Most number of consecutive odd primes in a Collatz sequence

Question

The Collatz function applied to a number $x$ gives $3x+1$ when $x$ is odd and $x/2$ when $x$ is even. Consider the following Collatz sequence:

97760291, 293280874, 146640437, 439921312, 219960656, 109980328, 54990164, 27495082, 13747541, 41242624, 20621312, 10310656, 5155328, 2577664, 1288832, 644416, 322208, 161104, 80552, 40276, 20138, 10069, 30208, 15104, 7552, 3776, 1888, 944, 472, 236, 118, 59, 178, 89, 268, 134, 67, 202, 101, 304, 152, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1

If we ignore even numbers then this sequence contains 14 consecutive odd primes (highlighted in bold). What is the most number of consecutive odd primes that can be found in a Collatz sequence?

This question was inspired by this one: Prime numbers in Collatz sequences

Answer 1

Your question is equivalent to find prime numbers : $$p,\frac{3p+1}{2^{a_1}},\frac{3\bigg(\frac{3p+1}{2^{a_1}}\bigg)+1}{2^{a_2}},\ldots$$ for arbitrarily long chains, say of length $n$. This is similar to Cunningham chains. According to Dickson's conjecture, $n$ can get arbitrarily large. Thus, one must expect there to be arbitrarily large chains of such primes in the Collatz Sequence.

Answer 2

There is almost surely an existing Mersenne-prime-like conjecture to the effect that for any prime $p\gt3$, there is a prime $q\gt3$ of the form

$$q={2^np-1\over3}$$

with $n\ge1$. (I'd appreciate any confirmation or correction on this conjecture that such a conjecture exists; I looked through Richard Guy's UPINT and didn't see it.) If we always take the smallest $n$ for which $(2^np-1)/3$ is a prime greater than $3$, we can form a Collatz chain backwards:

$$1\leftarrow2\leftarrow4\leftarrow8\leftarrow16\leftarrow5\leftarrow10\leftarrow20\leftarrow40\leftarrow13\leftarrow26\leftarrow52\leftarrow17\leftarrow34\leftarrow11\leftarrow22\leftarrow7\leftarrow14\leftarrow28\leftarrow56\leftarrow112\leftarrow37\leftarrow\cdots$$

(Suppressing the even entries, that's $1\leftarrow5\leftarrow13\leftarrow17\leftarrow11\leftarrow7\leftarrow37\leftarrow\cdots$.)

The existence of arbitrarily long chains of primes as consecutive odd entries in a Collatz sequence doesn't depend on this conjecture (that is, the conjecture need not hold for every prime $p$, just for enough primes to get arbitrarily long chains), but it would follow from it.

Answer 3

I believe I found a way to generate an infinite sequence of such primes. We proceed in reverse order. Start with a prime. Now multiply by 2. If the new number minus one is divisible by 3 and the result is a prime then output it (#). Otherwise continue multiplying by 2 until you reach condition (#), which could take a while. For example if we start with 19 then we will generate this sequence of primes: 19, 101, 67, 89, 59, 157, 13397, 9820104851543381, 1757376215861239855011157... There are more prime terms after this, but they get quite large. Also I am not sure if this will continue to find new primes.