A113434 Semi-Pierpont semiprimes which are also Pierpont semiprimes.

4, 9, 10, 25, 49, 65, 289

Offset: 1

Author

Jonathan Vos Post, Nov 01 2005

Keywords

• nonn
• fini
• full

Comments

Semiprimes both of whose prime factors are Pierpont primes (A005109), which are primes of the form (2^K)*(3^L)+1 and where the semiprime is itself of the form (2^K)*(3^L)+1.

No more under 10^50; what is the next element of this sequence?

No more terms <= 10^100. - Robert Israel, Mar 10 2017

This sequence is complete, see Links. - Charlie Neder, Feb 04 2019

Examples

			a(1) = 4 = 2^2 = [(2^0)*(3^0)+1]*[(2^0)*(3^0)+1] = (2^0)*(3^1)+1.
a(2) = 9 = 3^2 = [(2^1)*(3^0)+1]*[(2^1)*(3^0)+1] = (2^3)*(3^0)+1.
a(3) = 10 = 2*5 = [(2^0)*(3^0)+1]*[(2^2)*(3^0)+1] = (2^0)*(3^2)+1.
a(4) = 25 = 5^2 = [(2^2)*(3^0)+1]*[(2^2)*(3^0)+1] = (2^3)*(3^1)+1.
a(5) = 49 = 7^2 = [(2^1)*(3^1)+1]*[(2^1)*(3^1)+1] = (2^4)*(3^1)+1.
a(6) = 65 = 5*13 = [(2^2)*(3^0)+1]*[(2^2)*(3^1)+1] = (2^6)*(3^0)+1.
a(7) = 289 = 17^2 = [(2^4)*(3^0)+1]*[(2^4)*(3^0)+1] = (2^5)*(3^2)+1.
		

Links

• Chris Caldwell, "Pierpont primes." primeform posting, Oct 25, 2005.
• Chris Caldwell, "Pierpont primes." primeform posting, Oct 25, 2005. [Cached copy]
• Charlie Neder, Proof of the completeness of this sequence
• Eric Weisstein's World of Mathematics, Pierpont Prime
• Eric Weisstein's World of Mathematics, Semiprime

Crossrefs

Cf. A001358, A003586, A005109, A055600, A111153, A111206, A113432, A113433.

Programs

Maple

N:= 10^100: # to get all terms <= N
PP:= select(isprime, {seq(seq(1+2^i*3^j, i=0..ilog2((N-1)/3^j)),j=0..floor(log[3](N-1)))}):
SP:= select(t -> t <= N and t = 1+2^padic:-ordp(t-1,2)*3^padic:-ordp(t-1,3), [seq(seq(PP[i]*PP[j], j=1..i),i=1..nops(PP))]):
sort(convert(SP,list)); # Robert Israel, Mar 10 2017

Formula

{a(n)} = intersection of A113432 and A113433. {a(n)} = Semiprimes A001358 of the form (2^K)*(3^L)+1 both of whose factors are of the form (2^K)*(3^L)+1. {a(n)} = {integers P such that, for nonnegative integers I, J, K, L, m, n there is a solution to (2^I)*(3^J)+1 = [(2^K)*(3^L)+1]*[(2^m)*(3^n)+1] where both [(2^K)*(3^L)+1] and [(2^m)*(3^n)+1] are prime}.