50, Challenging Problems about Surprising Integrality of Sequences Defined by Linear Recurrences with Polynomial coefficients Problem Number, 1, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 (2 n + 9) (n + 6) (n + 1) a(n) - 4 (n + 4) (3 n + 24 n + 41) a(n + 1) + (2 n + 7) (n + 7) (n + 2) a(n + 2) = 0 subject to the initial conditions a(1) = 8, a(2) = 53 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 2, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 10 9 283 (n + 4) (n + 3) (n + 2) (n + 1) (14438110231 n + 662738225989 n 8 7 6 + 13628202601818 n + 165296759653302 n + 1309321019920818 n 5 4 3 + 7075607116983402 n + 26412410376615842 n + 67231179804110873 n 2 + 111648039786652251 n + 109192787510828994 n + 47741371418942640) a(n) 11 10 - (n + 2) (n + 3) (n + 4) (5399853226394 n + 255963876359477 n 9 8 7 + 5469464654997415 n + 69502634464127244 n + 583199539552983225 n 6 5 + 3390395670975402831 n + 13921629722462661667 n 4 3 + 40335251032026782992 n + 80709426177378224531 n 2 + 106065374370136617792 n + 82240570828372096368 n + 28438468003653128064 12 11 ) a(n + 1) + (n + 3) (n + 4) (534210078547 n + 26658154675781 n 10 9 8 + 606853371251622 n + 8332153192974396 n + 76838286386850921 n 7 6 5 + 501312555129504993 n + 2372250983901503981 n + 8202153203791175323 n 4 3 + 20560819405366671237 n + 36434812439757677391 n 2 + 43314258842936974332 n + 31010415982932836196 n + 10109500952507693280) 12 11 10 a(n + 2) + (144381102310 n + 7132716117975 n + 163281692766338 n 9 8 7 + 2292542874252729 n + 21989831681982015 n + 151678729265248020 n 6 5 4 + 770208020240645474 n + 2894658493107608757 n + 7971317835346312525 n 3 2 + 15646385116051071375 n + 20728716572945035818 n 2 + 16606657854076785624 n + 6072946824139243200) (n + 4) a(n + 3) + ( 14 13 12 -49739289745795 n - 2880004665481645 n - 76949853207826911 n 11 10 - 1257199715921288357 n - 14028121147823709860 n 9 8 - 113057352125688942264 n - 678476991312178426310 n 7 6 - 3078953469033127913405 n - 10613466771898237561473 n 5 4 - 27644059320308037362713 n - 53529566444159676376879 n 3 2 - 74685606372346722030768 n - 70931531140932775555332 n - 41016911634790835553648 n - 10886222994237643040640) a(n + 4) + 8 10 9 (4 n + 21) (2 n + 11) (4 n + 23) (n + 6) (14438110231 n + 518357123679 n 8 7 6 + 8313273528312 n + 78397141746642 n + 481195367364042 n 5 4 3 + 2007600217172430 n + 5762304671446688 n + 11227333748984283 n 2 + 14199855621949287 n + 10516587268476006 n + 3459251313041040) a(n + 5) = 0 subject to the initial conditions a(1) = 10, a(2) = 112, a(3) = 1318, a(4) = 15955, a(5) = 196731 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 3, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 6 5 62 (2 n + 3) (n + 3) (n + 2) (n + 1) (28999382 n + 577461015 n 4 3 2 + 4758631849 n + 20768185983 n + 50620094117 n + 65323567662 n 8 7 + 34864803472) a(n) - 2 (n + 3) (n + 2) (2899938200 n + 66445916100 n 6 5 4 + 655398655784 n + 3629639393839 n + 12323740018513 n 3 2 + 26219187324855 n + 34061780014603 n + 24647647699370 n + 7589609142096 9 8 7 ) a(n + 1) - 2 (n + 3) (24359480880 n + 619044397440 n + 6901939963747 n 6 5 4 + 44283873684575 n + 180071481573777 n + 480872586730973 n 3 2 + 842600344157736 n + 933228478491272 n + 592136308149536 n 10 9 + 163767837571584) a(n + 2) + (-98510900654 n - 2848233173841 n 8 7 6 - 36686930461687 n - 277089967306774 n - 1358263526234752 n 5 4 3 - 4512566908222669 n - 10284063369144719 n - 15864554531863180 n 2 - 15843193709431500 n - 9242668681750464 n - 2390360715596160) a(n + 3) 6 5 + 8 (4 n + 17) (2 n + 7) (4 n + 15) (n + 4) (28999382 n + 403464723 n 4 3 2 + 2306317504 n + 6928281097 n + 11527707842 n + 10066718764 n + 3603314160) a(n + 4) = 0 subject to the initial conditions a(1) = 7, a(2) = 72, a(3) = 825, a(4) = 9928 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 4, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 7 6 62 (2 n + 3) (n + 3) (n + 2) (n + 1) (28999382 n + 695490298 n 5 4 3 2 + 7096243323 n + 39917032546 n + 133641266479 n + 266189508568 n 9 + 291926909052 n + 135906538992) a(n) - 4 (n + 3) (n + 2) (1449969100 n 8 7 6 5 + 39124422200 n + 462246265722 n + 3134902458715 n + 13431359742310 n 4 3 2 + 37644597083534 n + 68902966174186 n + 79271420889727 n 10 + 51911097931578 n + 14711788524528) a(n + 1) - 4 (n + 3) (12179740440 n 9 8 7 + 359094497580 n + 4703624314143 n + 36020067201405 n 6 5 4 + 178448203943974 n + 597062554392378 n + 1364954262639439 n 3 2 + 2102823582386577 n + 2086455699080284 n + 1202079999032964 n 11 10 + 304803289055376) a(n + 2) + (-98510900654 n - 3249178648192 n 9 8 7 - 48204514086831 n - 424378205166463 n - 2461785267848236 n 6 5 4 - 9873153122363158 n - 27911838411507035 n - 55569744544680303 n 3 2 - 76270018823366956 n - 68640133610795004 n - 36397060935536880 n - 8597874869474688) a(n + 3) + 8 (2 n + 9) (4 n + 17) (n + 4) (4 n + 15) ( 7 6 5 4 28999382 n + 492494624 n + 3532288557 n + 13853192031 n 3 2 + 32040741935 n + 43628838625 n + 32314831670 n + 10015152168) a(n + 4) = 0 subject to the initial conditions a(1) = 8, a(2) = 85, a(3) = 987, a(4) = 11964 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 5, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 8 (n + 1) (4 n + 7) (2 n + 3) (4 n + 9) 4 3 2 8 (4633 n + 50511 n + 205872 n + 371772 n + 250976) a(n) + (-137604733 n 7 6 5 4 - 2463460342 n - 19116792428 n - 83962463622 n - 228208545115 n 3 2 - 392928901748 n - 418402906412 n - 251841652608 n - 65586097152) a(n + 1) + 64 (4 n + 11) (2 n + 5) (4 n + 13) (n + 3) 4 3 2 (4633 n + 31979 n + 82137 n + 93029 n + 39198) a(n + 2) = 0 subject to the initial conditions a(1) = 11, a(2) = 134 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 6, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 5 (n + 1) (5 n + 6) (5 n + 7) (5 n + 8) (5 n + 9) 4 3 2 (57708 n + 560592 n + 2037554 n + 3284037 n + 1980418) a(n) + ( 9 8 7 6 -49862943648 n - 858354958512 n - 6508353895504 n - 28523900473552 n 5 4 3 - 79616489815120 n - 146750605586848 n - 178599000400696 n 2 - 138373502531128 n - 61923820736832 n - 12194229852000) a(n + 1) + 80 (5 n + 11) (5 n + 12) (5 n + 13) (5 n + 14) (n + 2) 4 3 2 (57708 n + 329760 n + 702026 n + 659873 n + 231051) a(n + 2) = 0 subject to the initial conditions a(1) = 11, a(2) = 151 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 7, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 2 (n + 1) (n + 3) a(n) - (2 n + 5) (15 n + 75 n + 89) a(n + 1) 2 + (n + 4) (n + 2) a(n + 2) = 0 subject to the initial conditions a(1) = 29, a(2) = 806 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 8, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 