[seqfan] Repunit wrapped primes

[David Sycamore]

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.