20200920, 20:09  #1 
(loop (#_fork))
Feb 2006
Cambridge, England
1929_{16} Posts 
largest n such that n^2+1 has prime factors within a set
To find Machinlike formulae for pi, I want to find sets of N where N^2+1 has only small prime factors. Tangentially, it would be nice to have a proof that, for example, n=485298 is the largest number where n^2+1 has no prime factor greater than 53.
(asking the same question about n^21 gives you combinations of hyperbolicarccotangents which sum to the logarithms of small primes, which combined with efficient seriessumming tools give quite good expressions for said logarithms  I've used ycruncher to get ten billion digits of log(17) without difficulty) And I've not the faintest clue where to start for this sort of question. (annoyingly, ycruncher has quite a large perterm overhead, so getting faster convergence for the individual arctangents doesn't win if you have to sum more terms; so Code:
? lindep([Pi/4,atan(1/485298),atan(1/85353),atan(1/44179),atan(1/34208),atan(1/6118),atan(1/2943),atan(1/1772),atan(1/931)]) %64 = [1, 183, 215, 71, 295, 68, 163, 525, 398]~ ? lindep([Pi/4,atan(1/485298),atan(1/330182),atan(1/114669),atan(1/85353),atan(1/44179),atan(1/34208),atan(1/12943),atan(1/9466),atan(1/5257)]) %80 = [1, 808, 1389, 1484, 2097, 2021, 1850, 1950, 398, 2805]~ 
20200920, 20:17  #2  
"Robert Gerbicz"
Oct 2005
Hungary
2×3^{2}×83 Posts 
Quote:
This problem is solvable by https://en.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem using a Pell type equation, where on the right side there is a 1 not 1. 

20200920, 20:33  #3 
"Robert Gerbicz"
Oct 2005
Hungary
10111010110_{2} Posts 
See also for lots of big arctan formulas: http://www.machination.eclipse.co.uk/FSChecking.html
(last update was at 2013) 
20200921, 03:23  #4  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×3×797 Posts 
Quote:
Ah memories, memories... I remember that I submitted that computation to a school informatics (which was just beginning in the USSR) conference  and went on to present it in my first talk in my life in 10^{th} grade  and the other viral problem I learned from a talk next to mine was the 'couch problem' where another kid was just cutting pieces from a digital rectangle, and he didn't get much far but he knew (and put it in his talk) at that time the best known answer of \(S = {\pi \over 2} + {2 \over \pi}\) .and I remember being able to get to that answer analytically because at that time I already knew derivatives and some trigonometry. 

20200921, 09:52  #5  
(loop (#_fork))
Feb 2006
Cambridge, England
3·19·113 Posts 
Quote:
Some of the example Machin formulae on Wikipedia clearly use a onelargeprime method which I can't find a very good description for  I've added some paragraphs to make the examples on https://en.wikipedia.org/wiki/Machinlike_formula a bit less unmotivated. 

20200921, 17:39  #6 
Romulan Interpreter
Jun 2011
Thailand
2^{4}×13×47 Posts 
About that age (a bit younger actually), one colleague of mine and me, after we learned from the teacher that 22/7 is a "good" approximation of \(\pi\) (known to the old Greeks too, of course), we started a "quest" to find better approximations, by increasing the denominator little by little and looking for a "suitable" numerator. No computers, all on paper, and with "pocket" calculators (which also, were kind of huge and expensive toys at the time, we call them pocket today, but in those times they were not called so, and you needed a backpack to carry them, but well... both our fathers had some accountingrelated jobs...). We did this "research" for few weeks, actually, and we found few "better" approximations, of which we were very proud, and almost ready to show them to the teacher, when we realized suddenly that you can get more accurate approximations, in fact as accurate as you want, just by writing first n digits of \(\pi\) over the corresponding power of 10, and reducing the fraction.
