Poulet numbers with 3 distinct prime factors

flouran · 2009-04-18, 00:54

I know that in 2004 Galway conjectured a formula for the Poulet number counting function with 2 distinct prime factors. I was wondering if there are any conjectures for the Poulet number (not necessarily Carmichael numbers) counting function with 3 distinct prime factors or how Galway's formula could be extended for 3 distinct prime factors?

flouran · 2009-04-18, 03:57

Quote:

Originally Posted by flouran

I know that in 2004 Galway conjectured a formula for the Poulet number counting function with 2 distinct prime factors. I was wondering if there are any conjectures for the Poulet number (not necessarily Carmichael numbers) counting function with 3 distinct prime factors or how Galway's formula could be extended for 3 distinct prime factors?

For pseudoprimes with 3 distinct prime factors, I conjecture $\frac{40000{x^{1/3}}}{\log^3 x}$ . or not

flouran · 2009-04-18, 14:18

Quote:

Originally Posted by flouran

For pseudoprimes with 3 distinct prime factors, I conjecture $\frac{40000{x^{1/3}}}{\log^3 x}$

Just now I was looking at a paper of Erdos, and he states the pseudoprimes with k distinct factors are bounded by ${c_1}{\log x} < P_k(x) < {c_2}{\frac{x}{(\log^k x)}}$
Although these bounds are weak (and were improved by Pomerance), the upper-bound looks a lot like my formula, eh? Note the trivial fact that:
${\frac{x^{\frac{1}{k}}}{(\log^k x)}} < {\frac{x}{(\log^k x)}}$ .

flouran · 2009-04-19, 00:56

So does anyone have any ideas at all? It's fine if they are not good, all I just want are some responses at least (especially the experts)! C'mon guys, it's like a freaking ghost town on this thread!

cheesehead · 2009-04-19, 04:07

Quote:

Originally Posted by flouran

all I just want are some responses at least (especially the experts)! C'mon guys, it's like a freaking ghost town on this thread!

Perhaps you need some hobby to occupy your time while waiting s-o-o l-o-o-o-n-g for responses here, especially on a weekend.

flouran · 2009-04-19, 04:23

Quote:

Originally Posted by cheesehead

Perhaps you need some hobby to occupy your time while waiting s-o-o l-o-o-o-n-g for responses here, especially on a weekend.

Well, I am at my computer right now and I just finished listening to "Megadeth - Kill the King". Now, I'm listening to "Metallica - Master of Puppets" (one of the epic thrash metal songs!).
If only I had friends to hang out with on a Saturday night....

Anyways, cheesehead, do you have any ideas at all (since you were the first one who responded

)?

cheesehead · 2009-04-19, 05:21

Quote:

Originally Posted by flouran

Anyways, cheesehead, do you have any ideas at all (since you were the first one who responded

)?

I'm not qualified to comment knowledgeably on this thread's mathematical topic.

As for ideas about hobbies and/or friends: List your interests, then consider how meeting with other people would fit in.

Two quick examples (not claimed to be representative!):

Stamp collecting -- There are conventions and shows. Friend-making potential would lie in conversations one struck up before, during and after the main event.

Thrash metal songs -- I'm not especially familiar with this genre, but I suspect that all song genres would have in common (more or less ...) that (a) folks could get together to sing [hmm... a thrash metal choir

], (b) a small group could form a band to play the music for others to sing [but that's who you're listening to: small groups Megadeth and Metallica!], or (c) folks could get together to attend a performance. Friend-making potential would mainly lie in conversations one struck up before and after the main event.

There -- now extend the list!

flouran · 2009-04-19, 08:16

Quote:

Originally Posted by cheesehead

I'm not qualified to comment knowledgeably on this thread's mathematical topic.

Anyone else have any ideas?

Quote:

Originally Posted by cheesehead

As for ideas about hobbies and/or friends: List your interests, then consider how meeting with other people would fit in.
There -- now extend the list!

Thanks for the advice cheesehead.

Just kidding.

Anyways, so back on topic, anyone at all have any clues? At all? Even the slightest clue?

retina · 2009-04-19, 10:57

I know that 3*5*7 = 105. Three prime factors right there. Hope that helps.

So, remind me again, what is this thread about anyway?

flouran · 2009-04-19, 15:12

Quote:

Originally Posted by retina

I know that 3*5*7 = 105. Three prime factors right there. Hope that helps.

105 is not a base-2 pseudoprime (it needs to be for it to be at all relevant to this thread).

Quote:

Originally Posted by retina

So, remind me again, what is this thread about anyway?

This thread is about what the formula is which counts the number of base-pseudoprimes with 3 prime factors.

flouran · 2009-04-29, 03:57

Quote:

Originally Posted by flouran

Just now I was looking at a paper of Erdos, and he states the pseudoprimes with k distinct factors are bounded by ${c_1}{\log x} < P_k(x) < {c_2}{\frac{x}{(\log^k x)}}$

That article does not give the proof of the second inequality in (1). Does anyone know where this proof may be (I can't seem to find it anywhere online)?

2009-04-18, 00:54	#1
flouran Dec 2008 341₁₆ Posts	Poulet numbers with 3 distinct prime factors I know that in 2004 Galway conjectured a formula for the Poulet number counting function with 2 distinct prime factors. I was wondering if there are any conjectures for the Poulet number (not necessarily Carmichael numbers) counting function with 3 distinct prime factors or how Galway's formula could be extended for 3 distinct prime factors?

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2009-04-19, 00:56	#4
flouran Dec 2008 833₁₀ Posts	So does anyone have any ideas at all? It's fine if they are not good, all I just want are some responses at least (especially the experts)! C'mon guys, it's like a freaking ghost town on this thread!

2009-04-19, 10:57	#9
retina Undefined "The unspeakable one" Jun 2006 My evil lair 15027₈ Posts	I know that 357 = 105. Three prime factors right there. Hope that helps. So, remind me again, what is this thread about anyway?