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

[QUOTE=science_man_88;227051]good now I can check Mersenne exponents for things to eliminate.[/QUOTE]

On my main computer I can just type
[code]findrec(bfile(43))[/code]
where 43 means A000043. :smile:

[QUOTE=CRGreathouse;227052]On my main computer I can just type
[code]findrec(bfile(43))[/code]
where 43 means A000043. :smile:[/QUOTE]

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

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

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.

[QUOTE=science_man_88]I was trying to find patterns in what to eliminate but cool.
[/QUOTE]

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

[QUOTE=3.14159;227056]Mersenne numbers do not have any patterns, they're randomly distributed and have no covering set of divisors (Besides having Proth numbers as factors)[/QUOTE]

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.

[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=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]

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

Like this?
[code]
114491994149119144419491191491149119149149194941149914914141141919149419199199991914491999419494949149194994194114411411144999419941994144919941491199494114914949199914444949941911494919414914949941449941944191494919144919194949491491949141941144114111449194191419111
[/code]

But it's not 266 digits!

[QUOTE=science_man_88;227057]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.[/QUOTE]

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.)

[QUOTE=kar_bon;227064]Like this?
[code]
114491994149119144419491191491149119149149194941149914914141141919149419199199991914491999419494949149194994194114411411144999419941994144919941491199494114914949199914444949941911494919414914949941449941944191494919144919194949491491949141941144114111449194191419111
[/code]

But it's not 266 digits![/QUOTE]

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

[QUOTE=CRGreathouse;227065]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.)[/QUOTE]

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 ?