[seqfan] Re: Repunit wrapped primes

[Hans L]

Hi,

For p=1823, I have k=9565 as probable prime using BPSW.

Hans Loeblich

On Sat, Apr 27, 2019 at 3:26 AM David Sycamore via SeqFan <
seqfan at list.seqfan.eu> wrote:

> Dear Seqfans,
> Let kpk be the number obtained by placing k leading and trailing 1s around
> prime p. Thus, for p=7 and k=1 we get the “repunit wrapped” prime 171,
> which is composite (3*57)
>
> With k=2 we get 11711; also composite (7*1673). However, k=3 gives kpk =
> 1117111, which is prime, and 3 is the smallest k >0 for which this is the
> case.
>
> For many primes p we can find a k>0 such that kpk is prime, and define a
> sequence where a(n) is the least k such that kprime(n)k is prime, or -1 if
> no such k exists. The sequence is easy to compute in cases where a k exists
> within reach of modest computational means, but there are cases where it is
> not clear that such a k exists or not. The  -1 terms are the challenge here.
>
> Examples where a(n)=-1:
>
> p=2: k2k is composite for all k>0 because it is divisible by the (k+1)-th
> repunit. Ditto for p=101.
>
> p=11: k11k is always divisible by 11 for all k>0.
>
> p=37: k37k is composite for all k>0 since its factor cycle comprises a
> covering congruence for k in which each residue class is associated with a
> particular prime divisor (3, 13 or 37). Therefore any choice of k >0 leads
> to a divisor.
>
> The trouble starts when p= 397, 563, 739, 1249, 1823...
>
> In these cases the factor pattern reveals a specific prime divisor for all
> but one single residue class of the covering congruence of k. For p=397 we
> find that k== 1(mod 3)—> 3/kpk, k== 2(mod 3)—>37/kpk, k==3(mod 6)—>7/kpk.
> But for the remaining k==6 (mod 6) there is always (k<=30000) a prime
> divisor of k397k,  but it’s never the same one, so there is apparently no
> possibility to predict a divisor for any k divisible by 6.  Therefore all
> we can say is that if there is a k that makes k397k prime, then it must be
> a multiple of 6 and >30000. Alternatively a proof is needed that for all
> k==6 (mod 6) k397k is composite.
>
> To make matters worse,  there are cases like the above (a residue class
> with no distinct prime divisor associated with it) for which we do
> eventually reach a k such that  kpk becomes prime. An example of this is
> p=61, for which k=repunit of length 14 gives k61k prime, and 14 belongs to
> k=={0,2,5} (mod 6) which has no fixed prime divisor associated with it,
> whereas the other two, k==1 (mod2) and k== {3,4} (mod 6) are associated
> with prime divisors 3 and 13 respectively. There are plenty of other
> examples like this one.
>
> It’s a bit like the de Polignac problem, and whilst the use of covering
> congruences is promising, there are cases where It is hard to be sure for a
> given p, whether a k>0 exists or not.
>
> Can anyone offer suggestions to resolve this question for the troublesome
> primes mentioned above?
>
> I have entered draft sequence A306861 in an attempt to record  the least k
> for which kpk is prime,  and -1 if no such k exists. The current draft,
> including data up to n=86 includes prime(78)=397, for which an uncertain
> (but perhaps highly likely) term -1 has been recorded.
> The other four “-1” terms in the present data (p=2,11,37,101) can be
> confirmed as explained above. Assuming the sequence is accepted there will
> be a need to confirm values for n> 86, which would include the
> indeterminate cases mentioned above (eventually for a b-file).
>
> Included in the Comments of draft A306861 is a conjecture that there are
> an infinite number of -1 terms (primes p for which kpk is composite for all
> k>0).  Can anyone suggest a proof, or argue that the list of such primes is
> finite?
>
> I would like to improve draft A306861 and submit it for review sometime
> soon.  Any assistance gratefully received.
>
> Best regards
>
> David.
>
>
>
>
>
>
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