Rational generating functions for the Certain Stanely-Stern Sums By Shalosh B. Ekhad Theorem Number, 1 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- \ j 1 ) Z[j] t = ----------- / 2 ----- -t - t + 1 j = 0 Let n - 1 --------' ' | | Z[i + 1] (Z[i] + Z[i + 1]) F[n](x) = | | (1 + x + x ) | | | | i = 0 Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 1 and v is a certain vector of length, 1 that are too big to display. At any rate we can use them to find the first, 42, terms starting at n=0, for the sake of the OEIS. [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801, 36472996377170786403] ----------------------------- This took, 0.011, seconds. Theorem Number, 2 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- \ j 1 ) Z[j] t = ----------- / 2 ----- -t - t + 1 j = 0 Let n - 1 --------' ' | | Z[i + 1] (Z[i] + Z[i + 1]) F[n](x) = | | (1 + x + x ) | | | | i = 0 Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ 2 H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 9 and v is a certain vector of length, 9 that are too big to display. At any rate we can use them to find the first, 42, terms starting at n=0, for the sake of the OEIS. [1, 3, 15, 79, 433, 2393, 13289, 73893, 411145, 2288081, 12734645, 70878521, 394500829, 2195750337, 12221338077, 68022847185, 378609022277, 2107303781033, 11729063786477, 65282917454017, 363358867159109, 2022422889211849, 11256624551926317, 62653363449827121, 348723006063657557, 1940960999778000729, 10803214979192674877, 60129726408905266881, 334676668471180024261, 1862780343574291404297, 10368068453227287593453, 57707739842555446334257, 321196108297252004006133, 1787748753750740701515065, 9950449348469504297701469, 55383309331766437238417377, 308258536385587055973303909, 1715739387932915960518125289, 9549651671680537079702679629, 53152505381543738714113682705, 295842081519384100431249520213, 1646630513110410967949318055767] ----------------------------- This took, 0.010, seconds. Theorem Number, 3 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- \ j 1 ) Z[j] t = ----------- / 2 ----- -t - t + 1 j = 0 Let n - 1 --------' ' | | Z[i + 1] (Z[i] + Z[i + 1]) F[n](x) = | | (1 + x + x ) | | | | i = 0 Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ 3 H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 35 and v is a certain vector of length, 35 that are too big to display. At any rate we can use them to find the first, 42, terms starting at n=0, for the sake of the OEIS. [1, 3, 27, 255, 2559, 26157, 269133, 2776767, 28674477, 296222925, 3060507969, 31622105337, 326734415487, 3375995779845, 34882675134387, 360427597299525, 3724143414213591, 38479976748616845, 397597108240464603, 4108200565870109895, 42448276378247529201, 438599854265890177137, 4531864391670786975465, 46825813247478685056621, 483831067480852838710929, 4999219140577695076339887, 51654789647067772840363011, 533726811817790316656633271, 5514770490776029008220728165, 56981760879391125847430085717, 588768123414411937503895588317, 6083488783061991858755887727157, 62858083346999319095951452359177, 649485645976571653455484950253293, 6710856931493418295329173629224681, 69340409651790296277004972401046239, 716464746568230785709614742572138367, 7402923283159829028434904573484272519, 76491234773029874581989310474928108859, 790351159036379972183686028777329721211, 8166359929260769620964676536019118480369, 84379498570667924607880915301165592107643] ----------------------------- This took, 0.175, seconds. Theorem Number, 4 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- \ j 1 ) Z[j] t = ----------- / 2 ----- -t - t + 1 j = 0 Let n - 1 --------' ' | | Z[i + 1] (Z[i] + Z[i + 1]) F[n](x) = | | (1 + x + x ) | | | | i = 0 Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ 4 H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 95 and v is a certain vector of length, 95 that are too big to display. At any rate we can use them to find the first, 42, terms starting at n=0, for the sake of the OEIS. [1, 3, 51, 871, 15901, 301781, 5753009, 110221965, 2113573489, 40554660269, 778238375765, 14935431364805, 286634581508833, 5501020400081349, 105574379018989929, 2026162291995057861, 38885704466342620025, 746286786052626616661, 14322589129378193335961, 274876313410075515514969, 5275372130301634595168741, 101243904072480173854797877, 1943053087534432263993892557, 37290692559762532646762815101, 715675634559835944497245270181, 13735105966703944156981815795441, 263601451274721175037410194703913, 5058987188157013035903024738216081, 97091086737433826642158369452336493, 1863353033569459914276999491719652805, 35761104797392402052739111724564871717, 686320087117344028828827899533071579037, 13171720075467413784699381928217944476093, 252789060094639480827835322852590010376653, 4851477903979261882185759712231328438871701, 93108609383598817769836915601878330268877793, 1786922111762466143137080127172734896955198041, 34294257584176692279771816973767302080697031097, 658168644009816548165520904220501521124346761441, 12631443118266884041072480968252281998753443286937, 242420171033903108543554493740160311230323127906037, 4652480225249996819599663754102569621153643283687871] ----------------------------- This took, 2.026, seconds. Theorem Number, 5 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- \ j 1 ) Z[j] t = ----------- / 2 ----- -t - t + 1 j = 0 Let n - 1 --------' ' | | Z[i + 1] (Z[i] + Z[i + 1]) F[n](x) = | | (1 + x + x ) | | | | i = 0 Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ 5 H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 210 and v is a certain vector of length, 210 that are too big to display. At any rate we can use them to find the first, 42, terms starting at n=0, for the sake of the OEIS. [1, 3, 99, 3087, 101751, 3602733, 127251429, 4529545887, 161369015901, 5754036811533, 205192230061449, 7317994828787577, 260990929387738911, 9308145475084216173, 331971647724043959339, 11839658958806007164397, 