Hoggatt and King [lo] defined a complete sequence of natural numbers

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1 REPRESENTATIONS OF N AS A SUM OF DISTINCT ELEMENTS FROM SPECIAL SEQUENCES DAVID A. KLARNER, Uiversity of Alberta, Edmoto, Caada 1. INTRODUCTION Let a, I deote a sequece of atural umbers which satisfies the differece equatio \+o ~ \+i + a k ^ o r k = 1,2, '. ^ is easy to prove by iductio that a x + a 2 + " ' + a = a +2 " a 2 f o r = 1,2?«; we use this fact i defiig (1) P(x) = I I (1 + x ~ l = V A(k)x k k=l k= ad a +2 a 2 - \ * * (2) P ^«=( 1+^)= E A «= J I II,+ x - j = ^ A^(k)^ k=l k= It follows from these defiitios that A(k) eumerates the umber of r e p r e- setatios (3) a i t + ai a^. = k with < i < < i. f ad that A (k) eumerates the umber of these represetatios with i. <. Hoggatt ad Basi [9] foud recurrece formulae satisfied by {A (k) ad {A(k) J whe a I is the Fiboacci sequece; i Sectio 2 we give geeral - izatios of these results,, Hoggatt ad Kig [lo] defied a complete sequece of atural umbers ia I as oe for which A() > for = 1,2,«-- ad foud that (i) JF I is complete, (ii) JF I with ay term deleted is complete, ad (iii) F I with ay two terms deleted is ot complete. Brow [l] gave a simple ecessary ad 289

2 29 REPRESENTATIONS OF N AS A SUM OF [Dec. sufficiet coditio for completeess of a arbitrary sequece of atural umbers ad showed that the Fiboacci sequece is characterized by properties (ii) ad (iii) already metioed. Zeckedorf [l 3] showed that if F 1 is deleted from the Fiboacci sequece, the the resultig sequece has the property that every atural umber has exactly oe represetatio as a sum of elemets from this sequece whose subscripts differ by at least two. Brow [2] has give a expositio of this paper ad Dayki [4] showed that the Fiboacci sequece is the oly sequece with the properties metioed i Zeckedorf s Theorem. More o the subject of Zeckedorf s Theorem ca be foud i aother excellet paper by Brow [3]. Fers [5], Lafer [ l l ], ad Lafer ad Log [] have discussed various aspects of the problem of represetig umbers as sums of Fiboacci umbers. Graham [6] has ivestigated completeess properties of jf + (-1) ^ ad proved that every sufficietly large umber is a sum of distict elemets of this sequece eve after ay fiite subset has bee deleted. I Sectio 3 we take up the problem of determiig the magitude of A() whe la 1 is the Fiboacci sequece; i this case we write A() = R(). Hoggatt [7] proposed that it be show that R(F 2-1) = ad that R(N) > if N > F 2-1. We will show that ad that F < N < F J L l» l implies - +1 ^ [ H 1 ] ^ ROT < 2 F (+1)/2 if is odd ad J-^_2j < R(N) < F(+4>/2 if is eve. I Sectio 4 we ivestigate the umbers of represetatios of k as a sum of distict Fiboacci umbers, writig a = F. ad T() for A() i this case. The behavior of the fuctio T() is somewhat differet from that of