This is not a joke, haha, we were soooooooo disappointed! Last fiddled with by LaurV on 20200921 at 17:48 
20200927, 08:10  #7 
(loop (#_fork))
Feb 2006
Cambridge, England
6441_{10} Posts 
An oddity of this form
It's absolutely obvious in retrospect, but, looking at the results of a matchuplargeprimes run, if we have t^2+1=pq then (t+kq)^2+1 will also be divisible by q; and we may occasionally be lucky enough for (t+kq)^2+1 to have only small prime factors other than q.
eg t=18514 has t^2+1=29*11819593 and (t+11819593)^2+1/11819593 is divisible only by 2,5,13,17,29,37, so we can use 18514 and 11838107 instead of 29. Otherwise the large prime approach isn't all that fertile: p=137593 is the largest to appear twice as the largest factor of an otherwise17smooth n^2+1 for n<10^7 (n=1173 and 3853777) ; p=68761153 (n=18542 and 343824307) for 37smooth and n<10^9 Last fiddled with by fivemack on 20200928 at 19:57 
20200929, 17:20  #8 
"Robert Gerbicz"
Oct 2005
Hungary
2×3^{2}×83 Posts 
Don't know why not use this colossal 57 terms(!) arctan formula what I've found, its efficiency isn't competing with Chudnovsky, but using a bunch of computers we could get the result quicker in Wall time, since all terms can be computed paralel, and using num=57 arctan terms we can get the final sum basically using only O(log(num)) big number additions time (and multiplying by a "small" constant integer in c[]).
For the smallest term in the sum(arctan(n)) we have n>10^20, so we get more than 20 digits per term, better than Chudnovsky. Code:
c=[212346171621984379202607910, 141986132978022176645261831, 19188947083479808676847750, 72976183437824305758327029, 90487680380658315708343594, 311666636439147152580655021, 164886394092602675087156920, 277675188573061374603700591, 328918225746279750915200446, 115228975332701852008106265, 91695937787306382534509492, 274816818440262075651640693, 15278882553455239903148046, 12345819849357697383382739, 157487043779043149246827005, 165649173043654388123361981, 268335731435818979971293832, 74376668880349669845919200, 136746152203763097091123740, 106971635007532827887780437, 105558968251925687253287026, 82410840215405324255866021, 296511073341938960924412299, 79414242640647747520579196, 26575505338669030526976157, 42582496871221199838103045, 431056435239179795449215, 322897940072599977312538725, 5058363070279926676095997, 138150568339123858964501887, 100697699225584221681812015, 40360165609976142590233256, 32168480347955895535959243, 265774660195351767787225477, 19927028999571156486476849, 9604760200607125790956233, 388197118646979923787984357, 342841339813260618645453450, 178967052427653826777184469, 243278199242825683334544770, 32735042905672245875593049, 380865428210048809909621749, 215830479721667949495349715, 4859860087340886670720953, 306797318475862261176909614, 253850710497913248888215152, 99683692694159392561113651, 171658379893183050731039940, 89191773347130083621036329, 172802843689931488647278961, 96455659594184508250880480, 107781332940660320060027780, 195623057567176555762442409, 82697175828995156518216025, 8171045079609761562016517, 35001730194790928786362252, 28720454810100608397545222]; t = [100706129803452075294, 114063843547135341423, 117387028264098620557, 118182626635495860199, 120422248000399031137, 120870463680930344868, 120927259045345571307, 121843699825114397177, 125275271043850344818, 144552427347806978193, 151179894086004836582, 155531320434402458222, 171268442677083970343, 186312414964043780693, 200597192291437604193, 229771399727574656128, 242114657461222775367, 242526457156343868609, 252241001866777989537, 276914859479857813947, 279268215504325418912, 293274837014756552545, 306254909186162917405, 311286554505870488322, 321507762595941798843, 395467645802520991318, 397699150117042862902, 400464045964625262913, 408987081828419988057, 424370650490416068993, 