2 (n + 1) (n + 3) a(n) - (2 n + 5) (17 n + 85 n + 101) a(n + 1) 2 + (n + 4) (n + 2) a(n + 2) = 0 subject to the initial conditions a(1) = 33, a(2) = 1041 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 9, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 2 (n + 1) (n + 3) a(n) - (2 n + 5) (19 n + 95 n + 113) a(n + 1) 2 + (n + 4) (n + 2) a(n + 2) = 0 subject to the initial conditions a(1) = 37, a(2) = 1306 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 10, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 6 5 4 -29 (n + 1) (n + 2) (n + 3) (n + 6) (66348 n + 1750446 n + 19150210 n 3 2 + 111170485 n + 361061977 n + 621826774 n + 443468610) a(n) + ( 8 7 6 5 4 2653920 n + 89922240 n + 1315629346 n + 10847284212 n + 55067984575 n 3 2 + 176035151185 n + 345463573374 n + 379730808808 n + 178511006040) 9 8 7 (n + 2) (n + 3) a(n + 1) - (265392 n + 9655704 n + 157415134 n 6 5 4 3 + 1513142042 n + 9467824897 n + 40016669016 n + 114181499081 n 2 + 211717786138 n + 230877711776 n + 112461969720) (n + 3) a(n + 2) + 9 8 7 6 (n + 4) (36358704 n + 1304652096 n + 20656904462 n + 189339467122 n 5 4 3 + 1106685460533 n + 4275551204004 n + 10911986897263 n 2 + 17729410088768 n + 16628676272688 n + 6853911764760) a(n + 3) - 3 6 5 4 (3 n + 17) (3 n + 16) (n + 5) (n + 4) (66348 n + 1352358 n + 11393200 n 3 2 + 50747145 n + 125942542 n + 164967577 n + 88999440) a(n + 4) = 0 subject to the initial conditions a(1) = 19, a(2) = 353, a(3) = 6599, a(4) = 124489 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 11, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 8 7 -259 (n + 3) (n + 2) (n + 1) (125253901652 n + 4632561910856 n 6 5 4 + 74541232590301 n + 681402021725697 n + 3869501250901221 n 3 2 + 13974547881313991 n + 31334851503868282 n + 39872092580673440 n 10 + 22035526641309440) (n + 6) a(n) + 2 (20416385969276 n 9 8 7 + 908230486239098 n + 18005217865688122 n + 209340042522366943 n 6 5 + 1579622737899782435 n + 8076224368279708297 n 4 3 + 28306299255392605443 n + 67077599143952503142 n 2 + 102710246116573610164 n + 91611083207303636520 n + 36069488369144771520 11 10 ) (n + 3) (n + 2) a(n + 1) - (1628300721476 n + 76506312055888 n 9 8 7 + 1651372205958791 n + 21613324651622224 n + 190419471335255404 n 6 5 + 1183961367730826478 n + 5290564393781123469 n 4 3 + 16952211191847680106 n + 38083944451319800700 n 2 + 57005318524736733544 n + 51066481341372255200 n + 20705380749824067840) 12 11 (n + 3) a(n + 2) + (-1252539016520 n - 63234895831580 n 10 9 8 - 1489207240059700 n - 21656678082805460 n - 216500884607135940 n 7 6 - 1564452533540235700 n - 8353764408611581620 n 5 4 - 33093036247818362580 n - 96170517648488237740 n 3 2 - 199259529957293994760 n - 278576424065688681440 n - 235378255792005580800 n - 90719580650957832960) a(n + 3) + ( 12 11 10 3562847232491140 n + 181653084809174880 n + 4215941839016849245 n 9 8 + 58878220897431384565 n + 550893028225001829140 n 7 6 + 3636724297934683102610 n + 17362147895633173163225 n 5 4 + 60372580622981027778845 n + 151681908063672693225610 n 3 2 + 268390426138901295201660 n + 317283862686923083473320 n + 224851379699915394874240 n + 72183212567645093280000) a(n + 4) - 8 8 (4 n + 25) (2 n + 11) (4 n + 23) (n + 5) (125253901652 n 7 6 5 + 3630530697640 n + 45620408460565 n + 324424207819355 n 4 3 2 + 1427237737362931 n + 3974863891547505 n + 6838536946049460 n + 6639216828523420 n + 2781870836946912) a(n + 5) = 0 subject to the initial conditions a(1) = 91, a(2) = 8704, a(3) = 860545, a(4) = 86941495, a(5) = 8916622705 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 12, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 (2 n + 5) (n + 1) a(n) + (-20 n - 80 n - 74) a(n + 1) + (n + 3) (2 n + 3) a(n + 2) = 0 subject to the initial conditions a(1) = 7, a(2) = 57 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 13, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 (n + 5) (n + 4) (n + 1) a(n) - (2 n + 7) (7 n + 49 n + 80) a(n + 1) + (n + 3) (n + 2) (n + 6) a(n + 2) = 0 subject to the initial conditions a(1) = 19, a(2) = 295 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 14, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 (2 n + 9) (n + 6) (n + 1) a(n) - 10 (n + 4) (2 n + 16 n + 29) a(n + 1) + (2 n + 7) (n + 7) (n + 2) a(n + 2) = 0 subject to the initial conditions a(1) = 15, a(2) = 177 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 15, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 (2 n + 9) (n + 6) (n + 1) a(n) - 4 (n + 4) (7 n + 56 n + 104) a(n + 1) + (2 n + 7) (n + 7) (n + 2) a(n + 2) = 0 subject to the initial conditions a(1) = 22, a(2) = 373 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 16, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 -17 (n + 1) (n + 2) (938 n + 5303 n + 7263) a(n) 3 2 + 2 (12194 n + 87230 n + 196391 n + 136809) (n + 2) a(n + 1) 4 3 2 + (426790 n + 4120025 n + 14469130 n + 21776093 n + 11723994) a(n + 2) 2 - 8 (n + 3) (2 n + 7) (938 n + 3427 n + 2898) a(n + 3) = 0 subject to the initial conditions a(1) = 17, a(2) = 377, a(3) = 9097 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 17, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 -21 (n + 1) (n + 2) (1442 n + 8156 n + 11175) a(n) 3 2 + 2 (23072 n + 165104 n + 371856 n + 259191) (n + 2) a(n + 1) 4 3 2 + (1015168 n + 9802496 n + 34435424 n + 51842716 n + 27922566) a(n + 2) 2 - 10 (n + 3) (2 n + 7) (1442 n + 5272 n + 4461) a(n + 3) = 0 subject to the initial conditions a(1) = 21, a(2) = 576, a(3) = 17191 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 18, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 3 2 -9 (n + 1) (n + 2) (n + 6) (254 n + 3159 n + 13034 n + 17809) a(n) + 2 5 4 3 2 (1778 n + 35448 n + 272741 n + 1010656 n + 1795593 n + 1213848) (n + 2) 6 5 4 3 2 a(n + 1) + (30226 n + 678181 n + 6301131 n + 31003595 n + 85122599 n + 123527904 n + 73943244) a(n + 2) 3 2 - 4 (2 n + 11) (n + 5) (n + 3) (254 n + 2397 n + 7478 n + 7680) a(n + 3) = 0 subject to the initial conditions a(1) = 17, a(2) = 257, a(3) = 3765 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 19, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 29 (n + 1) (n + 2) (n + 3) 5 4 3 2 (33174 n + 595944 n + 4239524 n + 14920437 n + 25960397 n + 17850324) 6 5 4 a(n) - (n + 3) (n + 2) (1326960 n + 25828200 n + 205202402 n 3 2 + 849376420 n + 1925039221 n + 2255103619 n + 1061113388) a(n + 1) + 7 6 5 4 3 (n + 3) (132696 n + 2914560 n + 27241892 n + 140418470 n + 430744611 n 2 8 + 785383767 n + 786837706 n + 333540904) a(n + 2) + (-18179352 n 7 6 5 4 - 462922452 n - 5089071778 n - 31515390068 n - 120111076173 n 3 2 - 288099695304 n - 424053372707 n - 349522255918 n - 123222341400) a(n + 3) + 3 (3 n + 14) (3 n + 13) (n + 4) 5 4 3 2 (33174 n + 430074 n + 2187488 n + 5445789 n + 6620189 n + 3133610) a(n + 4) = 0 subject to the initial conditions a(1) = 13, a(2) = 209, a(3) = 3621, a(4) = 65177 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 20, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence -28 (n + 1) (n + 2) (n + 3) ( 5 4 3 2 121363 n + 2182324 n + 15540534 n + 54749509 n + 95363264 n + 65647344) 6 5 4 3 a(n) + (4611794 n + 89846003 n + 714504836 n + 2960651124 n 2 + 6718428477 n + 7882404204 n + 3716308960) (n + 2) (n + 3) a(n + 1) - ( 7 6 5 4 3 364089 n + 8003328 n + 75111443 n + 390123106 n + 1210149238 n 2 + 2238505011 n + 2281594383 n + 986249756) (n + 3) a(n + 2) + ( 8 7 6 5 128644780 n + 3278099290 n + 36063781289 n + 223509942214 n 4 3 2 + 852563012331 n + 2046869800323 n + 3015895181976 n + 2488725108049 n + 878570672244) a(n + 3) - 3 (3 n + 14) (3 n + 13) (n + 4) 5 4 3 2 (121363 n + 1575509 n + 8024868 n + 20008221 n + 24363367 n + 11554016) a(n + 4) = 0 subject to the initial conditions a(1) = 25, a(2) = 777, a(3) = 26033, a(4) = 906321 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 21, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 7 6 -1819 (n + 1) (n + 2) (n + 3) (n + 4) (101469628444924 n + 2841064564427428 n 5 4 3 + 33766915323133159 n + 220643288481065992 n + 855157734493703557 n 2 + 1963367457418044700 n + 2468555532695031240 n + 1308551826284533800) 8 7 a(n) + 2 (116385663826327828 n + 3433279551137751658 n 6 5 + 43572344699441812908 n + 310056624290956277893 n 4 3 + 1349364603861707124516 n + 3664817982583106771749 n 2 + 6038143963919157800768 n + 5483153245306008760980 n + 2082392739699349436100) (n + 2) (n + 3) (n + 4) a(n + 1) - ( 9 8 7 9842553959157628 n + 314953478586091028 n + 4360403139065601841 n 6 5 + 34139594179380373073 n + 165632376627716087257 n 4 3 + 511988361341892386867 n + 994268235437484035754 n 2 + 1139326264120653522432 n + 657729810765349998120 n 9 + 117837796839633979200) (n + 3) (n + 4) a(n + 2) - 28 (253674071112310 n 8 7 6 + 8751542873298585 n + 133212126150832653 n + 1174357707335025978 n 5 4 + 6608531516294222388 n + 24620976965161838517 n 3 2 + 60736651149490308605 n + 95672532148270358394 n + 87325523880648116670 n + 35189748485719232400) (n + 1) (n + 4) a(n + 3) 11 10 + (15764828823345617260 n + 630579941932414758340 n 9 8 + 11350135366078916363719 n + 121255529183565607916694 n 7 6 + 853479371195291359580670 n + 4151252888546083343081352 n 5 4 + 14218183651716380490414071 n + 34234002971644577706559766 n 3 2 + 56665860182685823335313880 n + 61241103948195130692470808 n + 38747575228890622695002640 n + 10816190746089150346444800) a(n + 4) - 56 7 (2 n + 9) (4 n + 21) (n + 5) (4 n + 19) (101469628444924 n 6 5 4 + 2130777165312960 n + 18851390133911995 n + 90873243336239277 n 3 2 + 256983879507794959 n + 424569937861066407 n + 377219144044004238 n + 137821984607759040) a(n + 5) = 0 subject to the initial conditions a(1) = 43, a(2) = 2794, a(3) = 201951, a(4) = 15326130, a(5) = 1196224856 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 22, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 8 7 -2075 (n + 4) (n + 3) (n + 2) (n + 1) (21490654648928 n + 709091352566160 n 6 5 4 + 10153557714531983 n + 82374787352894049 n + 413946063356689574 n 3 2 + 1318686665202513678 n + 2599194210373296639 n + 2896260579678165849 n 9 + 1395838558405628820) a(n) + (56176571252297792 n 8 7 + 1937829652486388928 n + 29310400752211191322 n 6 5 + 254815075520456907829 n + 1401115510219577217040 n 4 3 + 5044083457127399500771 n + 11863116070739019634559 n 2 + 17528749446571567934980 n + 14714936050409510761359 n + 5323913683213941860220) (n + 4) (n + 3) (n + 2) a(n + 1) - ( 10 9 8 2299500047435296 n + 85070774914320304 n + 1397624032336737253 n 7 6 + 13414825217884021455 n + 83203106707174444941 n 5 4 + 347867182723560575229 n + 990645449131280258822 n 3 2 + 1891323469526070847100 n + 2305939246354808491568 n + 1609826297873111640432 n + 483214362341151562560) (n + 4) (n + 3) 10 9 a(n + 2) - 8 (214906546489280 n + 8272899531352640 n 8 7 6 + 143355994776968726 n + 1473159884612289533 n + 9943847222262350765 n 5 4 + 46064310060144876653 n + 148253268256569027285 n 3 2 + 327096854639222596206 n + 473020009682214358656 n + 404354906361926993376 n + 154934988081005317680) (n + 4) (n + 2) 12 11 a(n + 3) + (4353147005686855680 n + 195871308444043637760 n 10 9 + 4001425532686271930560 n + 49048611619885928492992 n 8 7 + 401534442957452542596096 n + 2311166308934075970440832 n 6 5 + 9582867993429324557814528 n + 28813836551182060209874752 n 4 3 + 62289474375659605098564800 n + 94301589503867752263138752 n 2 + 94767066492120859519744128 n + 56662166847875392102806720 n + 15211361057915356649568000) a(n + 4) - 64 (4 n + 21) (2 n + 11) (n + 5) 8 7 6 (4 n + 19) (21490654648928 n + 537166115374736 n + 5791656576738847 n 5 4 3 + 35140882809251543 n + 131061640795808434 n + 307193851693531844 n 2 + 441076907021212928 n + 353892205820405352 n + 121122756918656208) a(n + 5) = 0 subject to the initial conditions a(1) = 57, a(2) = 4353, a(3) = 363353, a(4) = 31668673, a(5) = 2831739961 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 23, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 10 9 265 (n + 1) (n + 2) (n + 3) (n + 4) (11353590151 n + 521887929869 n 8 7 6 + 10747422902178 n + 130551539670260 n + 1035716342951408 n 5 4 3 + 5606153142048296 n + 20962860711322390 n + 53455865318258145 n 2 + 88941829799482289 n + 87163832787929526 n + 38193774304865040) a(n) - 11 10 (n + 4) (n + 3) (n + 2) (3837513471038 n + 182154390502279 n 9 8 7 + 3897904266210717 n + 49607938492816016 n + 416942522243886719 n 6 5 + 2428138561898039449 n + 9989514858211410357 n 4 3 + 29003709037529243880 n + 58171750554645932657 n 2 + 76650389967255241256 n + 59614779940592227392 n + 20688687616273770240) 12 11 a(n + 1) + (n + 3) (n + 4) (215718212869 n + 10778743518987 n 10 9 8 + 246638791998454 n + 3417551817397522 n + 31935885609322713 n 7 6 5 + 211980409224774031 n + 1024504286012072621 n + 3631013815321540957 n 4 3 + 9361350523012443683 n + 17112419975750850147 n 2 + 21040398340329401100 n + 15614032981033177956 n + 5285681600370593760) 12 11 10 a(n + 2) + (113535901510 n + 5616254953975 n + 128740983157898 n 9 8 7 + 1810031811770089 n + 17384672975023289 n + 120066003121755410 n 6 5 4 + 610412277931559026 n + 2296687644943442737 n + 6331392688245213015 n 3 2 + 12440329496092200949 n + 16498100175644555310 n 2 + 13231015530508180008 n + 4843682348970312000) (n + 4) a(n + 