3 1966] DISTINCT ELEMENTS FROM SPECIAL SEQUENCES 291 R() of Sectio 3. For example, we show that there exist ifiitely may for which T() = k for a fixed k 5 ad i particular we fid the solutio sets for each of the equatios T() = 1, T() = 2, T() = 3. By defiitio T() < R() so that T(N) < - 1 if N < F We show that T < F +1 > - [ 4-1 ] ad T ( F ) = [/2] for = 3, 4, ". Hoggatt [8] proposed that oe show that M(), the umber of atural umbers less tha which caot be expressed as a sum of distict Lucas umbers 1^(1^ = 1, L 2 = 3, L + 2 = L L Q ) has the property M(L ) = F -; also, he asked for a formula for M(). I Sectio 5, we give a solutio to the same questio ivolvig ay icomplete sequece satisfyig a «= a - + a with a 1 < a^ <. I a paper ow i preparatio we have show that the oly complete sequeces of atural umbers which satisfy the Fiboacci r e - currece are those with iitial terms (i) a~ = a«= 1, (ii) a- = 1, a^ = 2, or (iii) a- = 2, a,, = RECURRENCE RELATIONS See Sectio 1 for defiitios ad otatio. Lemma 1. A (k) = A ( a + 2 " a 2 " k^ f o r k =» 1 >""> - Proof. Usig the product otatio for P we see The symmetric property of A ow follows o equatig coefficiets of the powers of x i (4). Lemma 2. (a) A +1 (k) = A (k) if < k < a (b) A +1 (k) = A (k) + A (k - a + 1 ) if a + 1 < k < a ^ (c) A +1 (k) = A (k - a + 1 ) if a a < k < a Proof. Each of these statemets is obtaied by equatig coefficiets of x i the idetity v

4 292 REPRESENTATIONS OF N AS A SUM OF [Dec. (5) P +1 (*) = (l + x + 1 j P Q (X) +r Lemma 3. (a) A +1 (k) = A(k) if 4 k 4 a (b) A +1 (k) = A(a a 2 - k) + A(k - a + 1 ) if a k 4 a a 2. (c) A +1 (k) = A(a a 2 - k) if a a k 4 a ^ - v Proof, (a) This follows by iductio o part (a) of Lemma 2. (c) Usig Lemma 1 we have A +1(k) = A + 1(a a^ - k) ad assumig a a < k < a a 2 we have 4 a Q+3 - a - k 4 a Q+1-1, so that we ca apply (a) of this lemma to get A + 1 (a - su - k) = A(a «- a 2 - k) for k i the rage uder cosideratio ad this is (c). (b) Statemet (b) of Lemma 2 asserts A _ (k) = A (k) + A (k - a..) for A 1 4 k 4 a a 2 ; but by (c) of this lemma wehave A (k) = A(a? - a 2 - k) for k i the rage uder cosideratio. Also, if a - 4 k 4 a 2 - a. we have 4 k - a a 2? so by (a) of this lemma we have A (k - a -) = A(k " a +i)' Combiig these results gives part b. Lemma 4. (a) A(k) = A(a +2 - a 2 - k) + A(k - a + 1 ) if a + 1 < k < a a 2 ad = 2, 3, " -. (b) A(k) = A(a a 2 - k) if a a + 1 < k < \ ad = 2, 3, (c) A(a +2 - a 2 + k) = A(a - a 2 + k) if 1 < k < a 2-1. (Note that i (b) ad (c) the rage of k is the empty set uless a 2 > 2.) Proof. (a) This is merely a combiatio of (a) ad (b) i Lemma 3. (b) If a a 2 + l < k < a l? the a a^ + 1 < a a - k - a +l ~ ±9 S t h a t b y ( a ) o f L e m m a 3 > A ( a +3 " \ ~ k ) = A +l ( a +3 ~ a 2 " k ^ A By Lemma 1, + i ( a a 2 - k) = 'A -(k) ad usig (a) of Lemma 3 agai we see that A +1(k) = A(k) for k i the proposed rage. (c) Writig k = a a 2 + j with 1 < j < a 2-1 i (b) we get (6). A(a +2 - a 2 + j) = A(a a 2 - a Q+2 + ^ - j) = A(a j) ;

