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.
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
ReplyDeletehttps://www.youtube.com/watch?v=NT4UCRoQEQ4
ReplyDeleteHey Hans, I found a solution to Base 2!
(Also a solution to base 6, 8, 29, etc. via XB+Y=X^Y)
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.
DeleteMy 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?
DeleteAlso, my Python code shows it accounts for this?
DeleteBonus Clarification:
ReplyDelete(1269)1011110101=11^11*101111(3^3*47)
(1007)1111101111=10011*110101(19*53)
Got it. (Your 1269 binary has a zero missing.) I got my one-cycle from David Seal's post, here:
Deletehttp://list.seqfan.eu/pipermail/seqfan/2017-June/017685.html
David also gives a couple of two-cycles, including yours.
Speaking of which, I don't know David's contacts or whether he considered this, but it is possible to set up equations like
ReplyDeletex^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.
You are right! I'm not much of a mathematician. Rather, I consider myself an empiricist.
DeleteI know it is fun to optimise code to find specific fun numbers, but you did not answer my question.
DeleteI 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.
DeleteIf 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)
ReplyDeleteIt's only fun when you find solutions. I found none.
Delete