ADVANCED PROBLEMS AND SOLUTIONS

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1 Edited by Rayond E* Whitney Please send all counications concerning ADVANCED PROBLEMS AND SOLUTIONS to RAYMOND E. WHITNEY, MATHEMATICS DEPARTMENT, LOCK HAVEN UNIVERSIIY, LOCK HAVEN, PA This departent especially welcoes probles believed to be new or extending old results. Proposers should subit solutions or other inforation that will assist the editor. To facilitate their consideration, all solutions should be subitted on separate signed sheets within two onths after publication of the probles. PROBLEMS PROPOSED IN THIS ISSUE H-550 Proposed by Paul S. Bruckan, Highwood, IL Suppose n is an odd integer, p an odd prie ^ 5. Prove that L n = \ (od p) if and only if either (i) a n = a, J3 n^ fi (od/?), or (ii) a n = fi, (3 n = a -(od/?). H-551 Proposed by N. Gauthier, Royal Military College of Canada Let k be a nonnegative integer and define the following restricted double-su, a-\ b-\ r=0 s=0 br+as<ab where a, b are relatively prie positive integers. a. Show that S k _ x - 2, T kb b t d {{ab + rf-a k r k )-f j { k n\b S k _ n for k > 1. The convention that (*) = 0 if > k is adopted. b. Show that & = [3a 2 h 2 +2a 2 b + 2ah 2 -a 2 -b 2 2-9ah 12 L + a + b + 2l H-552 Proposed by Paul S. Bruckan, Highwood, IL Given > 2, let {UJ =0 denote a sequence of the following for: where the a f \ and 0/s are constants, with the $/s distinct, and the U n 's satisfy the initial conditions: [/ w = 0,w = 0,l,...,fw-2; U ^ = l Part Ao Prove the following forula for the C/ W 's: 5(n-/w+l, ) 1999] 185

2 where S(N,) = {(i h i 2,...,i ):i l + i i = N,0<i J <N,j = l,2,...,}. (b) Part B. Prove the following deterinant forula for the U 's: U \ 02 0, 0 (? ; ) 2 {6 2 f (0,f... (0J 2 {oi - 2 {o 2 r 2 {0i) - 2 ' {0 r 2 w {02) n {0,r {0 r i i 0\ 02 0, {0 }? {02? {03? ; {0? {oir 2 {0 2 r- 2 {0 3 r- 2 ' {0J - 2 {0d ~ l {02r l W 1 - {0 ) SOLUTIONS Su Proble H-535 Proposed by Piero Filipponi & Adina Di Porto, Roe, Italy (Vol 35, no. 4, Noveber 1997) For given positive integers n and, find a closed for expression for Z =1 k F k. Conjecture by the proposers: 2*» = tk F k = p[ \n)f n+l + p^\n)f n+ C, (1) where p[ \n) and p^iri) are polynoials in n of degree, A () («) = Z ( - l ) ' ^ > ", ^ ) (» ) - I ( - l ) ' e > - ' > (2) 7=0 7=0 the coefficients a w) and b^ {k = 0,1,..., ) are positive integers and C is an integer. On the basis of the well-known identity Z h ={n-2)f +1 + (n-l)f + 2, (3) which is an alternate for of Hoggatt's identity I 40, the above quantities can be found recursively by eans of the following algorith: 1. p[ ^\n) = { + \)\p[ \n)dn + (-\) +y 4 +x \ p( +i \n) = ( + 1)J pi \n) dn + (- l) +1^+1). +l » = J(a/ w + 1 ) +^+1 >). +l 3. bt +V) = a< ffl+1 >. 4. C +l = (-\ya^\ 186 [MAY

