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This may help: A080670 A195265

Define $f(n)$ as this:

Take a number $n$, and split it into its prime composition using $^$ and $×$. Now remove all $^$ and $×$, you get a new number, this is $f(n)$ (Example: $f(20)=225$ because $20$=$2$^$2$×$5$ and $f(13)=13$ because $13=13$)
Now repeat $f(n)$, does it ever repeat or become prime?
The answer is yes for every number under $20$, but if you do start with $20$, you get numbers over: $$7619860311238375857894724798334256170952046407785111199930594835524474243089$$
I spent over half an hour determining if this sequence ever ends, and I haven't found any solution. If this ever ends, do other sequences like this don't?

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    $\begingroup$ Just to make sure that the sequence is precisely defined: You take the prime decomposition, order the occuring primes increasingly from left to right, omit the exponent 1 whenever it occurs, and of course 1 itself is not a prime number. Furthermore you work in base 10 (which also could affect the result). $\endgroup$
    – HenrikRüping
    Commented May 14 at 6:52
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    $\begingroup$ 3 votes to close as not research level? Is this question trivial for experts with the relevant knowledge base? It doesn't look trivial to me, but I have no expertise in this area. $\endgroup$
    – Alec Rhea
    Commented May 14 at 15:29
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    $\begingroup$ @StanleyYaoXiao But should we be closing questions just because 'most research mathematicians in that field wouldn't be interested in this problem'? What people are interested in (even active researchers in a given field) doesn't dictate what is and isn't mathematics, nor does it determine what is/isn't research level imo. For a really extreme example, most of Cantor's contemporaries were uninterested in his work. $\endgroup$
    – Alec Rhea
    Commented May 14 at 16:26
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    $\begingroup$ @AlecRhea sure, I didn't vote to close, just to say that problems of this sort are easy to generate and usually go nowhere. Number theorists in particular have to be vigilant about catching "mathematical diseases" (Tim Gowers used this phrase to describe the Collatz conjecture, which is actually somewhat reminiscent of this question), i.e., getting stuck thinking about a simple-sounding problem that ultimately goes nowhere. $\endgroup$
    – Stanley Yao Xiao
    Commented May 14 at 16:51
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    $\begingroup$ @StanleyYaoXiao We can all be grateful that John Conway chose to ignore the conventional wisdom about what professional mathematicians are "supposed to" work on, and instead carved out his own path. $\endgroup$
    – Timothy Chow
    Commented May 15 at 12:45

2 Answers 2

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This is John H. Conway's Climb to a prime problem. Conway originally conjectured that no matter what number you start with, you will eventually converge to a prime. This conjecture was disproven by James Davis, who found a composite number that is a fixed point.

That 20 is not known to converge was noticed very quickly after Conway first asked the question, but Conway still believed it would eventually converge.

One could ask if there are any heuristic arguments for or against the conjecture; I don't know of any published analysis.

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Probably an interesting question is why it seems that with base $2$ instead of $10$ and the same procedure, we get large primes: (Notice that for $d=2$ the starting number $20$ gets a prime).

def ff(n,d=10):
    ll = []
    for p in sorted(prime_divisors(n)):
        dd = Integer(p).digits(d)
        dd.reverse()
        ll.extend(dd)
        if valuation(n,p)>1:
            dd = Integer(valuation(n,p)).digits(d)
            dd.reverse()
            ll.extend(dd)
    return sum(ll[len(ll)-1-i]*d**i for i in range(len(ll)))

def iter_ff(n,k,d):
    if k==0:
        return n
    else:
        return ff(iter_ff(n,k-1,d),d)

x = 2**2*5       
while not is_prime(Integer(x)):
    x = ff(x,d=2)
    print(x)    
    
maxIter = 55    
for d in range(2,11):
     for n in range(2,100):
        x = n
        i = 0
        while not is_prime(Integer(x)) and i < maxIter:
            x = ff(x,d=d)
            print("base = ",d,"starting number = ",n,"current number = ",x,"iteration count = ", i+1)
            i+=1
        #print(d,n,iter_ff(n,k,d),is_prime(Integer(iter_ff(n,k,d))))

Output:

85
177
251
base =  2 starting number =  4 current number =  10 iteration count =  1
base =  2 starting number =  4 current number =  21 iteration count =  2
base =  2 starting number =  4 current number =  31 iteration count =  3
base =  2 starting number =  6 current number =  11 iteration count =  1
base =  2 starting number =  8 current number =  11 iteration count =  1
base =  2 starting number =  9 current number =  14 iteration count =  1
base =  2 starting number =  9 current number =  23 iteration count =  2
base =  2 starting number =  10 current number =  21 iteration count =  1
base =  2 starting number =  10 current number =  31 iteration count =  2
base =  2 starting number =  12 current number =  43 iteration count =  1
base =  2 starting number =  14 current number =  23 iteration count =  1
base =  2 starting number =  15 current number =  29 iteration count =  1
base =  2 starting number =  16 current number =  20 iteration count =  1
base =  2 starting number =  16 current number =  85 iteration count =  2
base =  2 starting number =  16 current number =  177 iteration count =  3
base =  2 starting number =  16 current number =  251 iteration count =  4
base =  2 starting number =  18 current number =  46 iteration count =  1
base =  2 starting number =  18 current number =  87 iteration count =  2
base =  2 starting number =  18 current number =  125 iteration count =  3
base =  2 starting number =  18 current number =  23 iteration count =  4
base =  2 starting number =  20 current number =  85 iteration count =  1
base =  2 starting number =  20 current number =  177 iteration count =  2
base =  2 starting number =  20 current number =  251 iteration count =  3
base =  2 starting number =  21 current number =  31 iteration count =  1
base =  2 starting number =  22 current number =  43 iteration count =  1
base =  2 starting number =  24 current number =  47 iteration count =  1
base =  2 starting number =  25 current number =  22 iteration count =  1
base =  2 starting number =  25 current number =  43 iteration count =  2
base =  2 starting number =  26 current number =  45 iteration count =  1
base =  2 starting number =  26 current number =  117 iteration count =  2
base =  2 starting number =  26 current number =  237 iteration count =  3
base =  2 starting number =  26 current number =  463 iteration count =  4
base =  2 starting number =  27 current number =  15 iteration count =  1
base =  2 starting number =  27 current number =  29 iteration count =  2
base =  2 starting number =  28 current number =  87 iteration count =  1
base =  2 starting number =  28 current number =  125 iteration count =  2
base =  2 starting number =  28 current number =  23 iteration count =  3
base =  2 starting number =  30 current number =  93 iteration count =  1
base =  2 starting number =  30 current number =  127 iteration count =  2
base =  2 starting number =  32 current number =  21 iteration count =  1
base =  2 starting number =  32 current number =  31 iteration count =  2
base =  2 starting number =  33 current number =  59 iteration count =  1
base =  2 starting number =  34 current number =  81 iteration count =  1
base =  2 starting number =  34 current number =  28 iteration count =  2
base =  2 starting number =  34 current number =  87 iteration count =  3
base =  2 starting number =  34 current number =  125 iteration count =  4
base =  2 starting number =  34 current number =  23 iteration count =  5
base =  2 starting number =  35 current number =  47 iteration count =  1
base =  2 starting number =  36 current number =  174 iteration count =  1
base =  2 starting number =  36 current number =  381 iteration count =  2
base =  2 starting number =  36 current number =  511 iteration count =  3

For $n=217$ and for $d=2$ you get as an example a cycle which is not a prime:

SageMath-Script

$$217, 255, 945, 1007, 1269,1007,1269,1007\ldots$$

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