Decomposition of terms in Lucas sequences
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1 Journl of Logic & Anlysis 1: ISSN Decomposition of terms in Lucs sequences ABDELMADJID BOUDAOUD Let P, Q be non-zero integers such tht D = P 4Q is different from zero. The sequences of integers defined by { Un = PU n 1 QU n ; U 0 = 0, U 1 = 1; V n = PV n 1 QV n ; V 0 =, V 1 = P. re clled the Lucs sequences ssocited to the pir P, Q [6,7]. In this pper we prove the following result: If P, Q re such tht D is strictly positive, then, for unlimited n, ech of the integers U n nd V n differs by limited integer from product of two unlimited integers. 000 Mthemtics Subject Clssifiction 11B39, 11A51, 6E35 Keywords: Lucs sequences, Fiboncci sequences, nonstndrd nlysis 1 Introduction Let N be ny lrge integer. Proceeding directly to the fctoriztion of N is not n esy tsk, even unfesible unless N belongs to prticulr fmily of integers. Then to surmount this mjor difficulty we might choose to sk bout the fctoriztion of n integer in smll neighborhood of N insted of N. This is expressed through the following question: Is there smll integer s such tht N = s + µϑ, where µ nd ϑ re two lrge integers? The fct tht the integer N s is product of two lrge integers, gives n ide of its fctoriztion. In the existing literture, the decomposition of integers is n immense problem which hs been posed in severl wys nd treted by different methods for exmple [1, 3, 8]. This ide originted in [], where we chose to work in the frmework of nonstndrd nlysis [4, 5] to be ble to give sense to the words "smll", "lrge",... nd the question hs the formultion: Is every unlimited integer the sum of limited integer nd product of two unlimited integers? To give prtil nswer we provided some exmples [] nd we devote the present work to nother exmple concerning Lucs Published: 16 April 009 doi: /jl
2 Abdelmdjid Boudoud sequences. In the finl section we give the clssicl equivlence of the result obtined nd generl remrk. We strt with brief overview of Lucs sequences ssocited to pir of integers [6]. Let P, Q be non-zero integers. Consider the polynomil p x = x Px + Q; its discriminnt is D = P 4Q nd the roots re 1 1 α = P + D, β = P D. Suppose tht P nd Q re such tht D 0. The sequences of integers U n P, Q = αn β n with U 0 P, Q = 0 nd U 1 P, Q = 1 1 α β V n P, Q = α n + β n with V 0 P, Q = nd V 1 P, Q = P re clled the Lucs sequences ssocited to the pir P, Q. We will note by U n resp. V n the element U n P, Q resp. V n P, Q. It cn be proved tht for n 1 3 U n = PU n 1 QU n ; U 0 = 0, U 1 = 1, V n = PV n 1 QV n ; V 0 =, V 1 = P. In the prticulr cse P, Q = 1, 1, the sequence U n n 0 begins 0, 1, 1,, 3, 5, 8,13... nd ws first considered by Fiboncci. The compnion sequence of the Fiboncci numbers, still with P, Q = 1, 1, is the sequence of Lucs numbers V n n 0 nd it begins, 1, 3, 4, 7, 11, We give here some known results [6, 7] 1 4 V n = V n Q n. Let p be prime integer, then 1 5 D U p = mod p for p 3, p V p = P mod p for p, where D p represents the Legendre symbol which is, ccording to the reltion between p nd D, one of the vlues 1, 0, +1. In ddition, if n, k 1, then 1 6 U n U nk for ll k, V n V nk if k is odd. Moreover 1 7 { Un P, Q = 1 n 1 U n P, Q, V n P, Q = 1 n V n P, Q. Journl of Logic & Anlysis 1:4 009
3 Decomposition of terms in Lucs sequences 3 Fermt s Little Theorem. If p is prime number nd if is n integer, then 1 8 p [p]. In prticulr, if p does not divide then p 1 1 [p]. Externl recurrence principle [4]. For ll internl or externl formul F n, we hve 1 9 [ F 0 nd st n F n = F n + 1 ] = st n F n. Nottions. Let x, y be rel numbers not necessrily integers. i x 0 resp. x + denotes tht x is infinitesiml resp. x is positive unlimited. We hve n nlogous definition for x. ii x nd y re clled infinitely close, denoted by x y, if x y 0. iii iv v vi We sy tht x is pprecible if it is neither unlimited nor infinitesiml. The inequlity x y mens tht x > y nd x y. We hve n nlogous definition for. The Greek letter φ is used for n infinitesiml strictly positive. Two occurrences of φ re not necessrily equl. P represents the set of ll prime integers. Min result Theorem.1 If P, Q re such tht D > 0, then, for unlimited n, ech of the integers U n nd V n differs by limited integer from product of two unlimited integers. Let P nd Q be such tht D > 0 nd let n +. We put λ = prove the min result we need the following lemms. P D. In order to Lemm. α β, mx α, β 1 nd b 1 If P > 0, then β α = λ 1 λ + 1. i ii α > β, 1 β α is positive infinitesiml λ +, Journl of Logic & Anlysis 1:4 009
