122 PROBLEMS AND SOLUTIONS

The limit limk-.oo T* is the normalized dyadic product of the two eigenvectors:

0

0

0

1In

0

2In...

Also solved by A. A. JAGERS (Technische Hogeschool Twente, Enschede, theNetherlands), W. B. JORDAN (Scotia, NY) and E. L. WACHSPRESS (University ofTennessee), INGEMAR KINNMARK (Princeton University), O. P. LOSSERS (EindhovenUniversity of Technology, the Netherlands), A. SIDI (The Technion, Haifa, Israel)and H. VAN HAERJNGEN (Delft University of Technology, the Netherlands), and theproposer.

A Binomial Summation

Problem 83-3, by GENGZHE CHANG and ZUN SHAN (University of Science and Technol-ogy of China, Hefei, Anhui, China).Determine the sum

{(n) (7) ()}{( n ) ( n ) ()}0+ +’’" +

k++k+2

+’’" +kO

Solution by D. R. BREACH (University of Canterbury, New Zealand).The coefficient of xk in (1 + x)"(1 + x + x + is ki=0 (7). Since (,) (n-m),

the coefficient of x"-t-k in (1 + x)"(1 + x + x + ..-)is Y’.."i=k+ (7) The given sum istherefore the coefficient of x"-l in

(1 + x)2n(1 q- x q- X -}- .)2_ (t+x)" d (t+x)" (t+x):"-l

2n(1 x) dx (1 x) x

Hence the sum is the coefficient of x" in

n(1 + x)2n(1 x) -1 2n x(1 + x)2n-l(1 x) -1.

Now the coefficient of x" in (1 + x)2n(1 x) -1 is

,+ (?)= + (2/.n) q_ (2r/n) 22n_,:o. ++:o + n)

Likewise the coefficient of x" in x(1 + x)2n-l(1 x) -1 is 22n-2. Therefore the given sumis equal to

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PROBLEMS AND SOLUTIONS 123

Solution by H. PRODINGER (Vienna University, Vienna, Austria).The desired sum may be written as

n

O<=k<n O<=i<=k<j<__n (ni) (j)

=no,.<.()(n- n- i) n (2 11z. )=no=,.<n()(n- 1- )"Solution by D. E. KNUTH (Stanford University).

Equivalently, the given sum is

Thus

Sn+! -’"{(")j<k _qt_ (j n i)}{() ( n )}(k _j) ,, + ,2 + ,3 ++ k

where

and

Hence

n

This gives the recurrence

which is satisfied by

S.+ 4S. + (2nn), So=0

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124 PROBLEMS AND SOLUTIONS

Incidentally, this problem is closely related to the evaluation of

n nS--n= (j) (nk)max(j,k), S_n ..k (j) (nk)min(j,k).Indeed, since S---, _Sn 2Sn and S---, + _S (7)(7,)(J + k) n. 2, we have -nn. 22n-l + Sand_S- n. 22-1 Sn.

Editorial note. Can one explicitly sum the analogous extensions of Sn and Sn to ruthorder summations? [M.S.K.]

Also solved by J. C. BUTCHER (University of Auckland, Auckland, New Zealand),C. GEORHIOU (University of Patras, Patras, Greece), H. VAN HAERINGEN (DelftUniversity of Technology, Delft, the Netherlands), S. D. HENDRV (Baltimore, MD), M.HOFFMAN (Memorial University of Newfoundland), E. Hou (no affiliation or addressgiven), A. A..lAGERS (Technische Hogeschool Twente, Enschede, the Netherlands),M. S..IANKOVlC (PHH Engineering, Calgary, Alberta), R. A. JOHNS, (E-Systems,Garland, TX), W. B. JORDAN (Scotia, NY), I. KINNMARI< (Princeton, N‘l), O. P.LOSSERS (Eindhoven University of Technology, Eindhoven, the Netherlands), W. A. J.LUXEMBURG (California Institute of Technology), ‘l. PIEPERS (Castricum, the Nether-lands), M. RENARDV (University of Wisconsin), ‘l. RtJPPERT (Wirtschafts UniversititWien, Wien, Augasse), O. G. RUEHR (Michigan Technological University), A. SIDI(NASA, Cleveland, OH), .l.A. WILSON (Iowa State University), P. Y. Wu (NationalChiao Tung University, Hsinchu, Taiwan) and the proposer.

A Covering Problem

Problem 83-4*, by J. VIJAY (University of Florida).Let M be a set of m fixed and distinct points in the plane. Define a hoop of radius

r > 0 to be any maximal subset ofM that is simultaneously coverable by a circle of radiusr. Obviously, the number of such hoops depends on both r and the locations of the mpoints. For example, if r is very small then each point is a hoop, and if r is very large thenthe only hoop is M itself.

It is conjectured that for a given m, the number of hoops cannot exceed 3mirrespective of the locations of the m points and the value of r. This bound is attained in thelimit when m-- and the points are in an equilateral triangular lattice with 2r being thedistance between any point and any of its six closest neighbors.

Solution by O. P. LOSSERS (Eindhoven University of Technology, Eindhoven, theNetherlands).The conjecture is false!Consider in the equilateral triangular lattice the following configuration.

T

It is a matter of straightforward checking that every hoop has the above form (apart froma rotation over a multiple of r/3). Since every hoop has a unique "top" T and any point isthe top of 6 hoops, it follows that 6m(l + a(1)) is the number of hoops for m .

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