PARI's commands

[science_man_88]

good now I can check Mersenne exponents for things to eliminate.

[CRGreathouse]

Originally Posted by science_man_88

good now I can check Mersenne exponents for things to eliminate.

On my main computer I can just type

findrec(bfile(43))

where 43 means A000043.

[science_man_88]

Originally Posted by CRGreathouse

On my main computer I can just type

findrec(bfile(43))

where 43 means A000043.

I was trying to find patterns in what to eliminate but cool.

though I'll have to look at this closer http://www.research.att.com/~njas/sequences/A055010

every prime above 5 in this sequence seems fit to eliminate.

and I tried the same basic thing (which gave me the same equation by your code) starting on 29 as well but I haven't checked that one either.

[3.14159]

Originally Posted by science_man_88

I was trying to find patterns in what to eliminate but cool.

Mersenne numbers do not have any patterns, they're randomly distributed and have no covering set of divisors (Besides having Proth numbers as factors)

[science_man_88]

Originally Posted by 3.14159

Mersenne numbers do not have any patterns, they're randomly distributed and have no covering set of divisors (Besides having Proth numbers as factors)

mersenne numbers have the equation a(n) = 3a(n-1) - 2a(n-2) which is weird as it's the same equation to fit the sequences I just talked of starting with the respective prime exponent that doesn't give a prime.

(14:39) gp > findrec([1,3,7,15,31,63,127])
Recurrence relation is a(n) = 3a(n-1) - 2a(n-2).
3 d.f.

[3.14159]

41441994149199491949199941949494914919499419411441141114499941994199

68 digits, and only has square numbers for digits!

Update:

41449199414911994941149149491999144449499419114949194149949499414499419449919999191449199941949494914919499419411441141114499941994199

134 digits, and only composed of square numbers! HA!

[science_man_88]

Originally Posted by 3.14159

41441994149199491949199941949494914919499419411441141114499941994199

68 digits, and only has square numbers for digits!

Update:

41449199414911994941149149491999144449499419114949194149949499414499419449919999191449199941949494914919499419411441141114499941994199

134 digits, and only composed of square numbers! HA!

if only there was a pattern you'd find a 266 digit one.

[kar_bon]

Like this?

114491994149119144419491191491149119149149194941149914914141141919149419199199991914491999419494949149194994194114411411144999419941994144919941491199494114914949199914444949941911494919414914949941449941944191494919144919194949491491949141941144114111449194191419111

But it's not 266 digits!

[CRGreathouse]

Originally Posted by science_man_88

mersenne numbers have the equation a(n) = 3a(n-1) - 2a(n-2) which is weird as it's the same equation to fit the sequences I just talked of starting with the respective prime exponent that doesn't give a prime.

(14:39) gp > findrec([1,3,7,15,31,63,127])
Recurrence relation is a(n) = 3a(n-1) - 2a(n-2).
3 d.f.

They do. But remember that this script only suggests what relations are possible, not that they apply to your sequence. If numbers of the form 2^n - 1 form a 3-term homogeneous linear recurrence relation (and they do!) then it is the one listed. But a sequence could fit a recurrence relation for some length, then deviate from it. For example, using the first six terms of A006261 generates a recurrence, but it's not the right one!

(It happens in that case that there still is a nice recurrence, but you need more terms to get it.)

[science_man_88]

Originally Posted by kar_bon

Like this?

114491994149119144419491191491149119149149194941149914914141141919149419199199991914491999419494949149194994194114411411144999419941994144919941491199494114914949199914444949941911494919414914949941449941944191494919144919194949491491949141941144114111449194191419111

But it's not 266 digits!

very close though doubt I'm able to find the length of the next one with this though.

[science_man_88]

Originally Posted by CRGreathouse

They do. But remember that this script only suggests what relations are possible, not that they apply to your sequence. If numbers of the form 2^n - 1 form a 3-term homogeneous linear recurrence relation (and they do!) then it is the one listed. But a sequence could fit a recurrence relation for some length, then deviate from it. For example, using the first six terms of A006261 generates a recurrence, but it's not the right one!

(It happens in that case that there still is a nice recurrence, but you need more terms to get it.)

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 ?