Arbitrarily Long Chains of Consecutive Almost-Primes with Prescribed Greatest Common Divisor

We derive an unconditional theorem on arbitrarily long blocks of consecutive almost-primes with prescribed greatest common divisor. Let \(Ω(n)\) denote the number of prime factors of \(n\), counted with multiplicity, and let \( A_r=\{n≥2:Ω(n)=r\}\), ordered as \(Ar,1<Ar,2<⋯\). For every integer \(g≥1\) and every \(m≥2\), put \(r=Ω(g)+1\). We prove that infinitely many blocks of \(m\) consecutive elements satisfy \((Ar,k+i,Ar,k+j)=g\) for all \(0≤ i<j<m\), together with the bounded-span inequality \(Ar,k+m-1-Ar,k≤ gH_m,\) where \(H_m\) is a finite constant depending only on \(m\). In particular, every prescribed prime occurs as the pairwise greatest common divisor of arbitrarily long blocks of consecutive ordinary semiprimes. The proof combines the Maynard--Tao theorem for admissible prime-producing linear forms with a Chinese remainder theorem construction that eliminates all unwanted intermediate almost-primes. This result is an unconditional consequence of established prime-tuple theory. It does not prove the twin-prime conjecture, and independent novelty is not claimed. As an explicit example, consider the three consecutive semiprimes \(3903=3·1301\), \(3909=3·1303\), and \(3921=3·1307\), where \(1301\), \(1303\), and \(1307\) are all prime. A direct factorization check confirms that no other integer between \(3903\) and \(3921\) is semiprime. Thus these three numbers are consecutive in the complete semiprime sequence. Their pairwise greatest common divisors satisfy \((3903,3909)=(3909,3921)=(3903,3921)=3.\) Furthermore, their normalized gaps are \(3909-3903/3=2,3921-3909/3=4,\) with total normalized span \(3921-3903/3=6.\) This example illustrates the theorem for \(p=3\) and \(m=3\). The general result guarantees infinitely many such consecutive semiprime chains for every prescribed prime \(p\) and every finite chain length \(m≥2\).