431899278472593106531, 440044425799491348789, 503324067165721943132, 571415097863763305482, 647982671411101494018, 754220218301026231032, 860057504564641127682, 895965022987753171419, 904744940324446807318, 1350650129695249176568, 1474841158733738137711, 1702259183351533337068, 1707392125695342504348, 1786595743440215727323, 1866004788235399428730, 2021521390014319431432, 2149280509511211774827, 2224183918046598697675, 2262767288002926709269, 2355639885555472733772, 2627598404185081429432, 3661364551741763772965, 4256797797404613635163, 6694462477782585046432, 9443926883403066025057, 10442269772936340101219, 14218352152467117817607]; \p 10000 775078*Pisum(i=1,57,c[i]*atan(1/t[i])) \p 28 sum(i=1,length(t),log(10)/log(t[i])) vecmin(t) Code:
? realprecision = 10018 significant digits (10000 digits displayed) %3 = 6.630182182933390232 E10012 ? realprecision = 38 significant digits (28 digits displayed) ? %4 = 2.747877508222941264640834032 ? %5 = 100706129803452075294 Used the first 64 primes that is p=2 or p==1 mod 4, the largest such prime is 757. Could be able to eliminate 7 of them from lots of smooth solutions to keep only 57 primes and got the 57 terms formula. Last fiddled with by R. Gerbicz on 20200929 at 17:21 
20201001, 14:36  #9 
"Robert Gerbicz"
Oct 2005
Hungary
2×3^{2}×83 Posts 
Found 47668 positive integers x for that x^2+1 is 757smooth, downloadable at:
https://drive.google.com/file/d/1etr...ew?usp=sharing An old link giving all 200smooth solutions from Filip Najman: https://web.math.pmf.unizg.hr/~fnajman/rezplus1.html . Estimating that there could be roughly 50 missed solutions, among them maybe 30 could be found by an extended search [and note that most of these missing solutions are large so interesting]. The search is exhaustive up to a trivial bound of x<2^32. Improving a little my above arctan formula, using "only" 55 terms. Code:
c=[38700408465202267483521896, 8074294657163898941499704, 18084751995220504885485716, 10617955537629685604246870, 19381616293392725222395433, 16189431078249810956810540, 2663025114949728099146368, 50821290463616663282060093, 30425287250801654451464701, 5340157628142302862996217, 22257467534574983490163989, 14144052764618268971400705, 34131146882264238757385174, 11517420707788714186671327, 8809498563247969729331508, 9575603319763050834812608, 15013272622091770042897526, 29140822288315176159081433, 26490842270571571240632907, 5379280183188158967246185, 5063193213575554175323319, 21300699540591335158945812, 8765725584890890359073192, 17326267072532326425177716, 22969712961519988945581982, 31015402417393992158722336, 10715218695035024392773646, 6113768192429322401565231, 4959518686523357079068044, 880411196551256194486712, 27222064789916150912426297, 7028400065270324699241782, 6395517824304850284055402, 18382148784793948016890109, 2855737811750830775510733, 1068772694019506440335964, 41696581550717723841749863, 16855984420970193840253782, 17092009580615526091509951, 23678688598446556163790278, 26729067208666826400667006, 5535790741910445875384556, 23253488897611068459648879, 24301719811330032094835131, 13721545967545530609700672, 13638635186955032439915208, 18955462927775883750148916, 15175566500788591331215485, 5657889873218460002259652, 15307860016596651362908660, 12509769441463825443360783, 11660996728649361829160489, 31487859219307743314750041, 10708478645651051077334419, 22413720577850484247874341]; t=[100706129803452075294, 106655703945746057991, 113990292132078182007, 140759366414993038318, 149715090987851395482, 151179894086004836582, 155531320434402458222, 171268442677083970343, 177235659472193946346, 186312414964043780693, 242114657461222775367, 242526457156343868609, 277741650285265886109, 279268215504325418912, 293274837014756552545, 302685178196926874954, 306254909186162917405, 311286554505870488322, 363062694467323053757, 397699150117042862902, 398125775635597684856, 