In fact, it doesn't look like 31031 at all fits that pattern. It's not the correct value mod 9.

You're here?

Damn, I was planning on winning.

Most interesting thing I can come up with about 31031 is that it equals 6^6-5^6.

Not related to the current problem, but zultar, were you inspired by this video by any chance? https://www.youtube.com/watch?v=ea7lJkEhytA

This one is hard. I have spent too much time trying to figure it out, so I am going to stop now.

@Vashta, no I haven't seen that video. The look and say sequence was something I used regularly in my Algebra 1 and 2 classes when I taught. I used them in conjunction with other basic patterns like arithmetic, geometric, if-then sequence, so that students can see that not all patterns are simple and formulaic. Plus, it's fun for students/people to figure that kinda stuff out.

Ghug's current puzzle is hard, particularly since he hasn't really send me the solution so I don't understand yet. :)

Yeah, a clue is probably in order: pseudoprimes.

Any other clues?

Forgive me; I apparently can't type numbers correctly.

Correct sequence:
1122, 3315, 8645, 17255, 31031, 85813, 151487

Nope.

1122=2*3*11*17
3315= 3*5*13*17
8645= 5*7*13*19
17255 = 5 * 7*17*29
31031 = 7*11*13*31
85813=7*13*23*41
151487= 7*17*19*67

All products of 4 primes.

(In particular, they aren't all pseudoprimes to base a for anything with any prime factor less than 37 as they have all the other prime factors somewhere and they aren't pseudoprimes for each other in some way because they share a bunch of factors)

So what are the next numbers in the sequence uclabb??

I have no idea! I was trying to help us as a group figure it out!

You're getting closer.

At least at this point, pseudoprime has proved to be a useless hint- I trust that it isn't though. Just checking- a pseudoprime w.r.t a is a number x such that a^x - 1 = 0 mod x, yes?

Just thinking out loud here, didnt find anything useful:

Parentheses are what position each prime number is on a prime number list (so 5 is the third prime number and gets a 3 in the parentheses).

1122=2*3*11*17 (1,2,5,7)
3315= 3*5*13*17 (2,3,6,7)
8645= 5*7*13*19 (3,4,6,10)
17255 = 5 * 7*17*29 (3,4,7,11)
31031 = 7*11*13*31 (4,5,6,11)
85813=7*13*23*41 (4,6,9,13)
151487= 7*17*19*67 (4,7,8,19)

Doesn't seem to be any pattern I can figure out.

Well, the lowest prime factor keeps increasing, so the lowest prime factor of the next number is either a 7 or 11. I'm leaning towards eleven.

I had a more general definition in mind, but that works.

That was @uclabb.

Im stumped... this is a good one ghug.

@uclabb: you're close on your definition of pseudoprimes. Wrt a, a pseudoprime is some x s.t. a^(x-1) = 1 under mod x, not 0. If mod x=0, then the number is composite.

At first glance, it looks like x=9 with the absence of 9 in the outcomes or it might be that these are pseudoprimes written in base 9, but I'm not finding anything that really puts together that pattern.

My guess is that these are the occurences of the base a, but I am not finding an appropriate value for x where this holds true.


I do have to say that it's a hard puzzle because you have to understand upper-level math in order to even follow it, whic limits who can participate.

Granted, I'm stupid, but I do know 2^1122 is a f'ing huge number, if smashing around with pseudoprimes is the toe-hold toward the solution. Also I don't see how 1122 comes out as x ^ a +- 1 for any x or a, so it's probably not the result of any exponentiation either. So give us another hint: does this number have to be "crunched" or is there an insight that we're all missing?

You don't have to know what any a^x is when dealing with module arithmetic. For example, lets look at base 2 under mod 3
2^0=1 mod 3=1
2^1=2 mod 3=2
2^2=4 mod 3=1
2^3=8 mod 3=2
2^4=16 mod 3=1

So for base 2, any x= even even, the result under mod 3 = 1. If x is odd, under mod 3, the result is 2.

Therefore, we know that 2^1661 mod 3 = 2 because x is odd

Told you I was stupid :)

For those that are reading this that don't know what module arithmetic is, it's when you divide one number by another and the answer is the remainder value. So, if we look at mod 5, we divide numbers by 5 and get our answer from the remainder. 1 mod 5 = 1 because 1 divided by 5 is 0 remainder 1. 2 mod 5 is 2, etc etc until we get to 5. 5 mod 5 is 0 because 5 / 5 is 1 and the remainder is 0. 6 mod 5 is 1 because 6/5 is 1 with a remainder of 1.

Therefore mod 5 has 5 different repeating answers 0,1,2,3,4 because these are the only possible remainder values when we divide any number by 5

All I can say is that this puzzle is a product of a deranged mind.

I think the next number in the sequence is the lowest product of unique primes (so not the same prime twice) that is larger than 151487. Don't know wether that works and I can't be bothered to find out what number that would be anyway.

Lowest product of *4* unique primes that is larger than 151487, that is. I should be clear. That doesn't seem to work though.