3) + ( 14 13 12 -110073056513945 n - 6380580158247295 n - 170677974432255581 n 11 10 - 2791879211162910457 n - 31191605923453076284 n 9 8 - 251715490653461613950 n - 1512692684777025415468 n 7 6 - 6874800165116117192357 n - 23735532077048971203823 n 5 4 - 61926922676178453548453 n - 120134630924484285218831 n 3 2 - 167950375080680535432976 n - 159860348102527204812708 n - 92667767839666808849712 n - 24662622184580028159360) a(n + 4) + 8 10 9 (4 n + 21) (2 n + 11) (4 n + 23) (n + 6) (11353590151 n + 408352028359 n 8 7 6 + 6561343090152 n + 61997691110000 n + 381329074343286 n 5 4 3 + 1594478509338782 n + 4587482234813606 n + 8961551395651393 n 2 + 11366680290697389 n + 8445258800924562 n + 2788015259277360) a(n + 5) = 0 subject to the initial conditions a(1) = 28, a(2) = 880, a(3) = 29098, a(4) = 990451, a(5) = 34352251 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 24, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 3 2 -9 (n + 1) (2 n + 3) (n + 2) (4494 n + 37380 n + 102563 n + 92783) a(n) + ( 5 4 3 2 1438080 n + 16275840 n + 71934214 n + 154965280 n + 162518151 n 6 5 4 + 66332151) (n + 2) a(n + 1) + (-5096196 n - 70417998 n - 399180474 n 3 2 - 1186916673 n - 1949559153 n - 1674222006 n - 585965520) a(n + 2) + 6 3 2 (3 n + 8) (3 n + 10) (n + 3) (4494 n + 23898 n + 41285 n + 23106) a(n + 3) = 0 subject to the initial conditions a(1) = 11, a(2) = 173, a(3) = 3003 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 25, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence -42 (2 n + 3) (n + 2) (n + 1) 4 3 2 (133494 n + 1559444 n + 6770125 n + 12940493 n + 9186528) a(n) + 2 ( 6 5 4 3 234148476 n + 3437710204 n + 20592286796 n + 64299107428 n 2 + 110157314451 n + 97959244680 n + 35248319829) (n + 2) a(n + 1) + ( 7 6 5 4 -4440544416 n - 76296339504 n - 553229840592 n - 2192257507716 n 3 2 - 5121215926890 n - 7043253317712 n - 5272706511894 n - 1654981200540) a(n + 2) + 75 (3 n + 11) (3 n + 10) (n + 3) 4 3 2 (133494 n + 1025468 n + 2892757 n + 3544599 n + 1590210) a(n + 3) = 0 subject to the initial conditions a(1) = 31, a(2) = 1206, a(3) = 50601 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 26, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 3 2 -9 (2 n + 5) (n + 2) (n + 1) (749 n + 7014 n + 21652 n + 21984) a(n) + ( 5 4 3 2 239680 n + 3203200 n + 16800761 n + 43119881 n + 53973564 n + 26232624) 6 5 4 3 (n + 2) a(n + 1) + (-849366 n - 13474755 n - 87906546 n - 301541655 n 2 - 572829246 n - 570319872 n - 231884640) a(n + 2) + 3 2 6 (3 n + 10) (n + 4) (3 n + 8) (749 n + 4767 n + 9871 n + 6597) a(n + 3) = 0 subject to the initial conditions a(1) = 15, a(2) = 257, a(3) = 4643 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 27, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence -25 (2 n + 5) (n + 2) (n + 1) 4 3 2 (57293 n + 747730 n + 3634375 n + 7789736 n + 6204408) a(n) + ( 6 5 4 3 142544984 n + 2430532176 n + 17017350175 n + 62511995393 n 2 + 126789326751 n + 134243094121 n + 57742972140) (n + 2) a(n + 1) + ( 7 6 5 4 -1635715150 n - 31979839975 n - 264995918068 n - 1205408667742 n 3 2 - 3247479305656 n - 5175253948495 n - 4509907644474 n - 1654310437560) a(n + 2) + 54 (3 n + 11) (3 n + 10) (n + 4) 4 3 2 (57293 n + 518558 n + 1734943 n + 2535004 n + 1358610) a(n + 3) = 0 subject to the initial conditions a(1) = 43, a(2) = 2089, a(3) = 106903 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 28, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence -58 (2 n + 5) (n + 2) (n + 1) 4 3 2 (362178 n + 4727349 n + 22980245 n + 49260826 n + 39240816) a(n) + 4 ( 6 5 4 3 606648150 n + 10344902175 n + 72436692371 n + 266118739928 n 2 + 539817121021 n + 571633630567 n + 245924579328) (n + 2) a(n + 1) + ( 7 6 5 4 -32685840144 n - 639091753488 n - 5296187238154 n - 24093463900128 n 3 2 - 64916805125854 n - 103465368898896 n - 90176354144056 n - 33083848405440) a(n + 2) + 147 (3 n + 11) (3 n + 10) (n + 4) 4 3 2 (362178 n + 3278637 n + 10971266 n + 16033671 n + 8595064) a(n + 3) = 0 subject to the initial conditions a(1) = 50, a(2) = 2822, a(3) = 167763 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 29, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence -42 (2 n + 5) (n + 2) (n + 1) ( 5 4 3 2 133494 n + 2324916 n + 16123699 n + 55638300 n + 95486585 n + 65169174) 7 6 5 4 a(n) + 2 (234148476 n + 5014496568 n + 45510159230 n + 226678801610 n 3 2 + 668475627607 n + 1165762524640 n + 1111692003063 n + 446544732438) 8 7 6 (n + 2) a(n + 1) + (-4440544416 n - 106199544528 n - 1101643994232 n 5 4 3 - 6470857876854 n - 23526756637932 n - 54183605671962 n 2 - 77136925371828 n - 62008866695256 n - 21528349179552) a(n + 2) + 75 (3 n + 11) (3 n + 13) (n + 4) 5 4 3 2 (133494 n + 1657446 n + 8158975 n + 19881759 n + 23948888 n + 11388612) a(n + 3) = 0 subject to the initial conditions a(1) = 41, a(2) = 1751, a(3) = 77006 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 30, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 7 6 14 (2 n + 3) (n + 3) (n + 2) (n + 1) (45591040 n + 1093965440 n 5 4 3 2 + 11167227984 n + 62844113900 n + 210486566144 n + 419414285141 n 9 + 460139090235 n + 214296733776) a(n) - (n + 3) (n + 2) (2006005760 n 8 7 6 5 + 54152496640 n + 640082252736 n + 4342930062800 n + 18616206525200 n 4 3 2 + 52205415583072 n + 95617625657156 n + 110095549487291 n 10 + 72170501059923 n + 20480580514512) a(n + 1) - (n + 3) (70574929920 n 9 8 7 + 2081620615680 n + 27277597993152 n + 208978630052016 n 6 5 4 + 1035762010936088 n + 3467117416386492 n + 7930174621342418 n 3 2 + 12223731460129494 n + 12135935644000247 n + 6996734789788998 n 11 10 + 1775516445902880) a(n + 2) + (-586665502720 n - 19357136806400 n 9 8 7 - 287287940020032 n - 2530149132636656 n - 14682777520252016 n 6 5 4 - 58908936695846168 n - 166603526959955380 n - 331825021870033755 n 3 2 - 455623563141501959 n - 410224682094438942 n - 217628594070463368 n - 51436232356451904) a(n + 3) + 8 (2 n + 9) (4 n + 17) (n + 4) (4 n + 15) 7 6 5 4 (45591040 n + 774828160 n + 5560847184 n + 21821769180 n 3 2 + 50498767984 n + 68799060029 n + 50985247345 n + 15810622854) a(n + 4) = 0 subject to the initial conditions a(1) = 29, a(2) = 1129, a(3) = 48021, a(4) = 2132433 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 31, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 7 6 494 (2 n + 3) (n + 3) (n + 2) (n + 1) (143224309910 n + 3437069714410 n 5 4 3 + 35089377668227 n + 197487085219170 n + 661516731625047 n 2 + 1318259532097208 n + 1446396624109020 n + 673680894469488) a(n) - 4 ( 9 8 7 55284583625260 n + 1492562660638040 n + 17643802943808362 n 6 5 4 + 119724628856657555 n + 513263244102817370 n + 1439525817226645994 n 3 2 + 2636978241489942622 n + 3036804878346955467 n + 1991153446480334466 n 10 + 565213000555116144) (n + 3) (n + 2) a(n + 1) - 4 (4413027436946920 n 9 8 7 + 130174642943608980 n + 1705966420025118719 n + 13070948488593657077 n 6 5 + 64789993134920942486 n + 216901086232832236514 n 4 3 + 496163034176904643471 n + 764888485545087230153 n 2 + 759496226647864855644 n + 437938049031422181156 n 11 + 111151619093260602000) (n + 3) a(n + 2) + (-330758354249786430 n 10 9 - 10914301185824242800 n - 161997043411475417151 n 8 7 - 1426826264149194540543 n - 8280736514394082673148 n 6 5 - 33226019508771148101894 n - 93976331649838705020555 n 4 3 - 187189560622608007566735 n - 257050391156151707944332 n 2 - 231459736576553185210236 n - 122804439091523231516784 n - 29027926284132708369792) a(n + 3) + 648 (2 n + 9) (4 n + 17) (n + 4) 7 6 5 (4 n + 15) (143224309910 n + 2434499545040 n + 17474669889877 n 4 3 2 + 68583391747335 n + 158733623989287 n + 216286407062857 n + 160306454137110 n + 49718623788072) a(n + 4) = 0 subject to the initial conditions a(1) = 64, a(2) = 5509, a(3) = 518131, a(4) = 50878108 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 32, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 8 7 14 (2 n + 5) (n + 3) (n + 2) (n + 1) (45591040 n + 1421479040 n 6 5 4 3 + 19300444496 n + 149023556572 n + 715524331336 n + 2187104438765 n 2 + 4155021395735 n + 4484227429768 n + 2104182277344) a(n) - (n + 3) 10 9 8 (n + 2) (2006005760 n + 70569100800 n + 1106651457984 n 7 6 5 + 10180915770832 n + 60805503230048 n + 246149193890648 n 4 3 2 + 683349542935968 n + 1283262660735917 n + 1558114347683063 n + 1102911093578596 n + 345006355708560) a(n + 1) - (n + 3) ( 11 10 9 70574929920 n + 2659186598400 n + 45153949329088 n 8 7 6 + 455929848759728 n + 3040419377903912 n + 14053907769397164 n 5 4 3 + 45925182234478398 n + 106038011147137782 n + 169431940529231243 n 2 + 178307197445166451 n + 111144250863193234 n + 31058443763856840) 12 11 10 a(n + 2) + (-586665502720 n - 24158247313920 n - 452737598302528 n 9 8 7 - 5104410216528368 n - 38549249428314336 n - 205373272869890496 n 6 5 4 - 791144705851101920 n - 2219473627028376069 n - 4498239428372778449 n 3 2 - 6419676721814340049 n - 6120168109896878487 n - 3497025402100077018 n - 904950873128112120) a(n + 3) + 8 (2 n + 9) (4 n + 17) (4 n + 19) (n + 5) 8 7 6 5 (45591040 n + 1056750720 n + 10626640336 n + 60518851196 n 4 3 2 + 213352822316 n + 476432457381 n + 657550658456 n + 512301370093 n + 172297135806) a(n + 4) = 0 subject to the initial conditions a(1) = 37, a(2) = 1545, a(3) = 67933, a(4) = 3075665 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 33, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 7 6 137 (2 n + 7) (n + 3) (n + 2) (n + 1) (13596631022 n + 384622751292 n 5 4 3 + 4640904041475 n + 30953879689351 n + 123214751117243 n 2 + 292617789894903 n + 383749868447010 n + 214303665172104) a(n) - ( 9 8 7 5819358077416 n + 193715327940056 n + 2839087477803330 n 6 5 4 + 24024066330040912 n + 129219542920774651 n + 457615409766927983 n 3 2 + 1065477398985551185 n + 1570054628567851933 n + 1325857918782697398 n 10 + 487593744716056536) (n + 3) (n + 2) a(n + 1) - (516481626001692 n 9 8 7 + 18483892025590602 n + 295243841125815270 n + 2770612177388138343 n 6 5 + 16907872956316710349 n + 70076395139707649175 n 4 3 + 199650746715464994265 n + 385846473282922959432 n 2 + 483741808248785665504 n + 354971412379549998168 n 11 + 115661557632528439200) (n + 3) a(n + 2) + (-10753983374230460 n 10 9 - 422503489795916620 n - 7497551293354516290 n 8 7 - 79301285070757348900 n - 555297400881877978630 n 6 5 - 2701992905153484117610 n - 9318585207994196778920 n 4 3 - 22767779866258178054850 n - 38600557428484156765660 n 2 - 43223401292628226972020 n - 28750198050005446544040 n - 8598854677098416724000) a(n + 3) + 200 (2 n + 9) (4 n + 21) (n + 4) 7 6 5 (4 n + 19) (13596631022 n + 289446334138 n + 2618696785185 n 4 3 2 + 13042818665586 n + 38591699834519 n + 67771586282448 n + 65334983368306 n + 26640837270900) a(n + 4) = 0 subject to the initial conditions a(1) = 101, a(2) = 10741, a(3) = 1181221, a(4) = 132763241 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 34, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 11 -530 (2 n + 5) (n + 3) (n + 2) (n + 1) (359486627691668 n 10 9 8 + 16802140265641532 n + 355270987551611903 n + 4485042565631426382 n 7 6 + 37554362118993680808 n + 218947793948969310918 n 5 4 + 906738510734838139489 n + 2666706400098008658208 n 3 2 + 5456620513846490850148 n + 7396067491087053405456 n + 5974382531520026072640 n + 2177948016545409411840) (n + 4) a(n) + 2 13 12 (n + 3) (n + 2) (277523676577967696 n + 14081346991387133488 n 11 10 + 327164386872431393472 n + 4606724354892814276988 n 9 8 + 43839322767914618642107 n + 297585197008022907907845 n 7 6 + 1481506125871574517441622 n + 5473078156161343710424778 n 5 4 + 14991471540197929133636655 n + 30042341408368835628114601 n 3 2 + 42763710627726007621222208 n + 40889344702034115341234460 n + 23499213960706373291903280 n + 6117270593510496051619200) (n + 4) 14 13 a(n + 1) + 2 (n + 3) (7684386153537095168 n + 409111060316344506624 n 12 11 + 10032941970929068749128 n + 150169280123642637964698 n 10 9 + 1532007610783723083449663 n + 11264476630338725188249839 n 8 7 + 61531042025737691672190412 n + 253527604151090821825360164 n 6 5 + 791344216009930077089135431 n + 1860840659135042374578315543 n 4 3 + 3242506319253026416865303182 n + 4056209283948324148450045212 n 2 + 3439755617694040711254269016 n + 1767633555233868157106624160 n + 414587014609563282167020800) (n + 4) a(n + 2) + 12 (n + 4) ( 15 14 13 11863058713825044 n + 673101215904420996 n + 17683636655171842753 n 12 11 + 285367334875723970998 n + 3163612396599579780494 n 10 9 + 25524447474728079224114 n + 154850140736457661406595 n 8 7 + 719451615512737882342179 n + 2581596652832382052145756 n 6 5 + 7156340182201969540338380 n + 15204732517492032587349202 n 4 3 + 24322596924135746978289433 n + 28364015850522717215120796 n 2 + 22767308149309042100661900 n + 11247298381460250363957840 n 16 + 2577031071042480802041600) a(n + 3) + (-155802942388079677872 n 15 14 - 9541257464317257864072 n - 272202537764728345921200 n 13 12 - 4801099742209601561828268 n - 58583488456059920024590710 n 11 10 - 524240220305522987143529112 n - 3557793607231650737201310474 n 9 8 - 18673099001039510299396283436 n - 76572337544123825373858755586 n 7 6 - 246047199088585983629348374056 n - 617187277753690149322466898582 n 5 4 - 1195245003778747659909555867936 n - 1750876526537613962823579007752 n 3 2 - 1874208881872414185721848780480 