A237115 - OEIS

%I #17 Feb 07 2014 10:15:58

%S 2,3,3,3,3,3,3,3,3,691,3,17,313,3,7,11,7,3,11,47,19,3,1499,17,71,3,97,

%T 7,13,823,3,97,1163,31,17,199,1907,53,3,17,1231,1013,3,13,53,3,67,47,

%U 23,1013,787,127,347,17,37,97,683,631,73,4549,173,11,17,1039,3,17,47,6389,3,461,23,673,37,29,331,7451,1433,4561

%N Lesser 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 p such that ((p-1)^prime(n)+1)/p is prime; the corresponding primes ((p-1)^prime(n)+1)/p are A237116(n) = 3, 11, 43, 683, 2731, 43691, 174763, 2796203, ... and the corresponding semiprimes (p-1)^prime(n)+1 are A237114(n) = 9, 33, 129, 2049, 8193, 131073, 524289, 8388609, ... .

%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)/A237116(n), for n > 0.

%F (a(n)-1)^prime(n) = A237114(n)-1, for n > 1.

%F a(n) == A237114(n) (mod prime(n)) (for a proof, see A237114).

%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 3^2+1 = 10 = 2*5, so a(1) = 2.

%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 = {2}; 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 + 1], {k, 2, 78}]; L

%Y Cf. A001358, A103795, A123627, A123628, A237040, A237114, A237116, A237117.

%K nonn

%O 1,1

%A _Jonathan Sondow_, Feb 04 2014