I'd guess that there are infinitely many pairs $(p,p+2)$ of twin primes such that $p+1$ has four prime factors (or at least that this can't be ruled out even on heuristic grounds) but that sometimes there are more than four. Hence that the answer is no.

There is no proof that there are infinitely many twin primes, but there is every reason to expect that there are and even that the $k$th pair grows something like $k \log^2(k)$. Certainly for such a pair $p+1$ will be a multiple of $6$ . Since the sum of the reciprocals of the twin primes converges I think it highly unlikely that $p+1$ has only three prime factors infinitely often. I don't have an analysis for four prime factors but that feels right.

I checked the last 100 out of 100000 twin prime pairs from the OEIS and found the number of prime factors to be $[[3, 7], [4, 48], [5, 34], [6, 9], [7, 2]]$.

The last two cases are $18408750=2\cdot 3 \cdot 5^4 \cdot 4904$ and $18408990=2 \cdot 3 \cdot 5 \cdot 613633.$

As I was typing this it was posted that it follows from a reasonable conjecture that three happens infinitely often. I still like having these examples around $18,000,000.$

I'd guess that there are infinitely many pairs $(p,p+2)$ of twin primes such that $p+1$ has four prime factors (or at least that this can't be ruled out even on heuristic grounds) but that sometimes there are more than four.

There is no proof that there are infinitely many twin primes, but there is every reason to expect that there are and even that the $k$th pair grows something like $k \log^2(k)$. Certainly for such a pair $p+1$ will be a multiple of $6$ .

I checked the last 100 out of 100000 twin prime pairs from the OEIS and found the number of prime factors to be $[[3, 7], [4, 48], [5, 34], [6, 9], [7, 2]]$.

The last two cases are $18408750=2\cdot 3 \cdot 5^4 \cdot 4904$ and $18408990=2 \cdot 3 \cdot 5 \cdot 613633.$

As I was typing this it was posted that it follows from a reasonable conjecture that three happens infinitely often. I still like having these examples around $18,000,000.$

I'd guess that there are infinitely many pairs $(p,p+2)$ of twin primes such that $p+1$ has four prime factors (or at least that this can't be ruled out even on heuristic grounds). Hence that the answer is no.

There is no proof that there are infinitely many twin primes, but there is every reason to expect that there are and even that the $k$th pair grows something like $k \log^2(k)$. Certainly for such a pair $p+1$ will be a multiple of $6$ . Since the sum of the reciprocals of the twin primes converges I think it highly unlikely that $p+1$ has only three prime factors infinitely often. I don't have an analysis for four prime factors but that feels right.

I checked the last 100 out of 100000 twin prime pairs from the OEIS and found the number of prime factors to be $[[3, 7], [4, 48], [5, 34], [6, 9], [7, 2]]$.

The last two cases are $2\cdot 3 \cdot 5^4 \cdot 4904$ and $2 \cdot 3 \cdot 5 \cdot 613633.$

I'd guess that there are infinitely many pairs $(p,p+2)$ of twin primes such that $p+1$ has four prime factors (or at least that this can't be ruled out even on heuristic grounds) but that sometimes there are more than four. Hence that the answer is no.

There is no proof that there are infinitely many twin primes, but there is every reason to expect that there are and even that the $k$th pair grows something like $k \log^2(k)$. Certainly for such a pair $p+1$ will be a multiple of $6$ . Since the sum of the reciprocals of the twin primes converges I think it highly unlikely that $p+1$ has only three prime factors infinitely often. I don't have an analysis for four prime factors but that feels right.

I checked the last 100 out of 100000 twin prime pairs from the OEIS and found the number of prime factors to be $[[3, 7], [4, 48], [5, 34], [6, 9], [7, 2]]$.

The last two cases are $18408750=2\cdot 3 \cdot 5^4 \cdot 4904$ and $18408990=2 \cdot 3 \cdot 5 \cdot 613633.$

As I was typing this it was posted that it follows from a reasonable conjecture that three happens infinitely often. I still like having these examples around $18,000,000.$