Recounting the Rationals

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Emily Atkinson
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1 Rconting th Rationals Nil Calkin and Hrbrt S. Wilf pril, 000 It is wll known (indd, as Pal Erd}os might hav said, vry child knows) that th rationals ar contabl. Howvr, th standard prsntations of this fact do not giv an xplicit nmration; rathr thy show how to constrct an nmration. In this not w will xplicitly dscrib a sqnc b(n) with th proprty that vry positiv rational appars xactly onc as b(n)=b(n + ). Morovr, b(n) is th soltion of a qit natral conting problm. Th list of th positiv rational nmbrs will bgin lik this: ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; 7 ; 7 ; 8 ; 8 ; 7 ; 7 ; 7 ; 7 ; : : : Som of th intrsting fatrs of this list ar. Th dnominator of ach fraction is th nmrator of th nxt on. That mans that th nth rational nmbr in th list looks lik b(n)=b(n + ) (n = 0; ; ; : : :), whr b is a crtain fnction of th nonngativ intgrs whos vals ar fb(n)gn0 = f; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; 7; : : :g:. Th fnction vals b(n) actally cont somthing nic. In fact, b(n) is th nmbr of ways of writing th intgr n as a sm of powrs of, ach powr bing sd at most twic (i.., onc mor than th lgal limit for binary xpansions). For instanc, w can writ = + = + +, so thr ar two sch ways to writ, and thrfor b() =. Lt's say that b(n) is th nmbr of hyprbinary rprsntations of th intgr n.. Consctiv vals of this fnction b ar always rlativly prim, so that ach rational occrs in rdcd form whn it occrs.. Evry positiv rational occrs onc and only onc in this list.

2 + s Th tr of fractions Figr : Th tr of fractions For th momnt, lt's forgt abot nmration, and jst imagin that fractions grow on th tr that is compltly dscribd, indctivly, by th following two rls: is at th top of th tr, and Each vrtx i j has two childrn: its lft child is i W show th following proprtis of this tr. i+j and its right child is i+j j.. Th nmrator and dnominator at ach vrtx ar rlativly prim. This is crtainly tr at th top vrtx. Othrwis, sppos r=s is a vrtx on th highst possibl lvl of th tr for which this is fals. If r=s is a lft child, thn its parnt is r=(s r), which wold clarly also not b a rdcd fraction, and wold b on a highr lvl, a contradiction. If r=s is a right child, thn its parnt is (r s)=s, which lads to th sam contradiction.. Evry rdcd positiv rational nmbr occrs at som vrtx. Th rational nmbr crtainly occrs. Othrwis, lt r=s b, among all fractions that do not occr, on of smallst dnominator, and among thos th on of smallst nmrator. If r > s thn (r s)=s dosn't

3 occr ithr, ls on of its childrn wold b r=s, and its nmrator is smallr, th dnominator bing th sam, a contradiction. If r < s, thn r=(s r) dosn't occr ithr, ls on of its childrn wold b r=s, and it has a smallr dnominator, a contradiction.. No rdcd positiv rational nmbr occrs at mor than on vrtx. First, th rational nmbr occrs only at th top vrtx of th tr, for if not, it wold b a child of som vrtx r=s. Bt th childrn of r=s ar r=(r + s) and (r + s)=s, nithr of which can b. Othrwis, among all rdcd rationals that occr mor than onc, lt r=s hav th smallst dnominator, and among ths, th smallst nmrator. If r < s thn r=s is a lft child of two distinct vrtics, at both of which r=(s r) livs, contradicting th minimality of th dnominator. Th cas r > s is similar. It follows that a list of all positiv rational nmbrs, ach apparing onc and only onc, can b mad by writing down =, thn th fractions on th lvl jst blow th top of th tr, rading from lft to right, thn th fractions on th nxt lvl down, rading from lft to right, tc. W claim that if that b don, thn th dnominator of ach fraction is th nmrator of its sccssor. This is clar if th fraction is a lft child and its sccssor is th right child of th sam parnt. If th fraction is a right child thn its dnominator is th sam as th dnominator of its parnt and th nmrator of its sccssor is th sam as th nmrator of th parnt of its sccssor, hnc th rslt follows by downward indction on th lvls of th tr. Finally, th rightmost vrtx of ach row has dnominator, as dos th lftmost vrtx of th nxt row, proving th claim. Ths, aftr w mak a singl sqnc of th rationals by rading th sccssiv rows of th tr as dscribd abov, th list will b in th form ff(n)=f(n + )gn0, for som f. Now, as th fractions sit in th tr, th two childrn of f(n)=f(n + ) ar f(n + )=f(n + ) and f(n + )=f(n + ). Hnc from th rl of constrction of th childrn of a parnt, it mst b that f(n + ) = f(n) and f(n + ) = f(n) + f(n + ) (n = 0; ; ; : : :): Ths rcrrncs, togthr with f(0) =, vidntly dtrmin or fnction f on all nonngativ intgrs. W claim that f(n) = b(n), th nmbr of hyprbinary rprsntations of n, for all n 0. This is tr for n = 0, and sppos it is tr for all intgrs 0; ; : : : ; n. Now b(n + ) = b(n), bcas if w ar givn a hyprbinary xpansion of n+, th \" mst appar, hnc by sbtracting from both sids and dividing by, w'll gt a hyprbinary rprsntation of n. Convrsly, givn sch an xpansion of n, dobl ach part and add a to obtain a rprsntation of n +. Frthrmor, b(n + ) = b(n) + b(n + ), for a hyprbinary xpansion of n + might hav ithr two 's or no 's in it. If it has two 's, thn by dlting thm and dividing by w obtain an xpansion of n. If it has no 's, thn w jst divid by to gt an xpansion of n +. Ths maps ar rvrsibl, proving th claim.