5 1966] DISTINCT ELEMENTS FROM SPECIAL SEQUENCES 293 b u t a +l " ^ = a +l " a 2 + ^a2 " ^ w h e r e 1 < & 2 " ^ - a 2 " X s o t h a t w e c a use (6) to obtai ( 7 ) A < a +1 " j ) = A < V l " a 2 + <«2 " j ) ) = A < a " < a 2 "»> Combiig (6) ad (7) we obtai (c). Lemma 5. A(a j) = A(a +2 - a 2 - j) for < j < a - a 2 ad = 2,3, Proof. For j i the rage uder cosideratio we have a, < a, < a +2 " a 2 s o ^ a t ^ v ^ ^ ^ e m m a ^ w e have (8) A(a +1 + j) = A(a +2 - a 2 - a +1 - j) + A(a +1 + j - a +1 ) = A(a - a 2 - j) + A(j). But we also have a.. < a - a - j < a - a~ for the assumed rage of j, so that we ca apply Lemma 4 agai to write ^ A < V 2 " a 2 ' j ) = A ( a a 2 " a a 2 + j ) + A ( a + 2 " a 2 ^ ~ a +1 ) = A(j) + A(a - a - j). Sice the right members of (8) ad (9) are the same, so are the left members. Usig Lemmas 4 ad 5 it is ot hard to calculate A(k) for a give s e - quece ja l. Of particular iterest to us are the cases whe a I is the Fiboacci sequece, the Fiboacci sequece with the first term deleted, ad the Lucas sequece; we write A(k) = R(k), T(k) ad S(k) respectively i these cases. A table Is provided for each of these fuctios i order to i l l u s - trate some of our results. 3. SOME PROPERTIES OF R(k) I light of Lemma 4, it is atural to cosider the behavior of H(k) i the itervals [F, F - 1]; thus, as a matter of coveiece we write (1) I = {R(F ),R(F + 1), - -, R(F + 1-1)}

6 294 REPRESENTATIONS OF N AS A SUM OF [Dec. ad ote that Lemma 4 implies (11) I + 1 = (R() + R(F - 1), R(l) + R(F - 2),, R(F - 1) + R()}. As we metioed i the itroductio, Hoggatt has proposed that oe prove R(F 2-1) = ad that R(k) > R(F - 1) if k > F - 1. This problem has led us to prove a result ivolvig special values of R(k) ad to fid the maximum ad miimum of R(k) i I. Theorem 1. (a) R(F ) = [ - f 1 ] for > 1, (b) R(F - 1) = p ^ p ] for >, (c) R(F - 2) = - 2 for > 2, (d) R(F - 3) = - 3 for > 4. Proof. We prove oly (b) (the other proofs are aalogous) which implies the first part of Hoggatt f s proposal. First, we observe that (b) is true for small values of by cosultig Table 1. Next, suppose R(F t - 1) = p * X 1 for t = ad + 1 ad take k = F ~ - 1 i (a) of Lemma 4 to obtai () R(F + 2-1) = R() + R(F - 1) = 1 + [ ' H f 1 ] m Thus, the assertio follows by iductio o. Theorem 2. is a miimum of R(k) i I. *<*»> [H*\

7 1966 ] DISTINCT ELEMENTS FROM SPECIAL SEQUENCES 295 Proof. We ca verify the theorem for small values of by ispectio i Table 1. Suppose the theorem holds for all < N - 1. We kow by Theorem 1 that «<v [H*\ so that we are assumig (13) PHr^l = R ( F ) - R ( k ) for F - k - F +1 " X ad = 1,2,, N to obtai Now suppose F N 4 k 4 F N ad write = N - 1 i (a) of Lemma (14) R(k) = R(F N k) + R(k - F N ) ; but F N 4 k 4 F N implies 4 F N k 4 F^_ ± - 1 ad 4 k - F N < F N _ 1-1. Suppose (15) F t < F N k < F t + 1-1, where of course F t < F N or < t N - 2 (we are takig F Q ). Now (16) F N - F m < k + F N - F N _ t < F N - F t - 1, but with < t < N - 2 we must have F N _ 2 < F N - F t + 1 ad F N - F t - 1 < F N 1-1 so that evidetly < 1? ) F N-2 ^ k " F N S F N-1-1 Usig (16) ad (17) alog with (13) we have (18) [ f ] < R(k - F N )

