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 2020-09-29, 17:20 #8 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 5BA16 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*Pi-sum(i=1,57,c[i]*atan(1/t[i])) \p 28 sum(i=1,length(t),log(10)/log(t[i])) vecmin(t) output: Code: ? realprecision = 10018 significant digits (10000 digits displayed) %3 = -6.630182182933390232 E-10012 ? realprecision = 38 significant digits (28 digits displayed) ? %4 = 2.747877508222941264640834032 ? %5 = 100706129803452075294 this comes from solving a system of linear equations, so the result is really 0, hence we got 775078*Pi. [we need a division by a small integer at the end to extract Pi]. 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 2020-09-29 at 17:21