3 Exaple: Rearks: The following results were obtained using the above algorith. i; 25 = (/i 2-4/i-f8)F +1 + (/i 2-2^ + 5 ) ^ - 8, E 3, n = O 3 ~ 6n2 + 24n ~ 50 ) F n+i + O 3-3n 2 +15n-3l)F n +50, %4,n = (n 4 -&n 3 +4%n 2-200n + 416)F n+l + (n 4-4n n 2-124w + 257)F -416, S 5> w = (n 5-1On 4 + 8(k 3-500w O? )i^+1 + («5-5«4 +50/i 3-310w «-2671)i^ (i) Obviously, these results can be proved by induction on n. (ii) It can be noted that, using the sae algorith, E ljw can be obtained by the identity (iii) It appears that and a( + k) i b( + k) = C ( ) n s t = a() / ^) (jt = 1, 2,... ) \i ajf> / bjp = a. W > 0 0 Solution by Paul S. Bruckan, Highwood 9 IL We begin by defining certain polynoials of degree n, as follows: We then see that k=\ Now let [/denote the operator xdldx. Note that Z, = 5- {F Ja)-F, n (fl)}- (2) It is easily seen that Repeated application of the recurrence in (3) yields the following: U(F Jx)) = F +ln (x). (3) F 0 Jx) = (x" +l -x)/(x-l). (4) F hn (x) = {nx" +2 -(n + \)x"+ l +x}l{x-\) 2 ; (5) F 2, (x) = {n 2 x" +3 - {In 2 +2n- l)x" +2 + {n + 1)V x 2 - x) I {x -1) 3 ; (6) More generally (by induction or otherwise), F, n (x) = {x" +l P Jx)-P ^(x)}/(x-ir\ >0, (7) where P t (x) is a polynoial in x of degree. We ay suppose />, (*) = 40",»K- (8) r=0 1999] 187

4 Note that P 0n (x) = l, P x n (x) = nx-(n + l\ P 2 n (x) = n 2 x 2 -(2n 2 +2n-l)x + (n + l) 2, etc. Repeated application of (3), using (7) and (8), yields the following general forula for ^(, n): Ari, w) = (-\y- r +ic s _ r (n - w + s). (9) Substituting the last expression into (8) yields the following developent in the "ubral" calculus, involving the finite difference operators E and A (with operand z at indicated values of z): r f 1 P,n(*) = I * " "! ( -! ) ' +lc s (n-r + sy = x"< ( - ) ' ^ i Q E ( 1 / W \(z ) r=0 s=0 U=0 r=s J fw+1 1 = x»- (-EY +lc s ((Ex)-"- 1 - (Ex)-') I ((ExT 1-1) (2 ) = {E~ I (1 - x) (1 - ) +1 -E/(l-Ex)-(x- l) +1 } (2 ) ^ Note that (E - l) +1 (z ) = A +1 (z ) = 0. Thus, since E = 1 + A, In particular, i > - (x) = - ( x - i r I { 0 - x E ) - 1 } ( O Z=»+l = (x - l) {(l + Ax / (x -1))" 1 } (z ) lz=«+l and likewise, P >» = a- {(l + a 2 A)- 1 }(z"') z=m+i = p, n (P)=p- ll(-vp ls K(n a- M, (riya 2 'A\zy z=«+l z=n+l Now, substituting these last results into the forula in (7), we obtain f F n (a) = a +l \a n+l - J^(-l) s a 2s A s (n + l) -a- X(-iya 2 W(0) ] I s=0 s= s=0 = S (-V) s {a" +2s+2 A s (n + \) - a 2s+1 A s (0) }, s=0 where, for brevity, we write A s (a) for the ore precise expression A s (z) \ z=a. Siilarly, we obtain the following expression: F,nifi) = E (-l) s {P n+2s+2 tf(ri + l) - {3 2s+l A s (0) }. We ay now substitute these last expressions into (2) and obtain ^.^ZHra^A^+ir-F^AW}. (10) 188 [MAY

