Beyond Zudilin s Conjectured q-analog of Schmidt s problem

Transcription

1 Beyond Zudln s Conectured q-analog of Schmdt s problem Thotsaporn Ae Thanatpanonda thotsaporn@gmalcom Mathematcs Subect Classfcaton: 11B65 33B99 Abstract Usng the methodology of (rgorous expermental mathematcs we gve a smple and motvated soluton to Zudln s queston concernng a q-analog of a problem posed by Asmus Schmdt about a certan bnomal coeffcents sum Our method s based on two smple denttes that can be automatcally proved usng the Zelberger and q-zelberger algorthms We further llustrate our method by provng two further bnomal coeffcents sums 1 Introducton In 1992 Asmus Schmdt [3 conectured that for any nteger r 1 the sequence of numbers {c (r } 0 defned mplctly by ( r ( r n n + = ( ( n n + c (r n = are always ntegers It too more than 10 years to completely solve ths conecture([5 Shorter proofs were found recently ([1 4 The ey step n [4 was the fact that a (r defned mplctly by ( r ( r n n + = ( ( a (r n n + n = (1 1

2 are all ntegers The ntegral property of the a (r n fact drectly solves the conecture It s very possble that had Schmdt conectured ths equaton n the frst place hs conecture would have been proved much sooner Inspred by (1 we nvestgate bnomal coeffcents summands of the form ( n + d b + c where d b c are fxed ntegers such that ther powers can be wrtten as an nteger lnear combnaton of themselves 2 Results In ths secton we search for f(n for whch a (r f(n r = a (r f(n n = (2 are all ntegers It was not an accdent that the a (r happen to be ntegers Experments show that there exsts an nteger-valued functon S f ( free of r such that ā (r defne by ā(1 = 1 ā(1 = 0 for and ā (r+1 S f ( ā (r (3 agrees wth a (r n (2 We prove ths fact by showng that ā (r on an nducton On the one hand also satsfy (2 The proof reles ā (r+1 f(n = ā (r S f ( ā (r f(n S f ( f(n (by defnton of ā(r+1 2

3 On the other hand f(n r+1 = f(n r f(n ā (r f(n f(n (by nducton hypothess Hence the proof bols down to the condton that S f ( must satsfy: f(n f(n = S f ( f(n n = (4 Let s state ths observaton as a theorem Theorem 21 Gven a par f(n and S( such that f(n f(n = S( f(n for all n 0 Defne a (r recursvely by a(1 = 1 a(1 = 0 for and a (r+1 S( a (r Then for 0 and r 1 f(n r = a (r f(n n = Once we now where to loo of course n ths case guded by Schdmt s problem t becomes a routne ob that computers are so good at We frst pc some bnomal term f(n Then we cran out some numercal values of S( accordng to (4 Once we have enough data we as our computer to guess the relaton or even the formula for S( Fnally needless to say the proof of the denttes can be routnely done by Zelberger s algorthm Here s the lst that we were able to fnd Result 211: These are the functons we used n Schmdt s conecture ( ( ( ( n n + + f(n = S( = 3 +

4 Result 212: For any fxed nteger c ( ( n n + f(n = + c + c Result 22: f(n = S( = ( ( n S( = ( + + c! ( + c!( + c! Result 23: ( ( n + f(n = S( = ( q-analog ( ( + c!! + We now present q-analogs of the results from secton 2 [ n The q-bnomal are defned by [ { n (qn = (q (q n f 0 n 0 otherwse + where (q 0 = 1 and (q n = (1 q(1 q 2 (1 q n for n = 1 2 The proof of Theorem 31 below s smlar to the proof of Theorem 21 We leave the detals to the reader + c + + c Theorem 31 Gven a trple f(n A( n and S( satsfyng f(n f(n = q A S( f(n for all n 0 (5 Let B( and C( r n be any functons such that B( + C( r + 1 n = A( n + C( r n and 4

5 C( 1 n = 0 Defne P (r (q recursvely by P (1 (1 (q = 1 P (q = 0 for and P (r+1 (q q B S( P (r (q Then for 0 and r 1 f(n r = q C f(n P (r (q n = The proofs of (5 of the results below can agan be done automatcally usng the q-zelberger algorthm Once we fnd functon A from (5 t s only a matter of smple calculatons to solve for functons B and C Result 311: [ [ [ [ n n + + f(n = S( = A( n = (n ( + B( = ( + C( r n = (r n Result 312: [ [ n n + f(n = S( = (q ++c (q +c + c + c (q +c (q +c (q A( n = (n ( + + c(n c B( = ( + c( + + c C( r n = (r n + rcn cn Result 32: f(n = [ [ n S( = A( n = ( ( B( = ( ( 5 + [ + +

6 C( r n = 0 Result 33: [ [ n + f(n = S( = ( 1 ++ A( n = (+ (2n++ +1 B( = (+ ( C( r n = (r n 2 + Note that 311 agrees wth the results n [1 Also the postvty of B and C n result 32 and 33 mply that P (r are ndeed polynomals 4 Conectures After some expermentaton wth all of the bnomal terms of the form ( n+d b +c we found out that the only such terms that satsfy the condton (4 seem to be of the form ( ( n n+ ( +c +c or n+d for any fxed nteger c and d We already saw that t was true for the former but for the latter t s only proved to be true for d = 0 1 but we conecture that t holds for all non-negatve ntegers d as follows Conecture 41 For any ntegers d 0 and r 1 there exst ntegers such that a (r d ( r n + d = Moreover a (r d and a (r d ( n + d can be defned as followng: a(1 d a (r+1 d S d ( a (r d for all n = where S d ( are ntegers ndependent of r for all d = 1 a(1 d = 0 for Conecture 42 For a fxed nteger d S d ( defned above are holonomc but not of the frst order e no closed form soluton except d = 0 1 6

7 5 Conclusons We presented a motvated and streamlned new proof of the man result of [1 as well as two new much deeper denttes and made two conectures But the man nterest of ths paper s n llustratng a methodology usng computers (va expermental mathematcs to generate data then formulate conectures and fnally havng the very same computer rgorously prove ts own conectures We beleve that ths methodology has great potental almost everywhere n mathematcs References [1 Vctor J W Guo and Jang Zeng On Zudln s q-queston about Schmdt s problem The Electronc Journal of Combnatorcs 19(3 (2012 #P4 [2 Asmus Schmdt Generalzed q-legendre polynomals J Comput Appl Math 49:1-3 ( [3 Asmus Schmdt Legendre transforms and Apéry s sequences J Austral Math Soc Ser A58:3 ( [4 Thotsaporn Ae Thanatpanonda A Smple Proof of Schmdt s Conecture Journal of Dfference Equatons and Applcatons 20( [5 Wadm Zudln On a combnatoral problem of Asmus Schmdt The Electronc Journal of Combnatorcs 11(2004 #R22 7