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Least m such that (p^(2*m+1)-1)/(p-1) is a prime, where p = prime(n).
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%I #13 Jun 17 2018 02:01:48

%S 1,1,1,2,8,2,1,9,2,2,3,6,1,2,63,5,1,3,9,1,2,2,2,1,8,1,9,8,8,11,2,1,5,

%T 81,3,6,8,3,1,1,9,8,8,2,15,288,20,119,2,5,56,2,8,3,11,2

%N Least m such that (p^(2*m+1)-1)/(p-1) is a prime, where p = prime(n).

%H Andy Steward, <a href="http://www.users.globalnet.co.uk/~aads/primes.html">Prime Generalized Repunits</a>

%e a(5) = 8 because ithprime(5) = 11, (11^(2*m+1)-1)/10 is not a prime for m = 1..7 and (11^17-1)/10 = 50544702849929377 is a prime.

%t Do[p=Prime[w]; s=DivisorSigma[1, (p^r)^2]; z=DivisorSigma[0, (p^r)^2]; If[PrimeQ[s], Print[{p, r, p^r, s, z}]], {w, 1, 100}, {r, 1, 100}] For w=12, this prints out first {37, 6, 2565726409, 6765811783780036261, 13}.

%t lm[n_]:=Module[{m=1},While[!PrimeQ[(n^(2m+1)-1)/(n-1)],m++];m]; lm/@Prime[ Range[ 56]] (* _Harvey P. Dale_, Feb 16 2014 *)

%o (PARI) { allocatemem(932245000); for (n=1, 100, x=prime(n); s=x^2; q=x - 1; m=1; while (!isprime(((x*=s) - 1)/q), m++); write("b065813.txt", n, " ", m) ) } \\ _Harry J. Smith_, Oct 31 2009

%Y Cf. A000005, A000203, A000040, A065403-A065405, A000043, A001348, A065854.

%K hard,nonn

%O 1,4

%A _Vladeta Jovovic_ and _Labos Elemer_, Nov 13 2001