8 296 REPRESENTATIONS OF N AS A SUM OF [ D e c. ad (19) 1 < p T ^ l ~ R ( F N + 1 " * " k ) s i c e t ~ ' Combiig (18) ad (19) i (14) gives (2) -«^ [I] + ' - [H*\ for F AT k F X T 1-1. Hece the theorem follows by iductio o N. J N N+l Corollary. R(k) > R ( F 2-1) = if k > F 2-1. Proof. We kow from T h e o r e m 2 that the miimum value of R(k) i I 2 ad I 2.-I is + 1 i each of them; hece the miimum of R(k) i J 2 U L is + 1. Thus, every value of R(k) i I? 2 U I 2,«i s a t least + 2 so that we ca coclude by iductio o that R(k) > R ( F 2 > F ) if k Theorem 3. The maximum of R(k) i I 2 is F 2 ad the maximum of R(k) i I is 2 F + 1 for = 1,2, ; also, (21) F 3 ^ 2 F 2 < F 4 < 2 F 3 < < F ^ < 2 F + 1 < F + 3 < for = 2, 3, - -. Proof. The result i (21) follows by a simple iductio. The r e s u l t s cocerig the maximum values of R(k) i I 2 ad I ca be verified for small by usig Table 1. Suppose these r e s u l t s hold for all ^ N; the we have by (a) of L e m m a 4, (22) R ( F *) = R ( F - t - 1) + R(t) for < t < F - 1. Also,.we kow by (b) of Lemma 4 that R(k) is s y m m e t r i c i I -, so it is eough to cosider the values of oly the first half of the elemets of I - i o r d e r to determie the maximum elemets. elemets of I M a r e cotaied i the sets +l More tha the first half of the (23) { R ( F t ) t =,!, -, F _ 1 - l } a d { R ( F t ) t = F ^, - -, F - l }.

9 1966] DISTINCT ELEMENTS FROM SPECIAL SEQUENCES 297 Cosider first the maximum of the first of the two sets i (23); evidetly, m a x R(F + t) = m a x {R(F^ - t - 1) + R(t)} i <t<f, - 1 <t<f, (24) ^ m a x R(F - t - 1) + m a x R(t) <t<f, - 1 V <t<f, =' 2 m a x!. -Z Next, we have for the secod set i (23) m a x R(F,. + t) + 1 F - t ^ F ^ m a x R(F - t - 1) + m a x R(t) (25) x F ^ t ^ F -1 F, ^ t ^ F -1 ' -1-1 = max I o + m a x I Together (24) ad (25) imply (26) max I + 1 < max {max I _ 1 + max I, 2 max I 2 } * Writig = 2N + 1 i (26) ad applyig the iductio hypothesis we have (27) max I 2 N + 2 max { F N + 3, 4F N > = F N + 3 ; similarly, = 2N + 2 i (26) gives (28) m a x I 2 N + 3 max {2F N + 2, 2F N + 2 } = 2F N + 2. I order to fiish the proof of Theorem 3 we eed to show that F N + 3 e T 2N+2 *** 2 F N+2 e Ws" Sice ^ F 2 N + t F 2 N for t =,1,, F, ^ - 1, we ca use (22) ad (b) of Lemma 4 to fid (29) R(F 2 N F 2 N + t) = R(F 2 N t - 1) + R(F 2 N + t) = 2R(F 2 N + t),