n - 1381496718103163119303705436544 n - 625908676253889598375467985920 n - 131159739026559198774596198400) a(n + 4) + 15 (5 n + 28) (5 n + 24) (n + 5) (5 n + 26) (5 n + 27) ( 11 10 9 359486627691668 n + 12847787361033184 n + 207021349418238323 n 8 7 + 1984384696051662975 n + 12566150901061564860 n 6 5 + 55168054633149255210 n + 171227055771084677887 n 4 3 + 375426735524922014495 n + 569350253728897885966 n 2 + 568162308356176513560 n + 335322846619849066032 n + 88519997690809807680) a(n + 5) = 0 subject to the initial conditions a(1) = 28, a(2) = 1012, a(3) = 39700, a(4) = 1622233, a(5) = 67931959 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 35, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 11 -2587 (2 n + 5) (n + 4) (n + 3) (n + 2) (n + 1) (1845778118037291700 n 10 9 + 86284074800337071950 n + 1824728715528584179705 n 8 7 + 23039933914924300568910 n + 192954587710926665671176 n 6 5 + 1125174192519995669279340 n + 4660694632011657954842135 n 4 3 + 13710040136208897729140984 n + 28060051451111693272696052 n 2 + 38042922749521087951486848 n + 30738486516079017180744000 n + 11208895643087691958195200) a(n) + (n + 4) (n + 3) (n + 2) ( 13 12 13784270985502494415600 n + 699506554550927230985000 n 11 10 + 16254778118042473382024340 n + 228917492636487989536851190 n 9 8 + 2178850574823274661442037913 n + 14793057295008147609108544137 n 7 6 + 73661488887131991902927800790 n + 272187488251226059648079129116 n 5 4 + 745746760057737867020051543709 n + 1494876752406831278812574635889 n 3 2 + 2128575722984091161395876044928 n + 2036051225652821292379875748428 n + 1170648778216490697646243240560 n + 304906035931339038980474361600) 14 a(n + 1) + (n + 3) (n + 4) (1287847383185687388674200 n 13 12 + 68573650364846951889768000 n + 1681932608444561640970943680 n 11 10 + 25178480446876351358404241805 n + 256910525937574030260414448486 n 9 8 + 1889337816051839336858804354529 n + 10322295868803791281614932649854 n 7 + 42539992936755505172624000588475 n 6 + 132811436212129264666293060855050 n 5 + 312382103330774844187618696202091 n 4 + 544471727725581430582572151374146 n 3 + 681310423623463048677604133395980 n 2 + 577959848046670835408917627480824 n + 297116243775833574663221445240480 n + 69716789440423817251754927116800) 15 a(n + 2) + 150 (n + 4) (210418705456251253800 n 14 13 + 11940571581800938740300 n + 313933400563860874266130 n 12 11 + 5071912886223840519495025 n + 56303428030791156171242210 n 10 9 + 454840838346444235790594111 n + 2761842850108501069454936118 n 8 7 + 12833616223989746650498601181 n + 46003554545609551845578129654 n 6 5 + 127186582730096056006578202001 n + 268932891308349870741655162480 n 4 3 + 426990714373867997015020615966 n + 492611240637759569945587210344 n 2 + 389696361190973164374279504120 n + 188921668793236008916161252960 n + 42281111798460036482764492800) a(n + 3) + ( 16 15 -43473612014132331410100000 n - 2662616187977257861084800000 n 14 13 - 75971521181932665677850502500 n - 1340164879103354393353539063750 n 12 - 16355172083962662629342244704250 n 11 - 146378260902659823007646028075750 n 10 - 993568535640950245073325661541250 n 9 - 5215661275924975523096390808098250 n 8 - 21391761216840262785178359154929750 n 7 - 68751360984477880153342265674073250 n 6 - 172494654863620250531976035701691250 n 5 - 334133623907026904435909691770025000 n 4 - 489591169544400926195139888550083000 n 3 - 524231316193925264021293777293816000 n 2 - 386541579267336264125778239798736000 n - 175192818624756688661769074088960000 n - 36727094894244147106982825356800000) a(n + 4) + 250 (5 n + 28) 11 (5 n + 24) (n + 5) (5 n + 26) (5 n + 27) (1845778118037291700 n 10 9 + 65980515501926863250 n + 1063405764017264503705 n 8 7 + 10195605451706058058815 n + 64580367953473149216276 n 6 5 + 283599922274525910759468 n + 880484500382832979678661 n 4 3 + 1931159398885327607658619 n + 2929755323157277761190586 n 2 + 2924838620621664522573000 n + 1727005494382790490082656 n + 456145177920456250318464) a(n + 5) = 0 subject to the initial conditions a(1) = 91, a(2) = 10721, a(3) = 1371191, a(4) = 182687201, a(5) = 24944360851 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 36, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 (3 n + 2) (n + 1) (3 n + 4) (1539 n + 6156 n + 6058) a(n) - (2 n + 3) 4 3 2 (20097801 n + 120586806 n + 258688098 n + 233423667 n + 74033678) a(n + 1) 2 + 49 (3 n + 5) (3 n + 7) (n + 2) (1539 n + 3078 n + 1441) a(n + 2) = 0 subject to the initial conditions a(1) = 31, a(2) = 1393 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 37, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 5 3 (n + 1) (3 n + 4) (3 n + 5) (550 n + 2625 n + 3122) a(n) + (-2839100 n 4 3 2 - 26326200 n - 96075914 n - 172440102 n - 152210240 n - 52864392) 2 a(n + 1) + 27 (3 n + 8) (3 n + 7) (n + 2) (550 n + 1525 n + 1047) a(n + 2) = 0 subject to the initial conditions a(1) = 13, a(2) = 217 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 38, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 2 5 (n + 1) (3 n + 4) (3 n + 5) (3078 n + 14649 n + 17378) a(n) + (-80391204 n 4 3 2 - 744363000 n - 2713197294 n - 4864840098 n - 4290631232 n - 1489230552 ) a(n + 1) 2 + 49 (3 n + 8) (3 n + 7) (n + 2) (3078 n + 8493 n + 5807) a(n + 2) = 0 subject to the initial conditions a(1) = 37, a(2) = 1729 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 39, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 3 2 3 (n + 1) (3 n + 5) (3 n + 4) (2015 n + 15145 n + 37808 n + 31348) a(n) + ( 6 5 4 3 -239244980 n - 2874798550 n - 14180313406 n - 36728479674 n 2 - 52655422182 n - 39601179296 n - 12203204952) a(n + 1) + 192 (3 n + 8) 3 2 (3 n + 7) (n + 3) (2015 n + 9100 n + 13563 n + 6670) a(n + 2) = 0 subject to the initial conditions a(1) = 50, a(2) = 2836 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 40, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 3 2 3 (n + 1) (3 n + 4) (3 n + 5) (2555 n + 19199 n + 47918 n + 39723) a(n) + ( 6 5 4 3 -436940770 n - 5249531251 n - 25890669007 n - 67052564785 n 2 - 96121586155 n - 72287208124 n - 22274739948) a(n + 1) + 243 (3 n + 8) 3 2 (3 n + 7) (n + 3) (2555 n + 11534 n + 17185 n + 8449) a(n + 2) = 0 subject to the initial conditions a(1) = 57, a(2) = 3681 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 41, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 6 5 -37 (3 n + 5) (n + 2) (n + 1) (3 n + 7) (1003368384 n + 19262372184 n 4 3 2 + 153398412858 n + 648502433819 n + 1534641656673 n + 1926981144394 n 9 8 + 1002768445344) a(n) + 4 (9554073752448 n + 235963713574512 n 7 6 5 + 2563690577056164 n + 16073127087793236 n + 64043935623825735 n 4 3 2 + 