20201105, 12:04  #1 
Jun 2009
683 Posts 
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 
20201105, 13:18  #2 
Sep 2002
Database er0rr
7041_{8} Posts 
Nice! Congrats!

20201105, 14:17  #3 
Feb 2017
Nowhere
4,457 Posts 
Excellent! Out of curiosity, I fed your number (I called it n) to PariGP 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 
20201105, 19:06  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9389_{10} Posts 
Great job! Congrats!

20201105, 21:52  #5 
"Daniel Jackson"
May 2011
14285714285714285714
2^{3}·3^{4} Posts 
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. 
20201106, 10:03  #6 
"Oliver"
Sep 2017
Porta Westfalica, DE
2^{2}×3^{2}×13 Posts 

20201106, 11:28  #7  
Jun 2009
683 Posts 
Quote:
Thanks! That explains why the n=0 number was provable via N1. 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 Last fiddled with by PuzzlePeter on 20201106 at 11:29 

20201111, 18:18  #8  
Aug 2020
2^{5}·3 Posts 
Quote:
Firstly: why choose the form x + y * z# +1? It's specifically the x + ... I don't get. If you want to go for N1, 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. Last fiddled with by bur on 20201111 at 18:18 

20201112, 11:37  #9 
Jun 2009
683 Posts 
1)
I didn't go for a number that is N1 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 ntuplets. 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 
20201112, 12:03  #10 
"Oliver"
Sep 2017
Porta Westfalica, DE
2^{2}×3^{2}×13 Posts 
This is the way, if FactorDB already knows a shorthand for that number:
But I do not know how to tell FactorDB a new shorter formula for a given number. 
20201112, 12:35  #11 
Jun 2003
2^{4}·307 Posts 