5 This is still not in the for that is envisioned by the proposers, but it only takes a bit ore effort to put it into such for; we resort once ore to the ubral calculus. Returning to our earlier notation, we proceed as follows: Then, «^,»(«)=E(- 1 ) I «2SAS («+i r = {(l-(-a 2 A) +1 )/(l + a 2 A)}(» + l) = {l/(l + a 2 A)}(«+ l). 5- {a^2p n{a) _ p+n+ 2p n finally, we ay recast (10) as follows: where = 5-1/2 {a" +2 /(l + a 2 A)-^" +2 /(l + y0 2 A)}(n + l) -{(F +2+J F A)/(l + 3A + A 2 )}(«+ i r = {(iv, +1 + (l + A)F n )/(l + 3A + A 2 )}(«+ ir; Z, = F n+l G(A)(n + \y + F G(A)(n + 2T-G(A)(lT, (11) Coparing this with the desired forat, we have G(A) = 1/(1 + 3A + A 2 ). (12) p[ \n) = G(A)(n +1)-, p^\n) = p[ \n +1), C = -p[ ) (0). (13) Therefore, we ay express p[ ) and p$"\ as well as C, in ters of essentially only one polynoial; thus, 2 * n = F n+ ip () (n) + F n p^\n + 1) - p0">(0), (14) where p { \n) = p[ \n), as given in (12) and (13). This is a soewhat stronger stateent than the initial conjecture, since it gives a ore specific expression, relating the three quantities p[ \n\ p { \n), and C. Now we need to deterine the coefficients of these polynoials. Let p( >(rt) = (-l) -'a,< V; note that p {) (0) = (-\) a^) = -C, which gives, essentially, Part 4 of the proble. Also, fro (13), p^\n) = G(A)(n + \). Now therefore, G(A) = l/(l + 3A + A 2 ) = 5-1/2 {a 2 /(l + a 2 A)-/? 2 /(l + /9 2 A)} s = K-O'Jw* =TnYF 2s+2 (E-iy = K-iyF 2j+2 X(-i) r,c r ~; 5=0 s=q s=0 r=0 P () («) = I(-l) s^+2z(-l) r sc r (" + l + 5-r)" 5=0 r=0 1999] 189

6 s=0 r=0 7=0 s = I"'»QE(-i)'^2Z(-iy,C' rj E:^(ir' 7=0 s=0 r=0 = I > MQl(-l)^2s+ 2A 5 (ir-' = I > Q G(A)(1)-'. 7=0 ^=0 7=0 Coparison of coefficients in the two expressions for p^\n) yields ^) =(-ir' QG(A)(ir-'. In siilar fashion, we ay obtain the following expression: */ w =Hr w QG(AX2r. (i6) +i) We then see fro (15) and (16) that aj /b^+i) = G(A)(l) / G(A)(2). However, we see that such result is independent of/, so we express such ratio as p(). Therefore, we have (is) a( + i) /b( + i) = am /b () = ^ ^ We now return to the proble of deterining p = li^^ p(). First, however, we deduce soe additional properties of the coefficients a\ >) and b\ \ Fro the expressions in (15) and (16), it readily follows that Then We see then that a\ )^lia\^l\ $ ) =/ibfr l \.i = ],2,...,. (18) p { \n) = ( - l ^ a ^ W = {-l) 4 ) + ( - l ) ^ W z ^ f V 7 = 0 7=1 w-1 - (-l) w a^> + w (-l) w " w^(w - 1 >(/i, " +1 // +1). 7=0 pi \n) = (-l) w a^} +fv w _ 1 ) (0^? (19) which is essentially Part 1 of the proble (stated soewhat ore precisely). We now return to the expression in (13), naely, p ( \n) = G(A)(n + l). (20) Note that we ay allow n = 0 and n = -1 in this last expression, but that if n < - 2, extraneous and unintended ters arise that ake the expression incorrect. Note t\tp ( Xn + l) = (l + A)p ( \n), while p () (n + 2) = (l + A) 2 p M (n); then p () (n + 2) + p ( \n +1) = {1 + A + l + 2A + A 2 }/? w («)={l + l + 3A + A 2 }G(A)(w + i r = {l + G(A)}(// + l)^or In particular, setting «= -1 (and assuing >0), Now observe that /> (w) (w + 2) + /? (w) (7f + l) = (/i +1) /? (w) (//). (21) p {) (l) + p { \0) = p ( \-l). (22) 190 [MAY