10 298 REPRESENTATIONS OF N AS A SUM OF [Dec. for t =,1,, F 2 N From this we gather that all of the elemets of I 2 N multiplied by 2 occur i I 2 N + T e c e > twice the maximum i I 2 N is i Iovr+o a( * this is precisely 2 F N + 2. It is ot so obvious that F N + ~ Io N + 2 ; t o P r ve this we let X. deote a iteger such that R ( F 2 N + X ) = F + 2 for ^ N. We will also iclude i our iductio hypothesis that a admissible value of \ - for < N is give by X - = F X - 1. Now cosider R(F 2N+2 + F 2N-1 ~ X N " 1 ) m ( } = R ( F 2N+1 " F 2N-1 + V + R ( F 2N-1 " X N " X) = R < F 2N + X N> + R < F 2N-2 + W F N+2 + F N+1 = F N+3 The secod equality i (3) follows from (22). It is ow clear that a admissible value for X N + 1 is F on-l " X N ~ 1 a d t a t F N + 3 I 2N+2* T h i s c o m P l e t e s the proof of Theorem T(), THE NUMBER OF REPRESENTATIONS OF AS A SUM OF DISTINCT FIBONACCI NUMBERS For the momet we are takig a = F,. i the lemmas of Sectio 2 +l ad write A(k) = T(k) i this case. The followig theorem ca be proved i the same way we proved Theorem 1, so we leave out the proof. Theorem 4. (a) P + 1> = [ ] * = 1 ' 2 '' (b) T(F ) = [ ] if = 3,4," Theorem 5. (a) T(N) = 1 if ad oly if N = F for = 1,2,. (b) T(N) = 2 if ad oly if N = F F - 1 or F R F - 1 for = 1,2,' ". (c) T(N) > if N 2=. (d) T(N) = 3if adolyifn = F ^ + F - 1, F F + 1-1, ^ 6 " F -^ Ve^Vl" 1 f o r = 1.2,---. Proof, (a) ad (c): We ca check Table 2 to see that T(F - 1) = 1 if = 1,2,3,4. Suppose T(F ) = 1 for all less tha N > 4. The

11 1966] DISTINCT ELEMENTS FROM SPECIAL SEQUENCES 299 by (c) of Lemma 4 we have T(F N - F 3 + 1) = T(F N - 1) = T(F N _ 3-1) which is 1 by assumptio. Next, the table shows that the oly values of N < F r for which T(N) = 1 are N = F 2-1, F 3-1, F 4-1 ad F 5-1. Suppose for all 4 ^ < N, where N > 5, that F < k < F - 1 implies T(k) > 1. The by (a) of Lemma 4 we have for F XT ^ k < F ATl1-1, T(k) = T(F XT. - JN J N + 1 N + l F 3 - k) + T(k - F N ) > 2. This completes the proofs of both (a) ad (c). (b) By Lemma 5, we have T ( F F Q - 1) = T ( F F - F 3 + 1), ad sice F + 3 ^ F F - 1 < F F 3 we ca apply (a) of Lemma 4 to get T ( F F - 1) = T ( F F 3 - F ^ - F Q + 1) + T(F V 1 - F + 3 ) = T(F + 1-1) + T(F - 1). By (a) of this lemma, the last sum is 2. To prove the "oly if' part of (c), we use iductio with (a) of Lemma 4 just as i the proof of the "oly if" part of (a). (d) The proof ca be give usig iductio ad (a) of Lemma 4 just as (a) ad (b) were proved. Theorem 6. For every atural umber k there exist ifiitely may N such that N has exactly k represetatios as a sum of distict Fiboacci umbers, i fact, (31) T ( F + k F + 2-1) = k for = 1,2, ad k = 4, 5, '. Proof. The theorem is true for k = 1,2, 3, by (a), (b),ad{d)of Theorem 5. We will verify the theorem for k = 4 ad leave the verificatio for k = 5 as a exercise. Sice F + 6 ^ F F < F F 3 we ca apply (a) of Lemma 4 to obtai T < F F + 2 " *> = T < F + 7 " F 3 ' F + 6 " 2 F X > < 32 > + T ( F F F + 6 ) = T ( F + 5 " 2 F + 2 " *> + T < 2 F + 2 " *> h o w e v e r, 2 F ^ = F ^ + F ^ + F = F ^ + F so that l +3 (33) T ( 2 F ) = T ( F F - l ) = 2,