168073931129374679 n + 290301069908880441 n + 317965502151333873 n + 200223594607979852 n + 55174573542433380) (n + 2) a(n + 1) + ( 10 9 8 6978483299349504 n + 189798743627358336 n + 2303327703425088096 n 7 6 + 16418796477217013232 n + 76102725802100166264 n 5 4 + 239567820632535085580 n + 518471351096065120124 n 3 2 + 761337999356677552948 n + 725553779813853326812 n + 404961379775715771024 n + 100451032770167894160) a(n + 2) - 1944 6 5 (4 n + 13) (2 n + 7) (4 n + 15) (n + 4) (1003368384 n + 13242161880 n 4 3 2 + 72137077698 n + 207465136547 n + 331951636284 n + 279903131689 n + 97065932862) a(n + 3) = 0 subject to the initial conditions a(1) = 82, a(2) = 7588, a(3) = 739855 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 42, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 8 (n + 1) (4 n + 9) (2 n + 3) (4 n + 7) 4 3 2 (44039 n + 476933 n + 1931980 n + 3469412 n + 2330384) a(n) + ( 8 7 6 5 -154486918323 n - 2754468884142 n - 21299430112332 n - 93267444779850 n 4 3 2 - 252874881027309 n - 434570663965608 n - 462131506925028 n - 277961130809520 n - 72381222453888) a(n + 1) + 1728 (4 n + 11) (2 n + 5) 4 3 2 (4 n + 13) (n + 3) (44039 n + 300777 n + 765415 n + 860095 n + 360058) a(n + 2) = 0 subject to the initial conditions a(1) = 51, a(2) = 2766 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 43, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 8 (2 n + 1) (4 n + 3) (n + 1) (4 n + 5) 3 2 7 (38675 n + 216325 n + 399604 n + 243948) a(n) + (-24714446575 n 6 5 4 3 - 237095734725 n - 951435607291 n - 2066462836047 n - 2617589287918 n 2 - 1928455291068 n - 762745130712 n - 124563999456) a(n + 1) + 512 (4 n + 7) (2 n + 3) (4 n + 9) (n + 2) 3 2 (38675 n + 100300 n + 82979 n + 21994) a(n + 2) = 0 subject to the initial conditions a(1) = 19, a(2) = 556 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 44, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence (2 n + 1) (4 n + 3) (n + 1) (4 n + 5) 3 2 7 (38048 n + 212872 n + 393352 n + 240225) a(n) + (-7795883008 n 6 5 4 3 - 74800153344 n - 300224909696 n - 652249237760 n - 826503505392 n 2 - 609201146956 n - 241105131354 n - 39408188265) a(n + 1) + 125 (4 n + 7) 3 2 (2 n + 3) (4 n + 9) (n + 2) (38048 n + 98728 n + 81752 n + 21697) a(n + 2) = 0 subject to the initial conditions a(1) = 25, a(2) = 961 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 45, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence (2 n + 1) (4 n + 3) (n + 1) (4 n + 5) 3 2 7 (105244 n + 588986 n + 1088732 n + 665187) a(n) + (-87390618328 n 6 5 4 3 - 838634066244 n - 3366771804832 n - 7316616598722 n - 9274986041308 n 2 - 6840007214532 n - 2708968253094 n - 443185314039) a(n + 1) + 343 (4 n + 7) (2 n + 3) (4 n + 9) (n + 2) 3 2 (105244 n + 273254 n + 226492 n + 60197) a(n + 2) = 0 subject to the initial conditions a(1) = 37, a(2) = 2101 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 46, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 5 (5 n + 3) (5 n + 4) (n + 1) (5 n + 6) (5 n + 7) 4 3 2 (481984 n + 3855872 n + 11499041 n + 15149188 n + 7438980) a(n) - 8 8 7 6 (2 n + 3) (1138078936192 n + 13656947234304 n + 70235030592278 n 5 4 3 + 201921437449926 n + 354439223683103 n + 388395293991456 n 2 + 259052005171417 n + 95997509158764 n + 15109946027460) a(n + 1) + 1280 (5 n + 11) (5 n + 12) (5 n + 8) (5 n + 9) (n + 2) 4 3 2 (481984 n + 1927936 n + 2823329 n + 1790786 n + 414945) a(n + 2) = 0 subject to the initial conditions a(1) = 25, a(2) = 898 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 47, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 5 (5 n + 3) (5 n + 4) (n + 1) (5 n + 6) (5 n + 7) 4 3 2 (19213174 n + 153705392 n + 458423117 n + 604049332 n + 296698548) a(n) 8 7 - 4 (2 n + 3) (10269959163547082 n + 123239509962564984 n 6 5 4 + 633851763549513025 n + 1822621308124820229 n + 3200258385004249405 n 3 2 + 3508364379615583983 n + 2341399376592333332 n + 868345620740299608 n + 136818928133625960) a(n + 1) + 50000 (5 n + 11) (5 n + 12) (5 n + 8) (5 n + 9) (n + 2) 4 3 2 (19213174 n + 76852696 n + 112585985 n + 71466578 n + 16580115) a(n + 2) = 0 subject to the initial conditions a(1) = 73, a(2) = 7606 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 48, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence (5 n + 4) (n + 1) (5 n + 6) (5 n + 7) (5 n + 8) 4 3 2 (1234375 n + 10911875 n + 36019850 n + 52607755 n + 28675278) a(n) + ( 9 8 7 -77092890625000 n - 1182604942187500 n - 7966336108312500 n 6 5 4 - 30915345476025000 n - 76132657741852500 n - 123312507632295000 n 3 2 - 131284291656905300 n - 88527368749802840 n - 34278655055703744 n - 5800739370480576) a(n + 1) + 1296 (5 n + 12) (5 n + 9) (5 n + 11) (5 n + 13) (n + 2) 4 3 2 (1234375 n + 5974375 n + 10690475 n + 8366180 n + 2409873) a(n + 2) = 0 subject to the initial conditions a(1) = 46, a(2) = 2766 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 49, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence (n + 1) (5 n + 6) (5 n + 7) (5 n + 8) (5 n + 9) 4 3 2 (4937500 n + 47795000 n + 173137250 n + 278172565 n + 167249442) a(n) + 9 8 7 (-308371562500000 n - 5297823443750000 n - 40096573436250000 n 6 5 4 - 175436193801450000 n - 488937443227410000 n - 899980379591460000 n 3 2 - 1093943614279559000 n - 846620022228407320 n - 378500954001573696 n - 74471188113280224) a(n + 1) + 1296 (5 n + 11) (5 n + 12) (5 n + 13) (5 n + 14) (n + 2) 4 3 2 (4937500 n + 28045000 n + 59377250 n + 55533065 n + 19356627) a(n + 2) = 0 subject to the initial conditions a(1) = 51, a(2) = 3151 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Problem Number, 50, : Let , a(n), be the sequence of rational numbers satisfying the linear recurrence 5 4 (n + 1) (5 n + 9) (5 n + 8) (5 n + 7) (5 n + 6) (37548750 n + 469754625 n 3 2 + 2345396200 n + 5841651605 n + 7258123288 n + 3598881396) a(n) + ( 10 9 8 -52147563192187500 n - 1043500185561562500 n - 9316280608318468750 n 7 6 - 48857654911997325000 n - 166641633440763360000 n 5 4 - 386171996903305744500 n - 615651903901414627350 n 3 2 - 666614367941555432640 n - 469087081656179364944 n - 193683743084449243584 n - 35627972061755858400) a(n + 1) + 14641 5 (5 n + 11) (5 n + 12) (5 n + 13) (5 n + 14) (n + 3) (37548750 n 4 3 2 + 282010875 n + 841865200 n + 1248503255 n + 919733928 n + 269219388) a(n + 2) = 0 subject to the initial conditions a(1) = 111, a(2) = 13921 Prove that , surprisingly, that for EVERY n, a(n) an integer! ------------------------------------------------------------------------ Solutions to the Problems Proof of Problem, 1 It so happened that a(n) equals n ----- \ ) binomial(n, k) binomial(n + k + 