[QUOTE=CRGreathouse;233087]That's hard because some definitions require them to have a prime exponent, in which case they're not recurrence relations.[/QUOTE]

couldn't there be a side note ? lol

now if only I could prove which primes using the Lucas formulas

[QUOTE=science_man_88;233088]couldn't there be a side note ? [/QUOTE]

You could add it in if you like. You don't even need to log into Wikipedia to make changes.

[QUOTE=CRGreathouse;233091]You could add it in if you like. You don't even need to log into Wikipedia to make changes.[/QUOTE]

added in I still want to figure out what the V sequence would represent.

[QUOTE=science_man_88;233094]added in I still want to figure out what the V sequence would represent.[/QUOTE]

2^n + 1, Sloane's [url=http://oeis.org/classic/A000051]A000051[/url].

[QUOTE=CRGreathouse;233095]2^n + 1, Sloane's [url=http://oeis.org/classic/A000051]A000051[/url].[/QUOTE]

if we could find a P,Q that would work for the exponents to each if possible the hard part is if it has to be integer then the only way I see to get the Mersenne prime exponents started is 3,2 though I'm likely wrong.

[QUOTE=science_man_88;233097]if we could find a P,Q that would work for the exponents to each if possible the hard part is if it has to be integer then the only way I see to get the Mersenne prime exponents started is 3,2 though I'm likely wrong.[/QUOTE]

What does "work" mean?

[QUOTE=CRGreathouse;233099]What does "work" mean?[/QUOTE]

succeed in creating the list in this case.

I found the p and q for 0 and positive integers numbers lol p=2 q= 1

[QUOTE=science_man_88;233130]succeed in creating the list in this case.

I found the p and q for 0 and positive integers numbers lol p=2 q= 1[/QUOTE]

OK, so you're saying
[INDENT]If we could find a P,Q that would succeed in creating the list for the exponents to each if possible the hard part is if it has to be integer then the only way I see to get the Mersenne prime exponents started is 3,2 though I'm likely wrong.[/INDENT]
So what is "the list", "succeed", and "started"?

[QUOTE=CRGreathouse;233173]OK, so you're saying
[INDENT]If we could find a P,Q that would succeed in creating the list for the exponents to each if possible the hard part is if it has to be integer then the only way I see to get the Mersenne prime exponents started is 3,2 though I'm likely wrong.[/INDENT]
So what is "the list", "succeed", and "started"?[/QUOTE]

the list = Mersenne prime exponents
succeed = successfulness in the creation
started = already underway

look if P and Q need to be integers then 3*3-2*2 = 9-4 = 5 the next one so since P needs to be 3 Q=2 is the only way to get them started but as we proved that's also the start of 2^n+1.

[QUOTE=science_man_88;233183]the list = Mersenne prime exponents
succeed = successfulness in the creation
started = already underway

look if P and Q need to be integers then 3*3-2*2 = 9-4 = 5 the next one so since P needs to be 3 Q=2 is the only way to get them started but as we proved that's also the start of 2^n+1.[/QUOTE]
[INDENT]If we could find a P,Q that would create the list of Mersenne exponents, ___. The hard part is if it has to be integer then the only way I see to get the Mersenne prime exponents started is 3,2 though I'm likely wrong.[/INDENT]
I can't entirely understand this, but it should be easy to show that there are no P, Q that yield the Mersenne prime exponents.

Ah yes, here we go. My program [url=http://oeis.org/wiki/User:Charles_R_Greathouse_IV/Pari#Recurrences]findrec[/url] says:
[code]Cannot be described by a homogeneous linear recurrence relation with 19 or fewer coefficients.[/code]
of the first 39 Mersenne exponents. So not only can you not find P, Q (a 2nd-order homogeneous linear recurrence relation with particular starting conditions), you can't find *any* HLRR of order 2, or even 3 through 19.

[QUOTE=CRGreathouse;233187][INDENT]If we could find a P,Q that would create the list of Mersenne exponents, ___. The hard part is if it has to be integer then the only way I see to get the Mersenne prime exponents started is 3,2 though I'm likely wrong.[/INDENT]
I can't entirely understand this, but it should be easy to show that there are no P, Q that yield the Mersenne prime exponents.

Ah yes, here we go. My program [url=http://oeis.org/wiki/User:Charles_R_Greathouse_IV/Pari#Recurrences]findrec[/url] says:
[code]Cannot be described by a homogeneous linear recurrence relation with 19 or fewer coefficients.[/code]
of the first 39 Mersenne exponents. So not only can you not find P, Q (a 2nd-order homogeneous linear recurrence relation with particular starting conditions), you can't find *any* HLRR of order 2, or even 3 through 19.[/QUOTE]

is there a formula for telling which in a lucas sequence are prime ?