2:57 PM
lets say we have a number of form $2048*x+7$ , if $x$ were even, we could write it with a higher power of two coefficient and half of $x$ . It's important to note, that parity of such a number falls to the constant term. The reason this is important , is that Collatz output depends on parity. Multiplying by 3 and adding one, multiplies both terms by 3, and adds 1 to the constant term ( just like the constant terms Collatz sequence would). dividing by two eithe
either ends up at an odd linear coefficient, or divides both terms by 2. This means as long as the linear parts coefficient has suffient factors of two to continue we follow the collatz path of the first remainder term , in terms of up and down motion.
so take $x=1$ we get $2048+7=2055$, whiich goes to $6144+22=6166$, leading on again, we get 3083 , 9250,4625, 13876,6938,3469,10408,5204,2602,1301,3904,1952,976,488,244 which are all of the forms predicted algebraically.
with algebra you can give me any $x$ and I can be confident $2048*x+7$ gets you down to $243*x+1$
with Goldbach linked in with examples of $10=3+7\implies 16=5+11$
my(a=randomprime(10^18));print(2048*a+7","243*a+1) plug away.