7 and /> (M) (-i)=(-irl>/' 7=0 yo) Pw(o)=c-ir <#»>=^(-i)=(-i)» bf \ p< >(l) = ^>(0) = (-l) $">. Fro (22), we then deduce the following: cf ) + tfp = 4 ) or $*> = a/ w >, valid for > 0; J=0 /=1 this is essentially Part 3 of the proble. Also a^ = ^-) + b W = ( a (»0 + i>(-)) 3 v a ii d for /w > 0; /=i /=i this is essentially Part 2 of the proble. That the a\ >) and b^ are integers is obvious fro the forulas in (15) and (16). That they are positive requires ore effort. An easy induction on (18) yields the following relations: 4 ) = npdt*, */ w) = jcfltk i = 0,1,...,. (23) Thus, aj \b^) > 0 if and only if a^\b^) ) > 0, = 0,1,... Fro (23) and Parts 2 and 3 of the proble, we see that, if a^) > 0, - 0,1,2,..., then a\ ) > 0 =>b^) > 0 =>b\ ) > 0 (/ = 0,1,..., ). Thus, it suffices to prove that a^) > 0, = 0,1,... Our proof of this assertion is by induction (on ). The inductive step depends on the following recursive forula: [il] 4 +1) = X ^ / ^ ' + l), = 0,1,2,... (24) If (24) is vailid, our inductive hypothesis is that a^ > 0 for soe > 0. Fro our foregoing discussion, this iplies that a^ > 0, / = 0,1,..., [il]. Then (24) iplies that a ( 0 +l) > 0. Since a^ = 1, this proves the hypothesis. Our task is thus reduced to proving (24). We return to the results in (19) and (22). Expressing the integral recurrence in ters of the coefficients (and replacing by +1), we obtain then which is equivalent to (24). D f l (-l) 4 +l) = (w + 1) X(-0 w ~'{l + ( - 0 > ^ ' ' d t ; Jo /=o /=0 a ^ = ( + l)j^{l + (-iy}aj ) /(i + l), Note: As rioted, we allowed the values n = 0 and n--\ although, in the original stateent of the proble, n was required to be positive. It ay be observed that if we substitute the values n - 0 or n = -1 in the forula given by (14), we find that the expression for S w dutifully vanishes, as we should expect fro its original definition. 1999] 191

8 The only reaining task is to establish that p = a, as conjectured by the proposers. We write a for a^ and h for h^\ Fro (18), we easily deduce the following recursive forulas: -\ -\ a = X Q ( a, + b t ), b = Y j C i a i, = l,2,... (25) (=0 1=0 We ay also express a and b in the ubral calculus as follows: a = (-1)"G(A)(1)", b = (-l) +1 {AG(A)}(0)'". (26) It is apparent that a and h are unbounded (as» oo). Also note that 0<b <a for all >\. Then (25) iplies that 1 <p() <2 for all > 1. Fro the expressions for a w and h in (26), it is not difficult to show (by expansion of the operators) that p exists. We ay express (25) in the following for: \ -\ a = X CXpQ) +1)*/, * = E Q P ( 0 * = \2,... (27) ;=0 /=0 Hence, p() = l + l/p(-l), where ^-(w-1) is a "weighted average" p, and the "weights" are the quantities ^Q^- in the sus. Letting w->oo, we ay therefore deduce that \<p<2 and p = 1 + p _ 1 This, in turn, iplies that p- a. For additional confiration, the siple continued fraction (s.c.f.) expansion for p() approaches the infinite s.c.f. [1,1,1,...], which is known to be the s.c.f. for a. Also solved by I Strazdins. > * > * > NEW PROBLEM WEB SITE Readers of The Fibonacci Quarterly will be pleased to know that any of its probles can now be searched electronically (at no charge) on the World Wide Web at Over 23,000 probles fro 42 journals and 22 contests are referenced by the site, which was developed by Stanley Rabinowitz's MathPro Press. Aple hosting space for the site was generously provided by the Departent of Matheatics and Statistics at the University of Mirrouri-Rolla, through Leon M. Hall, Chair. Proble stateents are included in ost cases, along with proposers, solvers (whose solutions were published), and other relevant bibliographic inforation. Difficulty and subject atter vary widely; alost any atheatical topic can be found. The site is being operated on a volunteer basis. Anyone who can donate journal issues or their tie is encouraged to do so. For further inforation, write to: Mr. Mark Brown Director of Operations, MathPro Press 1220 East West Highway #1010A Silver Spring, MD (301) (Voice ail) bowron@copuserve.co (e-ail) 192