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-   -   when is 24m+7 prime ? (https://www.mersenneforum.org/showthread.php?t=14853)

CRGreathouse 2011-01-25 01:00

[QUOTE=science_man_88;248959]For all[TEX]m\in \text{A000040} \cap \text{A135659}[/TEX] that also fit [TEX]m\in \text{A139483} \cap \text{A002450}[/TEX] and if [TEX]m = \text{A002450(n)}
[/TEX] for some n we can show that A083420(n+1) is prime I can extend this to other sequences of form [TEX]6*4^n + \text{A083420(n-1)}[/TEX] assuming numbering n starts at 0.[/QUOTE]

So if you have a number that, when multiplied by 24 and added to 7, yields a prime, then that number, multiplied by 24 and added to 7, is prime.

science_man_88 2011-01-25 01:22

[QUOTE=CRGreathouse;249077]So if you have a number that, when multiplied by 24 and added to 7, yields a prime, then that number, multiplied by 24 and added to 7, is prime.[/QUOTE]

not the only thing in there. I know that it's been proven the 24m+7 fits all Mersenne primes >7 so if that m is in A002450 and has index n then A083420(n+1) is prime.

CRGreathouse 2011-01-25 01:41

[QUOTE=science_man_88;249084]if that m is in A002450 and has index n then A083420(n+1) is prime.[/QUOTE]

Let's work through this. If m = A002450(n) then we have m = (4[SUP]n[/SUP] - 1)/3. That means 3m = 4[SUP]n[/SUP] - 1 and so 3m + 1 = 4[SUP]n[/SUP]. Taking the base-4 log we get log(3m + 1) = n. Substituting this into the formula for A083420 we have
2 * 4[SUP]log(3m + 1) + 1[/SUP] - 1. Pulling a 1 out of the exponent we get 8 * 4[SUP]log(3m + 1)[/SUP] - 1. Since '4 to the power' and 'base-4 log' are inverses, we get 8 * (3m + 1) - 1. Simplifying, this is 24m + 7.

So yes, that's all this is. You're not using properties like the form of Mersenne numbers, you're just saying that 24m + 7 is prime iff 24m + 7 is prime.

science_man_88 2011-01-25 01:46

[QUOTE=CRGreathouse;249087]Let's work through this. If m = A002450(n) then we have m = (4[SUP]n[/SUP] - 1)/3. That means 3m = 4[SUP]n[/SUP] - 1 and so 3m + 1 = 4[SUP]n[/SUP]. Taking the base-4 log we get log(3m + 1) = n. Substituting this into the formula for A083420 we have
2 * 4[SUP]log(3m + 1) + 1[/SUP] - 1. Pulling a 1 out of the exponent we get 8 * 4[SUP]log(3m + 1)[/SUP] - 1. Since '4 to the power' and 'base-4 log' are inverses, we get 8 * (3m + 1) - 1. Simplifying, this is 24m + 7.

So yes, that's all this is. You're not using properties like the form of Mersenne numbers, you're just saying that 24m + 7 is prime iff 24m + 7 is prime.[/QUOTE]

how it links to mersenne numbers is that 24(A002450) + 7 = a odd indexed mersenne number.

CRGreathouse 2011-01-25 02:08

[QUOTE=science_man_88;249088]how it links to mersenne numbers is that 24(A002450) + 7 = a odd indexed mersenne number.[/QUOTE]

Yes. So if you have a prime Mersenne number you know that it is prime.

science_man_88 2011-01-25 14:47

[QUOTE=CRGreathouse;249093]Yes. So if you have a prime Mersenne number you know that it is prime.[/QUOTE]

not quite what I was going for. I was more heading towards if we can use properties of m such that 24m+7 is prime we can check to see which m in 2450 have the properties needed to make 24m+7 prime and hence in this case [COLOR="Red"]be[/COLOR] [COLOR="Lime"]create[/COLOR] a Mersenne prime.

3.14159 2011-01-25 20:22

Wow.. There hasn't been much going on in the time that I've disappeared..

Some things never change..

davar55 2011-01-27 16:03

[QUOTE=gd_barnes;248906]Why do you refer to yourself in the 3rd person? I've seen this several times now. Do you not think highly enough of yourself to say "I" or "me"?

Just curious...[/QUOTE]

I (meaning me) refer to myself as davar55 in this way in this forum
only in certain contexts. In that post, since I referred to Dr.Silverman
by the initials RDS, I used davar55 to refer to me, rather than me or I.

Just courtesy.

davar55 2011-01-27 16:08

Don't forget that the elements of oeis are sequences, not sets.

Sequences are ordered and can have repeated values.
Sets are unordered and (in normal representation) non-repeated.

For sets of integers BTW an ordered representation is
"standard normal form".

c10ck3r 2011-04-02 02:36

Bob's lack of attention?
 
[QUOTE=R.D. Silverman;247041]24m+7 can be prime if m != 0 mod 7[/QUOTE]

To get this back to the actual topic, I would like to point out the fallacies of Robert Silverman's above quoted formula. This statement is, in and of itself, false. If m!=0 mod 7, then m MUST be greater than or equal to 7. This is not a requirement, but merely a poor musing that put no thought into the question. 24(1)+7=31
24(3)+7=79
24(4)+7=103
24(5)+7=127
24(6)+7=151
((24(7)+7=175)) though 7!=0 mod 7
24(8)+7=199
This does not appear to be a modular function in terms of m, but I very likely could be wrong.

Just to CMA, if Mr. Silverman was attempting to be clever, I apologize for my inability to acknowledge his sarcasm, particularly after being followed by abashment of a peer. However, he will be reminded that what we lack in mathematical knowledge initially, we make up for in technology and willingness to learn.

That is all.

R.D. Silverman 2011-04-02 06:26

[QUOTE=c10ck3r;257336]To get this back to the actual topic, I would like to point out the fallacies of Robert Silverman's above quoted formula. This statement is, in and of itself, false. If m!=0 mod 7, then m MUST be greater than or equal to 7.
[/QUOTE]

Huh???. m = 1,2,3,4,5, or 6 are not 0 mod 7, yet m < 7. Surely, this
can't be what you meant?

m!= 0 mod 7 is a necessary condition for 24m+7 to be prime except for the
special (degenerate case) m = 0. Otherwise 24m+7 will be divisible by 7.
I believe that m > 0 was part of the conditions.


[QUOTE]

This is not a requirement, but merely a poor musing that put no thought into the question. 24(1)+7=31
24(3)+7=79
24(4)+7=103
24(5)+7=127
24(6)+7=151
((24(7)+7=175)) though 7!=0 mod 7

[/QUOTE]

Complete idiot. 7 certainly does equal 0 mod 7.

[QUOTE]
24(8)+7=199
This does not appear to be a modular function in terms of m, but I very likely could be wrong.
[/QUOTE]

You have ZERO idea of what a modular function is. Try google. I will
give some hints:

(1) It is a meromorphic function in the upper half plane.
(yes, I know. You are lost already)
(2) The function involves the modular group; an instance of
SL(2,Z) (still lost......)
(3) Do you know what a linear fractional transformation is?


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