4 It follows that b(n) and f(n) satisfy th sam rcrrnc formlas and tak th sam initial vals, hnc thy agr for all nonngativ intgrs. W stat th nal rslt as follows. Thorm Th nth rational nmbr, in rdcd form, can b takn to b b(n)=b(n + ), whr b(n) is th nmbr of hyprbinary rprsntations of th intgr n, for n = 0; ; ; : : :. That is, b(n) and b(n + ) ar rlativly prim, and ach positiv rdcd rational nmbr occrs onc and only onc in th list b(0)=b(); b()=b(); : : :. Rmarks Thr is a larg litratr on th closly rlatd sbjct of Strn-Brocot trs [6, ]. In particlar, an xcllnt introdction is in [], and th rlationship btwn ths trs and hyprbinary partitions is xplord in []. Or sqnc fb(n)g is sqnc #0087 in []. W thank Nil Sloan for pointing ot that still othr ways of conting th rationals ar in his sqncs #0868 and #006. Or intrst in fb(n)g was piqd by a problm in antm, in Sptmbr 997, that askd for b(906), and which was postd by Stan Wagon as his \Problm of th Wk." In Strn's original papr [6] of 88 thr is a strctr that is ssntially or tr of fractions, thogh in a dirnt garb, and h provd that vry rational nmbr occrs onc and only onc, in rdcd form. Howvr Strn did not dal with th partition fnction b(n). Rznick [] stdid rstrictd binary partition fnctions and obsrvd thir rlationship to Strn's sqnc. Nonthlss it smd to s worthwhil to draw ths two aspcts togthr and xplicitly not that th ratios of sccssiv vals of th partition fnction b(n) rn throgh all of th rationals. qstion: What othr fnctions f(n) ar thr that hav natral and intitiv dnitions and also hav th proprty that ff(n)=f(n + )g taks vry rational val xactly onc? Rfrncs [] chill Brocot, Calcl ds roags par approximation, novll mthod, Rv Chronomtriq 6 (860) [] Ronald L. Graham, Donald E. Knth and Orn Patashnik, Concrt Mathmatics, ddison- Wsly, Rading, 989. [] D. H. Lhmr, On Strn's diatomic sris, this monthly 6 (99) [] Brc Rznick, Som binary partition fnctions, in nalytic Nmbr Thory, Procdings of a confrnc in honor of Pal T. Batman, Birkhasr, Boston (990) -77. [] N. J.. Sloan (999), Th On-Lin Encyclopdia of Intgr Sqncs, pblishd lctronically at <

5 [6] M.. Strn, Ubr in zahlnthortisch Fnktion, Jornal fr di rin nd angwandt Mathmatik (88) 9-0. Dpartmnt of Mathmatics, Clmson Univrsity, Clmson, SC 96 Dpartmnt of Mathmatics, Univrsity of Pnnsylvania, Philadlphia, P wilf@math.pnn.d

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The van der Waals interaction 1 D. E. Soper 2 University of Oregon 20 April 2012

The van der Waals interaction 1 D. E. Soper 2 University of Oregon 20 April 2012 Th van dr Waals intraction D. E. Sopr 2 Univrsity of Orgon 20 pril 202 Th van dr Waals intraction is discussd in Chaptr 5 of J. J. Sakurai, Modrn Quantum Mchanics. Hr I tak a look at it in a littl mor