12 3 REPRESENTATIONS OF N AS A SUM OF [Dec. (34) T(F F + 2-1) = T ( F F j i - 1) = 2 by (b) of Theorem 5; combiig (33) ad (34) i the last member of (32) gives the desired result. Now suppose (31) holds for all k < N where N > 5. Sice F, + N + o - F +N F +2 " X ~ F +N+3 ~ F 3 ' w e c a u s e ^ o f L e m m a 4 t o o b t a i T < F + N F + 2 " X > = T < F + N + 3 " F 3 " F + N F > < 3 5 > + T < F + N F + 2 " * ~ F W = T < F + N + l " 2 F + 2 "!> + T < 2 F + 2 " ^ Sice < 2F 2-1 ^ F + N F~ we ca use Lemma 5 to write T ( F + N + l " F 3 " 2 F *> = T < F + N + l " 2 F + 2 " X > (36) = T < V N + 2F + 2 -!) ' but, this last quatity is - 2 by assumptio ad recallig (33) we see that the sum i the last member of (35) is (N - 2) + 2 = N. This cocludes the proof. 5. INCOMPLETE SEQUENCES I what follows, N() deotes the umber of o-egative itegers k < for which A(k) =. Lemma 6. Let < v 1 < v 2 < deote the sequece of umbers k for which A(k) = ad suppose v, -, v, 9,,v, is a complete listig of the v T s betwee a ad a + k + i J < a,- for > 2; J +1 the v,,. t+j = a + v. j for j = l, 2,, #,, s ad v is the largest v ot exceedig k + 1. s Proof. The lemma ca be verified for = 2 ad 3 by determiig A(k) for < k < a. ^ usig P^(x), sice by (a) of Lemma 3 we have A(k) = A s (k) for k i the supposed rage. Suppose for some N ^ 3 that the v. T s betwee a ad a - are give by J a + v r j a + v OJ, a + v«where v. is the largest v. ot exceedig V 2' ' i I i a 1 ad ^ N. We will show that this implies the v. betwee a N ad a + k < a N + 1 are give by a N + v p a N + v 2,, largest v. ot exceedig k + 1. a N + v s w h e r e v s i s t h e

13 1966] DISTINCT ELEMENTS FROM SPECIAL SEQUENCES 31 have Case 1. Let a < a^ + v. ; ^NT+I ~ a o> the by (a) of Lemma 4 we A(a N + v.) = A(a N a g - a N - v.) + A(a N + v. - a N ) (37) = A(a N _ 1 - a 2 - v.) + A(v.) = A(a N _ 1 - a 2 - v.) ; but for a XT + v. i the rage beig cosidered we have ^ v. ^ a XT., - a so & & N 3 j N-l 2 that by Lemma 5 (38) A(a N a 2 - v.) = A ( a ^ 2 + v.) ad the right member is zero by assumptio, so that A(a~, + v.) = is a cosequece. Now suppose there is a t ot a v. such that a N ^ a N + t ^ ^. i ~ a o ad A(a N + t) = ; the by (a) of Lemma 4 we would have (39) A(a N + t) = A(a N 1 - a 2 ~ t) + A(t). But this is a cotradictio sice A(t) 4= (t is ot a v^ ad we assumed A(a N + t) =. Thus a^ + v v a N + v 2> ' 9, a N + v g < a^ + k <; a ^ ^ - a is a complete listig of the v. betwee a N ad a^ + k ^ a^+1 - a 2 - Case 2. Let a N a 2 < a N + v. < ^ 4 4» t h e b ^ ( c ) o f Lemma 4 we have (4) A(a^ + v.) = A(a N 2 + v.) which is zero by assumptio. If we suppose there is a t such that t is ot a v. ad a N a < a N + t < a ^ ^ implies A(a N + t) =, we obtai a cotradictio sice A(a^ + t) = A(a N 2 + t) = would imply t is a v.. Thus, a~, + v.,. -, a^ + v*, with v. the smallest v. ot less tha a M~i> c o m P r i s e s a complete listig of the v. betwee a N. - a~ ad 3^,-,.

