OFFSET
1,1
COMMENTS
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, ... .
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).
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.
LINKS
FORMULA
EXAMPLE
Prime(1)=2 and the smallest semiprime of the form k^2+1 is a(1) = 3^2+1 = 10 = 2*5.
Prime(2)=3 and the smallest semiprime of the form k^3+1 is a(2) = 2^3+1 = 9 = 3*3.
MATHEMATICA
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Feb 04 2014
STATUS
approved