New quintuplet
 

Hey folks,


I found a nice prime quintuplet with 2002 decimal digits:


208488047305875799520424701159167603377978422253392827160551251670223210710614891907691541641382959723840996115289
318958605601286742246600010993517553901917394791881242104509064316215125334249990797768654323426401905700185762312
973101502417203414627615021877578347111556187226343129839878723543239514028859452273425017503776772488342168523806
555337714869173621580572315965374487147261226083647531568414780490982853232593685829566630669932630633444470918211
999633928609312118325875576920813714971662525835773899876739935144406350787555225478897037692667108951185837843041
0838456903480935830717252140156552040585940635622248479907218514227211263116845106658500348448430245301193675808540
102312557308426189637949823955941331873526535989406821435935541901709197482536077556126683973700169420737704263717
7467437249933177017065556829972602737526861578434084970782610239038458552601165483935466757814251146751103455217907
938078726693588103543543215280762586648367358605400185429330843021363708934677909890094218218782170606923150871656
839935178315944646490016690650863363285274169871242010749105092851693551791970123763781980690849253348754715396479
343986706783977573037060120856202411942545583562477337489914582426493532570855073018623659858737360575480961687314
762149322530154936216219512117333717542891211630756537196951271136555779891502809256677682113348703145008753212048
363503561735943243131797934510724709931261748493082683279073546586797759546699446440960062799507217666079471552602
999052843969789148580931001292018496575476777793944186237715158967924650689029824387931347019236468446262496452213
812196871763313279190999542300857278929324627719371003047188374237595434032103664647852167959501577854479390284138
939195918299170673301325066341948377598278982844668697073382594339778117319834158507613939093712586551664750148505
937985253904759741161696571847949280459369937911020386653751854775217407899090324931530875891438004994439497028783
61580792169756655089750911343810427617912385001801+464583344041*4657#+d


with d=0, 2, 6, 8, 12

Nice! Congrats! :smile:

Excellent! Out of curiosity, I fed your number (I called it n) to Pari-GP to see if there were any other small d's for which n+d might be prime. I excluded multiples of 2, 3, and 5. Assuming my mindless script was writ right, the answer is no.

? forstep(d=-1680,1680,[2,4,2,4,6,2,6,4],if(ispseudoprime(n+d),print(d)))
0
2
6
8
12

Great job! Congrats!

That can be rewritten as:

126831252923413*4657#/273+1+n, n=0,2,6,8,12

I figured out (using FactorDB) that the long number in the expression is simply 220*4657#/273+1.

[QUOTE=Stargate38;562326]That can be rewritten [...].[/QUOTE]

Thanks, that's [I]way[/I] easier to handle. :smile:

[QUOTE=Stargate38;562326]That can be rewritten as:

126831252923413*4657#/273+1+n, n=0,2,6,8,12

I figured out (using FactorDB) that the long number in the expression is simply 220*4657#/273+1.[/QUOTE]


Thanks!

That explains why the n=0 number was provable via N-1. The long number fell out of a script I knocked together for CRT. I didn't investigate further as I was happy enough it ran without error messages :cmd:

[QUOTE=Puzzle-Peter;562403]The long number fell out of a script I knocked together for CRT. I didn't investigate further as I was happy enough it ran without error messages :cmd:[/QUOTE]Two questions... ;)


Firstly: why choose the form x + y * z# +1? It's specifically the x + ... I don't get. If you want to go for N-1, wouldn't that require a product + 1? So why not test x * z# + 1?



Secondly: if the long number x = 220*4657#/273+1, then why does the whole end up as 126831252923413*4657#/273+1+n?


(Thirdly: how could you figure out it is a term containing a 1/273 and +1 by using FactorDB?)


These are maybe obvious, but not to me. :smile:

1)
I didn't go for a number that is N-1 provable. I was looking for quintuplets, so Primo would be needed anyway. The problem is that the sieve files get very sparse when you search for n-tuplets. the bigger the n, the worse it gets. Using the Chinese Remainder Theorem (CRT) to find a suitable number "x" you can make sure none of your candidates will be divisible by a small prime (in this case no number smaller z=4657 will divide a candidate). You end up with a much denser sieve file, i.e. the number of remaining candidates per, say, 1G of the running variable is much higher and sieving efficiency is much better.


2)
N = 220*4657#/273+1 + 464583344041*4657#
factoring out 4657#/273 you get

(4657#/273) * (220+273*464583344041) +1
= (4657#/273) * 126831252923413 +1


3)
That's an answer I'd like to know myself :redface:

This is the way, if FactorDB already knows a shorthand for that number:[LIST=I][*]Take a number, let's say [C](287834098572304895723840598273045234-176435783429783541914131513478)/519589773265891366699777[/C] (yes, some random dumb number) or [URL="http://factordb.com/index.php?query=%28287834098572304895723840598273045234-176435783429783541914131513478%29%2F519589773265891366699777"]this[/URL].[*]The third column, "number", has the number displayed as link, followed by its factors as links. Click on the very first one, the one of the number itself.[*]We get to [URL="http://factordb.com/index.php?id=553963794028"]this[/URL] site. The search box now contains the known shorthand.[*]Profit![/LIST]
But I do not know how to tell FactorDB a new shorter formula for a given number.

[QUOTE=kruoli;562984]
But I do not know how to tell FactorDB a new shorter formula for a given number.[/QUOTE]
Just do a search for the shorter formula. If factor db thinks that it is truly a shorter formula, it will replace the current formula with the new shorter one.