14 32 REPRESENTATIONS OF N AS A SUM OF [Dec. iductio. Take together, the results proved i Cases 1 ad 2 imply Lemma 6 by Corollary. If A(k) > for k ^ a 9, equivalet to sayig.(a., a 2 ) = (2,1), (1,2) or (1,1). Proof. This follows from Lemma 6 ad iductio. dt the {a } is complete; this is Also, ote that if {a } is ot complete, the there exist ifiitely may k such that A(k) =. Lemma 7. (a) N(a + k) = N(a ) + N(k) if. < k < a ad = 2, 3, 4,. (b) N(k) = k if < k < a 1. (c) N(k) = k - 1 if a x ^ k < a 2. (d) N(k) = k - 2 if a 2 < k ^ a 3. (e) N ( a - 1 ) - N(a ) if = 1,2,-. Proof, (a) Suppose > 2, the by Lemma 6, the v. such that a < v. ^ a + k with ^ k -^ a are give & by J i -1 a + v 1', a + v 2' OJ ',a + j v., ' where v. is the smallest v. ot exceedig k. Hece there are N(k) v. i the supposed rage. By defiitio the umber of v. % a is N(a ) so N(a + k) = N(a ) + N(k). (b) (c) (d) follow from the fact that A(k) 4 with k < a 3 oly if k =, a l' V (e) Sice a is ever a v., N(a - 1) = N(a ). v 7 I x ' x ' Lemma 8. N(a-) = a- - 1, N(a 2 ) = a 2-2, N(a ) = a - 3 ad N(a -) = N(a ) + N ( a ^ ) if = 3,4,-. Proof. N(a 1 ) = N(a- - 1) = a 1-1 by (e) ad (b) of Lemma 7 respectively; the secod ad third statemets follow by (e) ad (c) ad (e) ad (d) of the same lemma respectively. The last statemet follows by writig k = a Lemma 7. Lemma 9. N(a ) = a - F,- if = 1,2,... ad F Fiboacci umber. - i (a) of deotes the Proof. The statemet is clearly true for = 1,2,3 ad ca be see by the first part of Lemma 8. If we suppose the statemet true for all < k(k >: 3) we ca write (41) N V i > = N ( a k > + N ( a k-i> = \ ' \+l ~ F k+2 F k + i + V i - F k

15 1966] DISTINCT ELEMENTS FROM SPECIAL SEQUENCES 33 by the last part of Lemma 8; so Lemma 9 follows by iductio o k. Lemma 1. Every atural umber ca be writte i the form (42) = a k l + a k 2 +" + a k. + t with k > k., - ad ^ t < a. Proof. The lemma is trivially true for all ^ a 2. Every atural umber betwee a~ ad a«ca be writte a~ + t with t < a- ; = a«is of the form (42). Suppose (42) holds for all < N, ad let a, deote the largest a i ot exceedig N ad cosider N - a,. We must have N = a, < a, -, sice N < a, ^ a, - implies N ^ a. + 1 which cotradicts the maximal property of a,. It follows that N - a, < N ca be represeted i the form (42) with k + 1 > k-, hece, N = a fe. + a k l + + a kj[ + t is also of the form (42). Theorem 7. Let be a umber represeted as i (42). The ~ {%+! + F k F k. + i} if =st ==. a x (43) N() - {1 + F k l F k F k. + 1 } if a x < t ^ a 2 to obtai Proof. Sice a k a k. + t < a k l - 1 we ca apply Lemma 7 (44) N() = N(a kl ) + N(a k2 + + a k. + t) ; applyig Lemma 7 repeatedly i (44) we get (45) N() = N(a k l ) + N(a kg ) N(a k. + t). Now if a k. = a 2, ^ t < a 1? sice if t ^ ^ we would have a k i = a g ad we ca write (46) N(a k. + t) = N(a k.) + N(t) ;

