Topic, Someone disprove my prime number theory I have...

  Just like a fermat prime. A prime function is something very hard to find, because a lot of mathematiciens are looking for it for ages.

  my formula aims to always give out primes though...it doesnt...

  I think there already such a prime function, this is probably a derivation of an existing function, still nice that you've found that.
every prime that's 2^n+1 is a fermat prime, your formula looks similar to it.

  the problem im facing is that the bigger the B the likelier the number isnt prime...due to increased potential divisors...

  "I'm pretty sure we're onto something here."
I found it D: lol
im doing some observational tests...and it would seem that I get a fair bit of useless values with a ton of factors...
where A is 1 to 15 and B is 3 or 5 I have these:
A=2 B=3 Result=163841
A=9 B=3 Result=737281
A=14 B=3 Result=1146881
A=5 B=5 Result=104857601
A=8 B=5 Result=167772161
all results for B=3 apart from ones listed arent prime...and im doing B=5 now...

[edit] A=11 B=5 returns prime...and my prime finder choked out on B=7 numbers because theyre too big...

[re-edit] what if I did -1 not +1...

[re-re-edit] alright this is weird, ive re-copied out the function and im getting different values :S
for instance A=2 B=3 now returns 16385 not my previously speculated 163841

[re-re-re-edit] wait a second...I think A must be at least 8...or 9...

[edit x4] just realised the time...5AM o.O
also, the problem here is that I only accounted for divisors that are 2^n...so, im gonna change my theory a lil bit...

  @Hexicube's first edit: Good observation! So does that mean [A*(16^B)*2+1]/3 would be the prime number function? Wait... A*(16^B)*2+1 wouldn't always get a number divisible by 3, would it? I'm on it.

If B is always even, then it can be 6. But I'm pretty sure that N isn't a natural number when 6=2^n. So those two statements can't go together, not unless you slightly change it. Either that or I'm an idiot and those statements weren't meant to be put together. :P

So let's try them out seperately. I'll be right back.

[Edit] @Hex's re-re-edit: LOL, when I was gone doing the maths I got that same result. But I agree, I'm pretty sure we're onto something here.

[Re-edit] *See first paragraph of this post first* I got a set for A and B where [A*(16^B)*2+1]/3 does not get a prime number. Try A=2 and B=5. That would turn out to be 139801.66666....... :P

  lol...im slightly annoyed because ive tried like 50 numbers and they all worked :P
ok try it if B=2^n where N is natural number...
[edit] also, if the number is divisible by 3, dividing it by 3 makes it prime...(2731=prime)
[re-edit] hmm...what if B is always even?
[re-re-edit] ok I think I may need to say this hasnt worked...but there must be a correlation between my results and some restrictions :P

  A=1
B=3
1*(16^3)*2+1=8193
8193 = 3*2731

Simple.

  where A is a whole number 1 to 15 and B is a natural number bigger than 2:
A*(16^B)*2+1 is either prime, or a prime number multiplied by 3

can anyone find a set for A and B where the result isnt prime? if not I win at life for making a prime number function...even though it wont find every prime :P