in chat. – Enzo Creti Feb 27 '18 at 10:24
  • This shows that we have numbers having a very small prime factor and a very large prime cofactor. This does not rule out "hidden forced factors" , but it makes it less likely. – Peter Feb 27 '18 at 10:24
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    @Peter I don't know...humanity is far away from understanding the distribution and the patterns of primes...maybe a theory still to come will explain all these things one day... – Enzo Creti Feb 27 '18 at 10:34
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    I finished the search upto $10^5$ without success. – Peter Feb 27 '18 at 11:41
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    @Peter mistery is getting thicker – Enzo Creti Feb 27 '18 at 11:48
  • Someone call Andrew Wiles...perhaps he has a solution... – Enzo Creti Feb 27 '18 at 13:11
  • @Peter are that numbers congruent to 6 (mod 7) also squarefree? – Enzo Creti Feb 28 '18 at 08:53
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    @Peter tomorrow I will do, for this month I have spent all my 50 questions. – Enzo Creti Feb 28 '18 at 09:00
  • @EnzoCreti Not necessary, $n=1981$ gives a number divisible by $25$ and congruent $6$ modulo $7$ – Peter Feb 28 '18 at 09:10
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    @Peter Ok conjecture failed! – Enzo Creti Feb 28 '18 at 09:12
  • @EnzoCreti Surely, Andrew Wiles is a genius! But I don't think that he is able to solve such problems. Note that Goldbach's conjecture is still open, and we don't know whether $n^2+1$ produces infinite many primes. Until we cannot solve such problems, we cannot expect to be able to deal with the questions like this one. Of course, I cannot rule out a proof, but the usual approaches like algebraic or aurifeuillan factors don't seem to apply. Since I haven't found a factor of $f(530)$ yet, I don't think that we can solve it without computer-help. – Peter Feb 28 '18 at 09:24
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    @Peter Infact Erdos said that when it comes to primes, even a child could ask questions such that neither the most brilliant minds in the world could answer... – Enzo Creti Feb 28 '18 at 09:32
  • What about $f(7)=55=5(11)$? – Steven Alexis Gregory Mar 01 '18 at 04:45
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    @Steven gregory that is not a number of the required form. – Enzo Creti Mar 01 '18 at 07:46
  • @Peter I don't understand how you are able to check up to 150000 my conjecture in only few hours whereas for me with an even faster computer it takes days! – Enzo Creti Mar 01 '18 at 08:08
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    @EnzoCreti Presieving is the key! And here we only have to consider numbers being divisible by $5$, reducing the number of candidates to test with factor $4$. – Peter Mar 01 '18 at 09:28
  • And I did not reach 150k, only 100k. And for your conjecture, the range 100k to 140k took me very long as well. – Peter Mar 01 '18 at 09:31
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    @Peter we arrived to around 103229...tomorrow at 10.00/11.00 Greenwich mean time we should have finished! – Enzo Creti Mar 01 '18 at 09:52
  • @Peter I was wrong, the search is slower than I thought...this morning I just arrived to 153167...I think that it will take another day for finishing the job. The search has slowed down because numbers get bigger and bigger... – Enzo Creti Mar 02 '18 at 08:14
  • @EnzoCreti I told you that it will slow down linear to the magnitude of the numbers. Yes, if it finds a prime, it will display it. To be sure, you can check the pfgw-file in which probable primes are written automatically. It should be in the directory where pfgw is. – Peter Mar 02 '18 at 08:29
  • @Peter Where I found the pfgw-file? I only see the command line and the grey windows which don't display any prime – Enzo Creti Mar 02 '18 at 08:31
  • Anyway, if it doesn't display something with a probable $3-PRP$ (If this is still the standard base), it didn't have found a prime. – Peter Mar 02 '18 at 08:36
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    @Peter Yes I didnt find any 3-PRP and I checked also pfwg files...nothing...I will edit my post – Enzo Creti Mar 02 '18 at 08:37
  • At this stage, a candidate will be prime with probability $1:8\ 676$ and it will further decrease to $1:11\ 268$. Needless to say that it will get worse, if we continue. Perhaps, the special structure makes a prime somewhat more likely though. – Peter Mar 02 '18 at 08:47
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    Let us continue this discussion in chat. – Enzo Creti Mar 02 '18 at 08:50

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