@steephie- It's def not that. A bunch of products of primes have bee skipped. Maybe we should look at the ones that have been skipped, though, for clues as to why they aren't in the sequence.

I agree with you. Whatever the base is, it can't be divisible by the exponent, so the base and exponent have to be relatively prime. I was thinking that the representation might be given in base 9 much like he was changing his values in his previous answer to hex.

Doesn't that throw away the whole product of 4 primes thing?

@Steephie: I agree with uclabb. If it were a pattern of products of primes, then there would have to be a pattern to those products, and there's not. That, coupled with ghug's pseudoprime statement doesn't really give anything.

I really think that we need to find that "a" value where a^1122 mod 1123 = 1 (or as uclabb stated a^1122 - 1 mod 1123 = 0)

Do you mean a^1121 mod 1122 = 1

I don't promise this is free from error. It might be all errors. But I *think*:

Taking the numbers given as the exponent and 8 as the base:

(8 ^ 1122) % 1123 = 7
(8 ^ 3315) % 3316 = 7
(8 ^ 17255) % 17256 = 7
(8 ^ 31031) % 31032 = 7
(8 ^ 85813) % 85814 = 7
(8 ^ 151487) % 151488 = 151479

Taking the numbers given, minus 1, as the exponent, and 8 as the base:

(8 ^ 1121) % 1122 = 7
(8 ^ 3314) % 3315 = 7
(8 ^ 17254) % 17255 = 7
(8 ^ 31030) % 31031 = 7
(8 ^ 85812) % 85813 = 7
(8 ^ 151486) % 151487 = 151478

Why is only the last one different?!?!?!? :( :( :( :(

Oh I think I made a mistake disregard previous...

@uclabb, yes

If the remainder value isn't 1, it's not pseudoprime. It's considered a composite number.

ohhh.... why does this seem to be true:

(X ^ 1121) % 1122 = (X % 1122)

(1110 ^ 1121) % 1122 = 1110
(1123 ^ 1121) % 1122 = 1

Actually to put it into the variables names we're accustomed to... and again, maybe I'm nuts, but it seems like

(a ^ 1121) % 1122 = (a % 1122)

Spot-checking:

(946 ^ 1121) % 1122 = 946
(947 ^ 1121) % 1122 = 947
(948 ^ 1121) % 1122 = 948

(1119 ^ 1121) % 1122 = 1119
(1120 ^ 1121) % 1122 = 1120
(1121 ^ 1121) % 1122 = 1121
(1122 ^ 1121) % 1122 = 0
(1123 ^ 1121) % 1122 = 1
(1124 ^ 1121) % 1122 = 2

My thought process:

1122 = 2 * 3 * 11 * 17 therefore it has
1122 * 1/2 * 2/3 * 10/11 * 16/17 = 320 number less than it relatively prime to it. These form a multiplicative group. In particular, 1121 = 320 * 2.5 + 1, so many numbers (in fact, half of them) that are relatively prime to 1122 satisfy a^1120 = 1 mod 1122.

I could be wrong, but I do think that the fact that 320 goes into 1121 fairly cleanly is notable. It might be worth checking the other number to see if they are similar.

So let's check 3315:

There are 3315 * 2/3 * 4/5 * 12/13 * 16/17 = 1536 numbers relatively prime to 3315. There is no clean relationship between 3315 and 1536. Dead end.

There should be a side-contest into how one question can seem to have so many red herrings. Is it the question? Or is it always like that?

God dammit... crunching a bunch of possible values of "a" for which 3315 might be pseudoprime, I see a shitstorm of squares:

(2 ^ 3314) % 3315 = 4
(3 ^ 3314) % 3315 = 9
(4 ^ 3314) % 3315 = 16
(5 ^ 3314) % 3315 = 25
(6 ^ 3314) % 3315 = 36
(7 ^ 3314) % 3315 = 49
(8 ^ 3314) % 3315 = 64
(9 ^ 3314) % 3315 = 81
(10 ^ 3314) % 3315 = 100
(11 ^ 3314) % 3315 = 121
(12 ^ 3314) % 3315 = 144
(13 ^ 3314) % 3315 = 169
(14 ^ 3314) % 3315 = 196
(15 ^ 3314) % 3315 = 225
(16 ^ 3314) % 3315 = 256
(17 ^ 3314) % 3315 = 289
(18 ^ 3314) % 3315 = 324
(19 ^ 3314) % 3315 = 361
(20 ^ 3314) % 3315 = 400

WTF?

Here's some code I'm using to bash out some numbers with absolutely no insight:

#include <stdio.h>
#include <stdlib.h>

void modulo(int a, int x)
{
int i;
int modulo = x + 1;
int tmp = 1;
for(i = 0; i < x; i++)
{
tmp *= (a % modulo);
tmp = tmp % modulo;
}
printf("(%d ^ %d) %% %d = %d\n", a, x, modulo, tmp);
}

int main(int argc, const char* argv[])
{
int limit = atoi(argv[1]);
int x = atoi(argv[2]);
for(int a = 2; a <= limit; a++)
{
modulo(a, x);
}
return 0;
}

So, without proof:

(a ^ 1121) % 1122 = (a % 1122)
(a ^ 3314) % 3315 = a ^ 2

What does it mean?

Actually for higher values the 3314 deal doesn't work:


(3256 ^ 3314) % 3315 = 166
(3257 ^ 3314) % 3315 = 49
(3258 ^ 3314) % 3315 = 3249
(3259 ^ 3314) % 3315 = 3136
(3260 ^ 3314) % 3315 = 3025

A lot of them are still squares, but it's weirder.

This is amazing to watch. Do y'all need another hint?

I have a question, ghug- is it like that hexadecimal pi thing or is it actually something that can be reasoned out?

I think it's something that can be reasoned. I created the sequence just kind of with the assumption that there'd be some interesting properties to it that would serve as clues. It's possible that there aren't, but I'm hoping somebody will catch something cool.

I will point out that the first number us is even, making it impossible for it to be a pseudoprime, so some additional thing has been done.

I have another question: is the solution going to be gratifying, or is it horse shit? :)

Just kidding it's actually been food for thought just with the squares thing. I have no idea why that's like that.

Right the number 1122, with the clue "pseudoprimes", leads us to the semi-interesting kinda-fact that this is pretty much true:

(x^19) ^ 59 MOD (2 * 3 * 11 * 17) = x

And the number 3315, with the clue "pseudoprimes", leads us to the semi-interesting kinda-fact that this is somewhat, but not very, true:

(x^2) ^ 1657 MOD (3 * 5 * 13 * 17) = x^2 until x gets pretty big...

So ghug does that mean anything or should am I just seeing irrelevant sorta-patterns?

Instead of giving clues maybe you can call "wrong turn"

The thing is that I don't want to call wrong turn on things, as you could arrive at the solution in any number of ways. You're on track with Fermat's little theorem, but what you have there doesn't seem to be useful.

"I will point out that the first number us is even, making it impossible for it to be a pseudoprime, so some additional thing has to be done."

The Carmichael numbers are 561 = 1122/2, 1105 = 3315/3 (and so on I assume)

...201115, 364343

^

Your turn. Please be gentle.

Looks like it's been 24 hours though, so I'm gonna pretend that I won and that means something.

Explanation: The nth number in the sequence is the nth Carmichael number (composite numbers x such that x|(a^(x-1)- 1 for all a not a multiple of x- this a stronger version of the pseudoprime definition we were playing around with) times the nth prime number.

@ghug- It hasn't been 24 hours since you posted the correct numbers. :-)