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Old 2013-05-26, 11:24   #8
efeuvete
 
"Fernando Villanueva"
May 2013
Madrid /Spain

5 Posts
Default

Yes, those pseudos "Of mine" start from A081264 integer sequence (Which I did'nt know then, in 1997). But in my quest of "Taking those pseudos out" I found a different period (In "Pisano's Style"), which I use on my third rule.

I go now on how to test second rule. It is as follows:

SECOND RULE: When N and F(N) satisfy first rule. So:

F( SH( F(N))) MOD F(N) = 0. But, according to Fibonacci sequence rule:

"If F(X) MOD F(Y) = 0, then X MOD Y = 0", we have:

SH( F(N)) MOD N = 0 wich is as easy to check as first rule.

And that rule took out 50% of the pseudos. So, when I found this, I thought... ┬┐And what would happen with pseudos if F(F(N)) satisfy first rule too?

I would bet a beer this rule would take out another 50% of pseudos, but my modest math knowledge did'nt allow me here to find an easy way to check it. Anyhow, is easy to check that numbers 3 and 7 satisfy it (To test 11 you need more than 16 digits)

This would be wonderful, because if it would be true (And you can find how to check it) then you can go stairs up with F(F(FN))) and so on "Killing" pseudos as a "Prime warrior"

Another curious thing to mention is that, in any case, "My" pseudos decrease very fast with N, I mean: From 1 to 200.000 there are 21 pseudos, in the next 200,000 N there are only 6 more... and from 800.000 to 1.000.000 only 4 more.

I would like to know how many pseudos are from 1.000.000 to 2.000.000; some day I will check that.

Well, if someone in the forum wants to continue this game, I will continue too (But it would take me some time to translate my third rule)

Thanks and regards,
Fernando.

Last fiddled with by efeuvete on 2013-05-26 at 11:55 Reason: My short english
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