[QUOTE=blob100;206771]No, it does not fail.[/QUOTE]How do you know that?

By the conjecture, it is real, and I agree with the conjecture.

[QUOTE=blob100;206773]By the conjecture, it is real, and I agree with the conjecture.[/QUOTE]Are you aware of what "conjecture" actually means?

The conjecture never states what forms of p it is likely/unlikely to hold true for. How can you simply assume it will hold true for all p of the form 2q+1?

[QUOTE=Uncwilly;204207]You have shown far more respect for the help that you have gotten than many others have in the past. I think that you have paid attention to the help of others.

I also think that Bob has done well.:bow: Often times those that are unlearned cause him to react a bit strongly.[/QUOTE][QUOTE=R.D. Silverman;204299]Yes. Go learn some number theory.[/QUOTE]
[QUOTE=blob100;205792]Silverman,
I can't read now so I conjecture..[/QUOTE]
[QUOTE=R.D. Silverman;205793]Making conjectures out of ignorance makes one look like a fool.

One can not make intelligent conjectures without being aware of already existing conjectures. One needs to know what is known and what isn't known.[/QUOTE][QUOTE=R.D. Silverman;206661]STOP MAKING CONJECTURES. You don't know enough mathematics to be able to say anything new or meaningful.[/QUOTE]Tomer, I think that it time that you now learn what you need to know. My first assessment was right to that point, but now it has changed.

[QUOTE=blob100;206656]I found a stronger one:
conjecture: there are infinitely many primes [tex] q=2^{n+1}Q+2^{n+1}-1[/tex] such that [tex]m=2^{n}Q+2^{n}-1[/tex] is also a prime, where Q is not neccesarily a prime number.[/QUOTE]

[QUOTE=R.D. Silverman;206726]The original conjecture was that p and 2p+1 are both prime i.o.

The new conjecture is that p = 2^n R -1 and 2p+1 are both prime i.o.[/QUOTE]

Agreed. And these are obviously equivalent:

2 => 1: If for p = 2^nR - 1 there are infinitely many prime pairs (p, 2p+1), then there are infinitely many prime pairs of the form p, 2p+1.
1 => 2: If there are infinitely many prime pairs (p, 2p+1), then let n = 0, R = p+1; then there are infinitely many prime pairs (p, 2p+1) with p = 2^nR - 1.

If n is constrained to be positive, it can be chosen to be 1 in all cases except (2, 5); but if there are infinitely many such cases, then there are infinitely many after excluding that one.

[QUOTE=R.D. Silverman;206726]n is a free variable and adds an additional degree of freedom. The conjecture
did not say that 2^n R - 1 and 2^(n+1)R - 1 are prime i.o FOR ALL n.[/QUOTE]

I agree that it does not say this; if it did, it would be "stronger" in the sense that I could not prove it from the base conjecture. (I hesitate to say that it is actually stronger, in case both are true.)

[QUOTE=R.D. Silverman;206726](which is why I said to be careful about quantifiers). It could be the case
that for SOME n, 2^n R - 1 and 2^(n+1) R - 1 are both prime only
finitely often, yet the conjecture could still be correct.[/QUOTE]

Yes, but that's all that's needed for the (p, 2p+1) conjecture to hold.

[quote=Uncwilly;206791]Tomer, I think that it time that you now learn what you need to know. My first assessment was right to that point, but now it has changed.[/quote]
The point is that I do learn, as I said: "by reading the book "solved and unsolved problems in number theory" I found this conjecture... and by playing with it..."
1) I think reading this book is learning.
2) I think playing with the materials help to understand it.
Please agree that I'm learning and I STOPPED conjecturing.
As I said, I didn't find any new conjecture.. I just played with an existed one.

BTW: retina, I know what conjecture means. A conjecture is a proposition which wasn't proven but seems to be true, and did not unproven.

[quote=blob100;205790]I have a new conjecture (it isn't about mersenne numbers),
I conjecture that: between n^2 to n^2+2n there is always a prime number.
When n is an odd number.
We can show n^2+2n as n(n+2).
So, [tex]\sum_{n=1}^{\infty}n^{2}=1+3+5....2n-1[/tex]
[tex]\sum_{n=1}^{\infty}n^{2}=1+3+5....2n-1+2n[/tex][/quote]

Ouch,
I just now saw an horrible mistake,
I meant:
[tex]\sum_{n=1}^{\infty}n^{2}+2n=1+3+5....2n-1+2n[/tex].
By the second one.

[QUOTE=blob100;207245]Ouch,
I just now saw an horrible mistake,
I meant:
[tex]\sum_{n=1}^{\infty}n^{2}+2n=1+3+5....2n-1+2n[/tex].
By the second one.[/QUOTE]

Clueless. Truly clueless. What makes you clueless is NOT the posting
of incorrect mathematics. What does make you clueless is your repeated
posts made from ignorance despite repeated advice from professionals not to
do so.

As I have said before: stop making conjectures and start LEARNING.
Start listening to your teachers.

The above "formula" is mathematical gibberish. The left side of
the equal side is an infinite sum THAT DOES NOT CONVERGE.

The right side bears no relation to the left side:

(1) The left side is NOT a function of n. n is an index in the sum, but
the sum is not a function of n. OTOH, the RHS most definitely IS a
function of n.

(2) The RHS is finite. The LHS is not.

You are making basic, fundamental mistakes. Go learn some math.
We will help in that learning process. If you have questions, ask them.
Indeed. Your posts should consist of nothing EXCEPT questions.

But right now, you are trying to run before you can walk.

[quote=blob100;205790][tex]\sum_{n=1}^{\infty}n^{2}=1+3+5....2n-1[/tex]
[/quote]
[quote=blob100;207245][tex]\sum_{n=1}^{\infty}n^{2}+2n=1+3+5....2n-1+2n[/tex][/quote]tomer, your other mistake, in both, is the summation symbol on the left. Just remove it from both equations.

[tex]n^{2}=1+3+5....2n-1[/tex]

is true all by itself. So is

[tex]n^{2}+2n=1+3+5....2n-1+2n[/tex]
.

Putting a summation symbol on the left side of either one makes it mathematical gibberish.

Perhaps you might wish to use a summation symbol on the [I]right[/I] side of one or more of those equations, like this:

[tex]n^{2}=\sum_{i=1}^{n}(2i-1)[/tex]

but it makes no sense to use a summation symbol on the left side.

[QUOTE=cheesehead;207334]but it makes no sense to use a summation symbol on the left side.[/QUOTE]

Perhaps the intent was [tex]\forall x:\,1\le x<\infty[/tex]? That's all I can figure. That, or cargo-cult mathematics...

[quote=CRGreathouse;207344]Perhaps the intent was [tex]\forall x:\,1\le x<\infty[/tex]? That's all I can figure.[/quote]Perhaps it's simply that tomer, not yet understanding the proper roles of the summation index and limits, thought he could combine the summation of 1+3+5+... with the condition [tex]\forall n:\,1\le n<\infty[/tex] rather than expressing them separately.