%I #36 Feb 08 2014 18:01:40
%S 10,9,33,129,2049,8193,131073,524289,8388609,
%T 21214052113249267732127817825945098816023915043832462900000000000000000000000000001,
%U 2147483649,356811923176489970264571492362373784095686657,1821119122882338858450163704901509732674059569636703920027007853793548503164173361298060584748698304513
%N Smallest semiprime of the form k^prime(n)+1, or 0 if no such semiprime exists.
%C For n > 1, smallest number k^p+1 with both (k^p+1)/(k+1) and k+1 prime, where p = prime(n); the corresponding primes (k^p+1)/(k+1) for n > 1 are A237116(n) = 3, 11, 43, 683, 2731, 43691, 174763, 2796203, ... and the corresponding primes k+1 are A237115(n) = 3, 3, 3, 3, 3, 3, 3, 3, 691, 3, 17, ... .
%C a(n) == its smaller prime factor A237115(n) (mod prime(n)). Proof: 10 == 2 (mod 2), so true for n=1. For n>1, true by Fermat's little theorem: k^p+1 == k+1 (mod p).
%C a(n) is in A006881 (squarefree semiprimes), except for a(2) = 9 = 3^2. Proof: True for n=1. For n>1, if k^p+1 = (k+1)^2, then k^(p-1) = k+2, so k*(k^(p-2)-1) = 2. Now k>1 implies k=2 and p=3, so that n=2.
%C It appears that a(n) mod p > 0 for all n > 2 (see A237117), where p = prime(n). If true, then the larger prime factor A237116(n) of a(n) is == 1 (mod p), since a(n) == its smaller prime factor (mod p).
%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) = A237115(n)*A237116(n), for n > 0.
%F a(n) = (A237115(n)-1)^prime(n)+1, for n > 1.
%F a(n) == A237115(n) (mod prime(n)), for n > 0.
%F a(n) mod prime(n) = A237117(n), if a(n) > 0.
%e Prime(1)=2 and the smallest semiprime of the form k^2+1 is a(1) = 3^2+1 = 10 = 2*5.
%e Prime(2)=3 and the smallest semiprime of the form k^3+1 is a(2) = 2^3+1 = 9 = 3*3.
%t L = {10}; 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, q^p + 1], {k, 2, 12}]; L
%Y Cf. A001358, A006881, A103795, A123627, A123628, A237040, A237115, A237116, A237117.
%K nonn
%O 1,1
%A _Jonathan Sondow_, Feb 04 2014