%I #9 Feb 06 2014 03:22:48
%S 5,3,11,43,683,2731,43691,174763,2796203,
%T 30700509570548867919143006984001590182379037690061451374819102749638205499276411,
%U 715827883,20988936657440586486151264256610222593863921,5818271958090539483866337715340286685859615238455923067178938830011337070812055467405944360219483401
%N Larger prime factor of the smallest semiprime of the form k^prime(n)+1, or 0 if no such semiprime exists.
%C For n > 1, smallest prime of the form ((p-1)^prime(n)+1)/p, where p is prime; the corresponding primes p are A237115(n) = 3, 3, 3, 3, 3, 3, 3, 3, 691, 3, 17, ... and the corresponding semiprimes (p-1)^prime(n)+1 are A237114(n) = 9, 33, 129, 2049, 8193, 131073, 524289, 8388609, ... .
%C It appears that a(n) == 1 (mod prime(n)), for all n <> 2. See 4th comment in A237114.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Semiprime.html">Semiprime</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Semiprime">Semiprime</a>
%F a(n) = A237114(n)/A237115(n), for n > 0.
%F a(n) = ((A237115(n)-1)^prime(n)+1)/A237115(n), for n > 1.
%e Prime(1)=2 and the smallest semiprime of the form k^2+1 is 3^2+1 = 10 = 2*5, so a(1) = 5.
%e Prime(2)=3 and the smallest semiprime of the form k^3+1 is 2^3+1 = 9 = 3*3, so a(2) = 3.
%t L = {5}; Do[p = Prime[k]; n = 1; q = Prime[n] - 1; cp = (q^p + 1)/(q + 1); While[! PrimeQ[cp], n = n + 1; q = Prime[n] - 1; cp = (q^p + 1)/(q + 1)]; L = Append[L, cp], {k, 2, 13}]; L
%Y Cf. A001358, A103795, A123627, A123628, A237040, A237114, A237115.
%K nonn
%O 1,1
%A _Jonathan Sondow_, Feb 05 2014