418334276059947230443, 422700922123169074432, 503324067165721943132, 506455457999459591693, 521654927153748407703, 525407238990959323474, 547748886534189022833, 647982671411101494018, 649758297700675498335, 650661171274734088043, 750314893492593005877, 754220218301026231032, 817599728075480257318, 821365850240730409698, 895965022987753171419, 903117827229218160068, 904744940324446807318, 1110592749392907292182, 1195553184514347168724, 1350650129695249176568, 1702259183351533337068, 3022165924225178134193, 4256797797404613635163, 5100058866488107804193, 5710642112294212610443, 6322604305061057220059, 6694462477782585046432, 7172059472210548010478, 9443926883403066025057, 14055524716836234863307, 16322365254295693911102, 63405805856857901256461, 68376738690185154260432, 372635354609714721488943]; \p 10000 430143*Pi/2sum(i=1,length(t),c[i]*atan(1/t[i])) \p 28 sum(i=1,length(t),log(10)/log(t[i])) vecmin(t) that gives: ? realprecision = 10018 significant digits (10000 digits displayed) ? %31 = 2.762575909555579263 E10012 ? realprecision = 38 significant digits (28 digits displayed) ? %32 = 2.627625297114408276801574520 ? %33 = 100706129803452075294 
20210817, 22:22  #10  
Sep 2016
2^{3}×43 Posts 
Quote:
Coming from the other thread on Pi... Wow at this formula! There is a way to (fairly) accurately estimate how fast a formula like this is for actual computer implementations. Say you have a series that you want to sum up to D digits of accuracy. First you calculate how many terms N you need. (this is fairly easy since these are usually linearly convergent) Suppose you sum up the N terms without rounding or truncation (keeping full integers). You will get a very large fraction where the numerator and the denominator are roughly of equal size (in digits). The # of digits in either the numerator or denominator is roughly proportional to the amount of computation that is needed to sum it up. Thus if you need to compare the speed of series, (such as Chudnovsky vs. a single ArcTan series), you can use this method to get how fast they are relative to each other.  Now the question is how do you actually calculate the size of the resulting fraction? You first have to derive the binary splitting recursion for it. http://www.numberworld.org/ycrunche...tml#CommonP2B3 Which will have the polynomials: P(x), Q(x), and R(x). Q(0, N) will be the denominator of your resulting fraction (before simplification). Thus the speed of the series is O( log(Q(0, N)) ). Where the bigO is dependent only on the hardware and software. It is roughly the same for different formulas.  How do you compute O( log(Q(0, N)) )? Q(x) will be a polynomial. (usually one that completely factorizes) So factorize it, then do a loggamma on each term individually. Complex pairs will cancel out. Last fiddled with by Mysticial on 20210817 at 22:35 

20210910, 13:23  #11  
Feb 2017
Nowhere
2·47·53 Posts 
Quote:
It occurred to me that some of the quadratic fields , d as above, have fundamental units with norm +1; and for these d, n^{2}  dy^{2} = 1 will have no solution. Of the 255 values of d which are (nonempty) products of the eight primes 2, 5, 13, 17, 29, 37, 41, and 53, there are 70 such values of d, given below. Not a huge proportion, but it might help a little: [34, 205, 221, 377, 410, 689, 1394, 1517, 1537, 1802, 1885, 1961, 3034, 4810, 4930, 5945, 6290, 7685, 10730, 11713, 15170, 19610, 19981, 23426, 25493, 26129, 27898, 30914, 33337, 45305, 56498, 56869, 81770, 99905, 117130, 139490, 141245, 197210, 257890, 282490, 335257, 339677, 433381, 439930, 739297, 747881, 804010, 960466, 1478594, 1638442, 1676285, 2166905, 2401165, 3396770, 4096105, 4833865, 7478810, 9667730, 9722453, 13668170, 17768621, 19444906, 23316290, 30311177, 39637693, 97224530, 125680490, 177686210, 303111770, 396376930] 

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