## Sunday, October 12, 2014

### Climb to a prime

I took on my table of A195264 three years ago. I am still chipping away at the currently-317 primary unknowns (out of 10000), extending ECM-factorization attempts from 2000 to 5000 curves. I'm not yet even half-way through the list. Occasionally factordb will have found for me (over time) the large factors for up to 116-digit composites! And in a handful of cases I have evolved an unknown into a prime — which finishes those unknowns from consideration, but there are plenty more.

I've been a little resentful that I hadn't picked the arguably aesthetically-better (one less arithmetic symbol with which to deal), historically better-known home prime sequence (A037274) on which to while away my time. No matter now: John Conway has offered \$1000 for a resolution (see the video here, from 19:20 to 22:30) to term #20 of A195264, not term #49 of A037274 — which he could easily have picked instead. I don't believe that (mathematically) it makes any difference: one is as contrived as the other. Conway's point (I think) is that there exist easily-stated but impossible-to-prove conjectures about such sequences. To collect the \$1000, you will almost certainly have to luck into a prime in the evolution, because proving that it doesn't evolve into a prime may well be impossible.

While the prize may spur some to further evolve A195265, it may also impede them from sharing any results of that effort. I hope that's not the case.

1. Hey Hans! You're the only person a google search indicates ever worked on this problem. I found a number you might be interested in, but I'll be damned if I can find your email address... send me a note: lorentztrans@gmail.com

Hey Hans, I found a solution to Base 2!
(Also a solution to base 6, 8, 29, etc. via XB+Y=X^Y)

1. Hey Keith. Your base-two solution appears to be 1269 = 3*3*3*47. Of course how factorizations are expressed is somewhat arbitrary, but the way Conway's climb-to-a-prime was played was not. More specifically, repeating factors are expressed as powers. For example, 255987 = 3^3*19*499 which is, by the way, a "proper" base-two solution.

2. My solution is actually a dual one which involves a 2-cycle. It does follow the rules and I checked it myself. However, how did you find that one-cycle?

3. Also, my Python code shows it accounts for this?

3. Bonus Clarification:
(1269)1011110101=11^11*101111(3^3*47)
(1007)1111101111=10011*110101(19*53)

1. Got it. (Your 1269 binary has a zero missing.) I got my one-cycle from David Seal's post, here:

http://list.seqfan.eu/pipermail/seqfan/2017-June/017685.html

David also gives a couple of two-cycles, including yours.

4. Speaking of which, I don't know David's contacts or whether he considered this, but it is possible to set up equations like
x^c=x*b^2+c / x^(c-1)-c/x=b^2
for xyz(b) where c=yb+z.
I was wondering if you had any advice how to approach equations like this. I know you are a brute force kind of guy, but you may know a way to prove it is impossible or something like that.

1. You are right! I'm not much of a mathematician. Rather, I consider myself an empiricist.

2. I know it is fun to optimise code to find specific fun numbers, but you did not answer my question.

3. I have no advice on how to approach equations like yours. That is because I do not know how to approach them myself. That is because I am not much of a mathematician. Needless to say (but I will in case there is any doubt), I do not know a way to prove anything as it relates to this.

5. If you enjoy fun challenges, here is an equation you can try to solve for the purposes of this problem: 4^m+6m=27^n+6n+1 (m and n are natural numbers)

1. It's only fun when you find solutions. I found none.