A237114 Smallest semiprime of the form k^prime(n)+1, or 0 if no such semiprime exists.

10, 9, 33, 129, 2049, 8193, 131073, 524289, 8388609, 21214052113249267732127817825945098816023915043832462900000000000000000000000000001, 2147483649, 356811923176489970264571492362373784095686657, 1821119122882338858450163704901509732674059569636703920027007853793548503164173361298060584748698304513

Offset: 1

Author

Jonathan Sondow, Feb 04 2014

Keywords

• nonn

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.

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).

Examples

			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.
		

Links

• Eric Weisstein's World of Mathematics, Semiprime
• Wikipedia, Semiprime

Crossrefs

Cf. A001358, A006881, A103795, A123627, A123628, A237040, A237115, A237116, A237117.

Programs

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

Formula

a(n) = A237115(n)*A237116(n), for n > 0.

a(n) = (A237115(n)-1)^prime(n)+1, for n > 1.

a(n) == A237115(n) (mod prime(n)), for n > 0.

a(n) mod prime(n) = A237117(n), if a(n) > 0.