422257470737951458164111, 15059671821896728159352013, 537098149865644422197366499, 19155425449896022041374987967, 683171820602081310672880962681, 24365093668454819055931389892833, 868972885815474600060909179382009, 30991626223604723682310279350996237, 1105305943803858528477478672399726641, 39420365374550631265223994389019143463, 1405914095457883256642616663892904179419, 50141454169527931256610007575490107806367, 1788278127617117996738995670942736298698981, 63778338994582256275565689603307186638143213, 2274633046216103570614095123877529636099437749, 81124023869230761884068498539019886522922204477, 2893261073333259707794779792465374796906498005001, 103187184747625393972922856642606574318657344635373, 3680136298199148315112481747043357594776690999935929, 131250825443558850397858877992109202246980461174233567, 4681016621054314814872921503109363263392493174840725151, 166946886105561379190513849847849658328512249558793979663, 5954104639360535812698374228365592237738312752408243101339, 212351142830172787249054595945616948720758642329412079977187, 7573432210644417049450144849754984721161971753322690381498601, 270103916959360923873214329765717237142803371795176390547017013] ----------------------------- This took, 21.019, seconds. Theorem Number, 6 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- \ j 1 ) Z[j] t = ----------- / 2 ----- -t - t + 1 j = 0 Let n - 1 --------' ' | | Z[i + 1] (Z[i] + Z[i + 1]) F[n](x) = | | (1 + x + x ) | | | | i = 0 Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ 6 H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 406 and v is a certain vector of length, 406 that are too big to display. At any rate we can use them to find the first, 42, terms starting at n=0, for the sake of the OEIS. [1, 3, 195, 11239, 663853, 44060333, 2884670009, 190789343853, 12634604224945, 837631165808261, 55534210517369165, 3682296808183928621, 244158511630004049529, 16189359306185456000277, 1073462719483485096551337, 71177783962304274355649685, 4719562967745356346450573737, 312938564528959468382043230573, 20749916101482974687772908180057, 1375857941232760977118231759928977, 91228563644912914126745627380837229, 6049062575558086068420991396105166029, 401093215472777817976993133662521098397, 26595156742818344685005015347029719751381, 1763436364796638024292550508439395768720757, 116927598595193555089613284063916245227363849, 7753080057322173360970168716291012550074909257, 514080944926575221402942021505543056532378111321, 34086997164996391530787427666405425375179048317341, 2260195378158825901686122972926777079365695901486197, 149866036093360633979088020129946793902507132330929693, 9937118264803569901690595972126205307106642509590230317, 658897252391203523338556242530279210848483015651853072573, 43689284724160105833721560612902151350288798997835412023685, 2896891120401000346705868051734863100093920326661149894228309, 192083212541526647449581877000971931970964579196680143424599857, 12736398782970461347100740953918702494972584392351199416834048129, 844508230638748772080496715066262654046939768446263304021965118369, 55996531183538718737876340065027972600362344079851103378695478678169, 3712943688207024256534056523844545590578053326526145139542523576647345, 246192943391626222374182466205898906062185546473843609890099490161352893, 16324235018253517224393735478424868856830079332968248777302704248703470157] ----------------------------- This took, 157.801, seconds. Theorem Number, 7 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- \ j 1 ) Z[j] t = ----------- / 2 ----- -t - t + 1 j = 0 Let n - 1 --------' ' | | Z[i + 1] (Z[i] + Z[i + 1]) F[n](x) = | | (1 + x + x ) | | | | i = 0 Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ 7 H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 714 and v is a certain vector of length, 714 that are too big to display. At any rate we can use them to find the first, 42, terms starting at n=0, for the sake of the OEIS. [1, 3, 387, 41775, 4391559, 548649477, 66656236893, 8189115848847, 1008565975859997, 124375167875786085, 15337286429480737089, 1891562000629566444177, 233277819961925646841767, 28769433504493967563887885, 3548040879533305687950056907, 437568163586773812978322082445, 53963796720157815107104391843391, 6655170866157616974809778538474725, 820759357381580156536094917707139203, 101221428111662170105283009987117170215, 12483290207895001845116885080197628862961, 1539521185833317921650984167968380804447977, 189863844583134448582702936159972810820866665, 23415253886217222962381469405544779886066729941, 2887722596134458068374223898588744716742820289169, 356132879493996850244666954132442898842026150823567, 43920641136058663325131310068863453349837360872885771, 5416581362794905939640639919734939810837349749565312151, 668008319108006332838986278163094921083754483936137891125, 82383164676696686319155671725960602532862521302408365478157, 10160031885831923134730205259479598623340632635655621597224157, 1253001730707737423255522789713251879932143696374901081496045757, 154528386800216053633915570233670257209984750682301161061158926457, 19057453586746986227802990568567181710641007918263321571817146077813, 2350290097058751155550541447104011814291093182405097231698539406445081, 289853180813928618786137349727040474999032528247407939289453805253625439, 35746594232385029819446982079719776819636998209425468135699617612812199207, 440850431803637752051829435966447234665227814429973026210259857159923216\ 8679, 5436856500454556479430379417761830860840585691149875206914807075201\ 17395548579, 670508781983029382089674803176476051713642249007713227188365\ 28718955277310996731, 826915381487037007257218286776309871459519276886407\ 8705724042527089247201102380209, 1019806252376808852815566127106142464205\ 339066939805336630733186463532204057199483757] ----------------------------- This took, 929.988, seconds. ----------------------------------------- This concludes this article that took, 1537.102, seconds to produce.