5, k) / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 2 It so happened that a(n) equals n ----- \ ) binomial(n, k) binomial(n + 4 k + 4, k) / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 3 It so happened that a(n) equals n ----- \ ) binomial(n, k) binomial(2 n + 3 k + 1, k) / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 4 It so happened that a(n) equals n ----- \ ) binomial(n, k) binomial(2 n + 3 k + 2, k) / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 5 It so happened that a(n) equals n ----- \ ) binomial(n, k) binomial(4 n + 5 + k, k) / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 6 It so happened that a(n) equals n ----- \ ) binomial(n, k) binomial(5 n + k + 4, k) / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 7 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 2 + k, k) 7 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 8 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 2 + k, k) 8 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 9 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 2 + k, k) 9 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 10 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 5 + 3 k, k) 2 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 11 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 5 + 4 k, k) 9 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 12 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + k + 1, k) 2 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 13 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + k + 4, k) 3 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 14 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + k + 5, k) 2 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 15 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + k + 5, k) 3 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 16 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 2 k + 1, k) 4 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 17 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 2 k + 1, k) 5 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 18 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 2 k + 5, k) 2 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 19 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 3 k + 2, k) 2 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 20 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 3 k + 2, k) 4 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 21 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 4 k + 1, k) 7 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 22 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 4 k + 2, k) 8 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 23 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(n + 4 k + 4, k) 3 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 24 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 2 k + 1, k) 2 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 25 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 2 k + 2, k) 5 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 26 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 2 k + 3, k) 2 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 27 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 2 k + 3, k) 6 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 28 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 2 k + 3, k) 7 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 29 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 2 k + 4, k) 5 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 30 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 3 k + 2, k) 4 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 31 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 3 k + 2, k) 9 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 32 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 3 k + 4, k) 4 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 33 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 3 k + 5, k) 10 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 34 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 4 k + 3, k) 3 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 35 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(2 n + 4 k + 3, k) 10 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 36 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(3 n + k + 1, k) 6 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 37 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(3 n + k + 2, k) 2 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 38 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(3 n + k + 2, k) 6 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 39 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(3 n + k + 3, k) 7 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 40 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(3 n + k + 3, k) 8 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 41 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(3 n + 2 k + 4, k) 9 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 42 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(4 n + 5 + k, k) 5 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 43 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(4 n + k + 1, k) 3 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 44 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(4 n + k + 1, k) 4 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 45 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(4 n + k + 1, k) 6 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 46 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(5 n + k + 2, k) 3 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 47 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(5 n + k + 2, k) 9 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 48 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(5 n + k + 3, k) 5 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 49 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(5 n + k + 4, k) 5 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ Proof of Problem, 50 It so happened that a(n) equals n ----- \ k ) binomial(n, k) binomial(5 n + k + 5, k) 10 / ----- k = 0 By the Zeilberger algorithm this sum satisfies the given recurrence and sinc\ e the initial conditions match (check!) and the sum is always an integer\ , the result follows by induction. ------------------------------------------------------------------------ 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