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All use subject to JSTOR Terms and Conditions 10988. Proposed by WuWei Chao, Guang Zhou University (New), Guang Zhou City, Guang Dong Province, China. Let al be a positive real number and define an+1 = al + n/an for n > 1. (a) Prove that limn,o an/,/- = 1. (b) Define bn = a,+2 - 2an+l + an. Prove that limnoon,bn+l/bn = 1. 10989. Proposed by Roland Bacher, Institut Fourier,France. Let z be a complex number other than 1, and let d be a natural number. Show that the function fz defined by fz(x) = x (Z l)d+l k=0 d k zk is a polynomial of degree d in x for x E N. That is, for each z and d, there is a polynomial p of degree d over the complex numbers so that p(x) = fz(x) for all x E N. 10990. Proposed by Rick Mabry, LSUS, Shreveport,LA.The n + 1 Bernstein polynomials of degree n are defined by bn,k(x) = (n)xk(1

)n-k forO < k < n.

When all n + 1 polynomials are plotted on the same graph for large fixed n over the interval 0 < x < 1, an 'upper envelope' begins to be seen. The figure below shows a scaled plot of the case n = 16, with the vertical scale multiplied by 4.

1 75 1.5 1 25 02 04 06 08 1 Let 1B(x)= limn-,,0 /maxo<k<n bn,k(x). Find a closed-form expression for P(x).

SOLUTIONS

Highly Variable Lists 10691 [1998, 859]. Proposed by Donald E. Knuth,Stanford University,Stanford, CA.

Denote by a mod n the remainder a - n La/n when the integer a is divided by the integer n. Given positive integers m and n, what is Solution by Allan Pedersen, S0borg, Denmark. The answer is

m max 0<al,....am<n min O<k<n

(\(aj + k) mod n)/? 1 1((m2 1)(n1) + gcd(m, n)1).

For a list a, the n sums over which the minimization is taken are the same as for the list b given by setting bj = (aj + 1) mod n for all j (with I fixed). Hence it suffices to January2003]

PROBLEMASNDSOLUTIONS 59 restrictourattentionto lists suchthat0 < aj < n foreveryj andsuchthEaj=ti k) modn is minimizedwhenk = O.Webeginby characterizinsguchlists.

For0 < k < n, thevalueof (aj + k) modn is aj + k ifaj < n - k andis aj + k n if aj > n - k. Thus aj +mk-(/ln-i j=l +... + in-k) n, where/ui is the multiplicityof i in the list a. Hencethe minimumis achievedwhen k = 0 if andonly if n-1.

(*)

Since ]=1 a = subjectto (*) and n ~l(n -i-)ni, ourproblemis to minai1ximize -i,_tn-i= m. For 1 < k < n, let ,n-k = Lmk/n] - Lm(k- 1)/nJ.

These values satisfy (*) with equality.Also feasible. Furthermore, whenever /tn-1, ..,/l satisfy (*), we have E Yi=ln-i k=l ji,-k = m, so these values are < Lmk/nj= E=1 in-i. Thus n-i yj(n -- =Z-i i=1 E Yn- i k=l i=1 Z k=l i=1

n-I - = L i(n )n-i

i=1

WWeheaavveepprroovveeddthaatttheeannsswweerris 5,n--E1k=1Lmk/n].Thissumhasthe valueclaimed, as can be seen by applyingformula(3.32) of R. L. Graham,D. E. Knuth,and 0.

PatashnikC,oncreteMathematics2,nd ed., Addison-WesleyR,eading,MA, 1994,p. 94.

Editorialcomment.A countingargumentgives a directproof of nk-- Lmk/nJ=

((m - )(n - 1) + gcd(m,n) - 1). Considerthe positiveintegerpoints with first coordinateless thann thatlie on or belowthe linejoiningthe originand (n, m). By symmetryh,alf the integerpointsin an (n - 1)-by-(m- 1) rectanglethatareoff the line arecaptureda,ndtherearegcd(m,n) - 1 pointsin thisrectangleon theline.

Also solved by D. Beckwith, R. Chapman(U. K.), D. Donini (Italy), J. H. Lindsey II, O. P. Lossers (The Netherlands),A. Nijenhuis, M. A. Prasad(India), J. Simpson (Australia),GCHQ Problems Group (U.K.), NSA ProblemsGroup,andthe proposer.

ExploringAll BinaryMazes 10720[1999,264].ProposedbyDonaldE.KnuthS,tanfordUniversityS,tanfordC, A.

A "binarymaze"is a directedgraphin whichexactlytwo arcsleadfromeachvertex, one labeled0 andone labeled1. If bl, b2,... , bmis anysequenceof Osand ls andv is anyvertex,let vblb2,... , b, be thevertexreachedby beginningat v andtraversing arcslabeledblb2,... , bminorder.A sequenceblb2,... , bmof OsandIs is a universal explorationsequenceof ordern if, for everystronglyconnectedbinarymazeon n verticesandeveryvertexv, the sequence( v, vbl, vblb2,.. , vbl ... bm)includes everyvertexof themaze.Forexample,01 is a universalexplorationsequenceof order 2, andit canbe shownthat0110100is universaolf order3. (a) Provethatuniversalexplorationsequencesof all ordersexist. (b)* Finda good estimatefor the asymptoticlengthof the shortestsuchsequenceof ordern. 60

() THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 110