A Pierpont number is a number of the form (2^k)(3^l)+1
A Pierpont prime is simply a prime Pierpont number.
A113434 is defined as the set of products of two Pierpont primes which are Pierpont numbers, and begins {4, 9, 10, 25, 49, 65, 289}.
We wish to prove that this list is complete.
Let N = pq be a member of this sequence, and let a and b be p-1 and q-1, respectively.
We split this proof into two cases:

Case 1: At least one of a or b (w.l.o.g. b) equals 1
	We have then that N-1 = 2a + 1 is 3-smooth.
	Trivially, if a is 3-smooth then 2a is, so the solutions to this are the numbers k such that 2k and 2k+1 are 3-smooth: 1 and 4.
	This gives N = 2*2 = 4 and N = 2*5 = 10.

Case 2: a and b are greater than 1
	Claim 1: One of a or b (w.l.o.g. a) divides the other.
	Proof: Assume otherwise, and let g = GCD(a,b).
	We can express a and b as g*2^k and g*3^l for some k and l, both nonzero
	We have that N-1 = g(2^k)(3^l) + 2^k + 3^l, but the first term is a multiple of 6 and the remainder is coprime to 6, so this number cannot be 3-smooth.
	Therefore, one of them divides the other. Q.E.D.
	Say b = ka, and so N-1 = k*a^2 + (k+1)*a.
	a is 3-smooth by assumption, so we can divide it out, and we are looking for a and k such that ka + (k+1) is 3-smooth.
	Remarks: k must be 3-smooth, since b = ka, and ka >= k+1 aince a > 1.
	Claim 2: k+1 is also 3-smooth.
	Proof: Assume k+1 is not 3-smooth. Then, k and a must be either powers of 2 or powers of 3, since otherwise ka + (k+1) would be a sum of a multiple of 6
		and a number prime to 6.
	We have two subcases:
		i) 2^x = 3^y + 3^z + 1. This can never happen, because x > 0 and the right-hand side is always odd.
		ii) 3^x = 2^y + 2^z + 1. This is only possible if x = 4 [citation needed], since y > z, giving (a,k,N) = (4,16,325),
				a Pierpont number but not a semiprime. Q.E.D.
	Because of this, k is restricted to the numbers m such that m and m+1 are both 3-smooth: 1, 2, 3, and 8.
	Claim 3: k+1 divides ka.
	Assume that ka is not divisible by k+1.
	Working as in the proof of Lemma 1 and using Lemma 2, we can express ka + (k+1) as g(2^k + 3^l), for some k and l both nonzero.
	Since the right-hand side is coprime to 6, this number cannot be 3-smooth. Q.E.D.
	Let ka = d(k+1).
	We must have that d and d+1 are both 3-smooth since ka = d(k+1) and N-1 = a(d+1)(k+1) are both 3-smooth, restricting d to {1,2,3,8}.
	Solving each choice of k and d in turn, we obtain (k,d,a,b,N) = (1,1,2,2,9), (1,2,4,4,25), (1,3,6,6,49), (1,8,16,16,289), (2,2,3,6,28)*, (2,8,12,24,325)*,
		(3,3,4,12,65), (8,8,9,72,730).

The only semiprimes among these N are exactly the known terms, and therefore A113434 is complete.