[QUOTE=3.14159;227058][CODE]41441994149199491949199941949494914919499419411441141114499941994199[/CODE]

68 digits, and only has square numbers for digits!

Update: [code]41449199414911994941149149491999144449499419114949194149949499414499419449919999191449199941949494914919499419411441141114499941994199[/code]

134 digits, and only composed of square numbers! HA![/QUOTE]

Hmm, I wonder how rare such numbers are. Heuristics suggest about
[TEX]\frac43\cdot\frac{3^n}{n\log10}[/TEX]
n-digit examples; how well does this hold in practice?

There are 0, 3, 11, 12, 43, 94, 239, 566, 1710 such primes with 1, 2, ..., 9 digits. The formula predicts 2, 3, 5, 12, 28, 70, 181, 475, 1266 primes; not too bad, perhaps, though biased on the low side. Maybe with the next correction term, the -1 in the denominator? That gives 3, 3, 6, 13, 31, 76, 193, 502, 1331.

This is an asymptotic density of something like n^0.477, so maybe somewhat rarer than palindromatic primes.

[QUOTE=science_man_88;227069]okay well since these other seem to fit the same relation and are all of form 2x+1,2(2x+1)+1,etc they should all deviate the same way in my eyes so what would the odds for an exception to give you a sequence that works in the same way as the original being inserted into 2^p-1 with p prime will actually be prime ?. can you give those odds ?[/QUOTE]

I don't understand.

[QUOTE=science_man_88;227060]if only there was a pattern you'd find a 266 digit one.[/QUOTE]

Here's a 266-digit one:
[code]11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111494999[/code]

[CODE](14:37) gp > findrec([11,23,47,95,191,383,767])
Recurrence relation is a(n) = 3a(n-1) - 2a(n-2).
3 d.f.
%63 = [3, -2]~
(14:39) gp > findrec([29,59,119,239,479])
Recurrence relation is a(n) = 3a(n-1) - 2a(n-2).
1 d.f.
%66 = [3, -2]~
(14:39) gp > findrec([1,3,7,15,31,63,127])
Recurrence relation is a(n) = 3a(n-1) - 2a(n-2).[/CODE]

what are the odds that 2^ (the primes in the first 2)-1 are prime (which would put them in the primes of the last sequence). answer this and not only can we predict Mersenne primes but double Mersenne primes.

[QUOTE=CRGreathouse;227072]Here's a 266-digit one:
[code]11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111494999[/code][/QUOTE]

is there a 530 digit one ? if so it would possibly be part of this idea i have lol.

[QUOTE=science_man_88;227075]is there a 530 digit one ? if so it would possibly be part of this idea i have lol.[/QUOTE]

If you look at post #881, we expect lots of 530-digit ones -- about 8 × 10[SUP]249[/SUP]. The first is
[code]11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111491141[/code]

[QUOTE=science_man_88;227073][CODE](14:37) gp > findrec([11,23,47,95,191,383,767])
Recurrence relation is a(n) = 3a(n-1) - 2a(n-2).
3 d.f.
%63 = [3, -2]~
(14:39) gp > findrec([29,59,119,239,479])
Recurrence relation is a(n) = 3a(n-1) - 2a(n-2).
1 d.f.
%66 = [3, -2]~
(14:39) gp > findrec([1,3,7,15,31,63,127])
Recurrence relation is a(n) = 3a(n-1) - 2a(n-2).[/CODE]

what are the odds that 2^ (the primes in the first 2)-1 are prime (which would put them in the primes of the last sequence). answer this and not only can we predict Mersenne primes but double Mersenne primes.[/QUOTE]

I don't understand your question at all. If I wanted to test if a small number was the exponent of a Mersenne prime, I would check if it's in the listed members of A000043 (by a binary search). If I wanted to test if a large number was a Mersenne exponent, I'd check if it was a prime, trial-divide, look for some small factors with ECM, and then do a Lucas-Lehmer test on it. I don't have a special test for numbers of a certain form unless I can show an algebraic factor, but that would be caught by my tests anyway. It wouldn't speed up the search at all.

if we prove all exceptions that aren't knocked out by others have a sequence of the same form and that series of that form starting with an exception can't have a 2^p-1 prime then we can knock out all primes that have form of that have sequence that start at an exception.

[QUOTE=science_man_88;227079]if we prove all exceptions that aren't knocked out by others[/QUOTE]

What's an exception? What does "knocked out" mean? What are "others"?

I assume that "knocked out" means that, in some sense, the number is guaranteed to not be the exponent of a Mersenne prime. But what determines if it's knocked out or not?

[QUOTE=science_man_88;227079]have a sequence of the same form and that series of that form starting with an exception can't have a 2^p-1 prime[/QUOTE]

The same form as what?

[QUOTE=science_man_88;227079]then we can knock out all primes that have form of that have sequence that start at an exception.[/QUOTE]

Without definitions for the terms, this doesn't tell me much of anything.

[i]But[/i] let me try something. Let's say that a number is "knocked out" if it is a member of one of the sequences S1, S2, S3, ..., Sk. Further, let's suppose that the sequences are, like A055010, exponential integer sequences, where each term is roughly twice the one before it. And let's look at large numbers, say exponents at least a million.

Suppose we want to find all Mersenne exponents between 1,000,000 and 1,100,000 by this method. (We know from Slowinski & Gage that there are none, but we're testing this!) Now, by our assumptions, a given sequence Sj will remove about lg (1100000/1000000) ≈ 0.13 primes from the list. Since there are about 100,000 members, we'll need a minimum of > 720,000 sequences to clear the list.

Not one sequence or three sequences; not even a hundred. Three-quarters of a million sequences! And that's just to clear a small part of a small range; a larger one would take many more. Further, if there's overlap, you'll need more.

So if you don't want to use millions of sequences, make sure your method doesn't follow my assumptions.

I realized that my idea takes a lot but if we can come up with something it's like a sieve.

Another try at predicting the next Mersenne? I think there are figures that place it at around 19M to 20M digits. I believe this was stated somewhere in the Prime Pages.