16 34 REPRESENTATIONS OF N AS A SUM OF [Dec. but if a k - = a-i». we would have a- + t < a 2 ad we res elect t as a 1 + t ; also, we ca coclude from this that a. # - 1 ^ a 3 so (46) still holds i this case. Thus (45) ca be writte i the form (47) N() = N(a k l ) + N(a k ) N(a k.) + N(t) Applyig Lemma 9 to the N(a k.) i the right member of (47) we get (48) N() = a k l - F k l a k 2 - F k a k. - F k N(t) = a k l + a k a k. + N(t) - { F k l F k. + 1 } ; but if t < a^, N(t) = t ad a k _. + + a k. + t = by assumptio so that the first part of (43) is true. If a 1 ^ t < a 2, N(t) = t - 1 ad we see that the secod part of (43) is also true. This completes the proof of Theorem 6. Hoggatt (Problem H-53, Fiboacci Quarterly) has proposed that oe show that M(), the umber of atural umbers less tha which caot be expressed as a sum of distict Lucas umbers L (L 1 = 1, L = 3, L = L +L x dt Tr^di + x has the property (49) M(L ) = F v 7-1 also, he asked for a formula for M(). The Lucas sequece ca be used i place of {a } i all of our lemmas ad theorems. I particular, Lemma 9 tells us M(L ) = N(L ) = L - F +l' it is a trivial matter to show L F - = F - by iductio so (49) is proved. Writig a k. = L^. i (42) ad Theorem 7 gives a formula for atural umbers. Table 1 R(k) for < k < 144 M() for all R() R()

17 1966] DISTINCT ELEMENTS FROM SPECIAL SEQUENCES 35 Table 1 (Cot'd) R() R() R() R() R() R() Table 2 T(k) for < k < 55. T() T() T()

18 36 REPRESENTATIONS OF N AS A SUM O F Dec DISTINCT ELEMENTS FROM SPECIAL SEQUENCES Table 3 S(k) for < k < S() S() S() REFERENCES 1. J. L. Brow, J r., ft Note o Complete Sequeces of Itegers," America Mathematical Mothly, Vol. 67 (196) pp , "Zeckedorf T s T h e o r e m ad S o m e Applicatios," Fiboacci Quarterly, Vol. 3, No. 1, pp , "A New Characterizatio of the Fiboacci N u m b e r s, " Fiboacci Quarterly, Vol. 3, No. 1, pp D. E. Dayki, "Represetatios of Natural Numbers a s Sums of Geeralized Fiboacci N u m b e r s, " J o u r a l of the Lodo Mathematical Society, Vol. 35 (196) pp H. H. F e r s, "O Represetatios of Itegers a s Sums of Distict Fiboacci N u m b e r s, " Fiboacci Quarterly, Vol. 3, No. 1, pp R. L. Graham, "A P r o p e r t y of the Fiboacci N u m b e r s, " Fiboacci Quarterly, Vol. 2, No. 1, pp V. E. Hoggatt, J r., P r o b l e m H-22, Fiboacci Quarterly, Vol. 2, No. 3, p. 47, also Vol. 3, No. 1, p. 45. p. 45., P r o b l e m H - 5 3, Fiboacci Quarterly, Vol. 3, No. 1, 9. V. E. Hoggatt, J r., ad S. L. Basi,,! Represetatios by Complete S e - queces, Fiboacci Quarterly, Vol. 1, No. 3, pp V. E. Hoggatt, J r., a d C. Kig, Problem E 1424, America Mathematical Mothly, Vol. 67 (196), p P. Lafer, "Explorig the Fiboacci Represetatio of Itegers," Fiboacci Quarterly, Vol. 2, No. 2, p (Cot f d o p. 322)