4 4 Abdelmdjid Boudoud iii 1 + β is positive infinitesiml λ is positive infinitesiml, α β iv ±1 if nd only if λ is pprecible nd positive. α c If P < 0, i ii iii iv α < β, 1 α β is positive infinitesiml λ, 1 + α is positive infinitesiml λ is negtive infinitesiml, β α ±1 if nd only if λ is pprecible nd negtive. β Proof α β becuse α = P + D possible cses. i P > 1. In this cse α 1. nd β = P D. The following re the ii P = 1. In this cse Q must be strictly negtive nd therefore α 1. iii P = 1. In this cse Q must be strictly negtive nd therefore β 1. iv P < 1. In this cse β 1. Hence mx α, β 1. By 1.1, β α = λ 1 λ + 1. b If P > 0, then it is immedite tht α > β. Furthermore, λ > 0 nd the reminder of the proof cn be deduced from the grph of the function β α λ = λ 1 λ + 1 β which is strictly incresing from [0, + [ onto [ 1, 1[, where lim λ = 1 nd λ 0 + α β lim λ = 1. λ + α c This is similr to b. Remrk. By 1 7 we need to prove the following lemms only for P > 0 in which cse α is positive nd ccording to Lemm. α > β. Consequently α 1. Lemm.3 Ech of U n nd V n is of the form ωn, where ω is unlimited ω is not necessrily the sme in ech cse. Journl of Logic & Anlysis 1:4 009
5 Decomposition of terms in Lucs sequences 5 Proof By 1 n 1 β/α U n = α n 1 1 β/α, V n = α n 1 + β/α n. We divide the proof into three cses. A β α = 1 φ. Then 0 < β/α n < β/α < 1. Hence 1 β/α n > 1 β/α > 0. By it follows U n = α n 1 c with c > 1 nd therefore U n = ωn, where ω +. Also V n = α n d with d > 1 nd consequently V n = ωn, where ω +. B β = 1 + φ. According to Lemm., λ is positive infinitesiml. There re two α subcses. 1 n is odd. By U n = α n φ n φ = α n 1, = α n φ n φ where is pprecible nd positive. Therefore U n = ωn, where ω +. Concerning λ 1 n V n by 1 nd, V n = α 1 n +. Hence λ λ n V n = α 1 n > α n 1 1 λ + 1 λ + 1 n nd then V n > α n nλ λ + 1 n which implies V n n > λ αn λ + 1 n. Moreover λ λ + 1 n αn +, indeed replcing α by P + D λ λ + 1 n αn = we get λ D n 1 + λn λ + 1 n 1 D, n = P where P n 1 D + becuse D +. Therefore V n is of the form ωn, where ω +. Journl of Logic & Anlysis 1:4 009
6 6 Abdelmdjid Boudoud n is even. By U n = α n 1 1 β/α n 1 β/α = αn 1 n 1 β/α φ = αn 1 n 1 β/α, where = φ is pprecible. By 1, β α = λ 1 where λ is positive infinitesiml, nd since n is even, λ + 1 U n = αn 1 1 λ n 1. λ λ n From λ + 1 n > 1 1 λ n it follows tht 0 < 1 λ + 1 λ + 1 n < 1, λ + 1 hence U n > αn λ + 1 n Replcing α resp. λ by P + D α n 1 nd then U n > αn 1 U n n > αn 1 λ λ + 1 n. resp. λ λ + 1 n = P D, we get D n P n λ. nλ λ + 1 n. So Finlly, from D +, λ 0 nd is pprecible we obtin αn 1 nd it follows tht U n n +. Therefore U n = ωn, where ω +. Concerning V n, the fct tht β n V n = α 1 n + = α n φ n α λ λ + 1 n + implies V n = ωn with ω + becuse φ n > 1. C β n 1 β/α α ±1: By, U n = α n 1 1 β/α hence U n = α n 1, where β n is pprecible. Therefore U n = ωn, where ω +. V n = α 1 n + = α α n where is pprecible, therefore V n = ωn with ω +, nd the proof is complete. Journl of Logic & Anlysis 1:4 009
7 Decomposition of terms in Lucs sequences 7 Lemm.4 If n is of the form n 1 n with n 1 > 1 nd n > 1, then U n 1 n nd U n1 V n1 n re unlimited. V n1 Proof Since n +, t lest one of n 1 nd n is unlimited. By 1 U n1 n = αn1n 1 β /α n 1 n U 3 n1 α n 1 1 β /α n, 1 V n1 n = αn1n 1 + β /α n 1 n V n1 α n β /α n. 1 We hve three cses. A β β n1 n β n1 α α = 1 φ. By 3 nd the fct tht 1 > 1 > 0 we hve α U n1 n = α n 1n 1 c = α n n 1 c, U n1 where c > 1. Since n n 1 + then U n 1 n V n1 n +. Also by 3, = U n1 V n1 α n 1n 1 c = α n n 1c, where c is positive nd pprecible. From the fct tht n n 1 + we hve V n 1 n V n1 +. B β α = 1+φ. By Lemm., λ is positive infinitesiml. There re three subcses. 1 n 1 is even. Then n 1 n is even nd U n 1 n = α n 1n 1 c with c > 1 becuse U n1 β n1 n β n1 1 > 1 > 0. Hence U n 1 n +. Concerning V n we hve α α U n1 n1 n V n1 n 1 + β/α = αn1n V n1 α n β/α n 1. Since n 1 nd n 1 n re even nd 1 < β/α < 0, then V n1 n = αn1n V n1 α n c 1 = α n 1n 1 c + becuse c is positive nd pprecible nd α n 1n φ n 1 n n 1 nd n re both odd. From U n 1 n U n1 = α n 1n 1 U n1 n U n1 = α n 1n 1 = α n 1n φ n 1 n φ n φ n 1 we hve Journl of Logic & Anlysis 1:4 009
8 8 Abdelmdjid Boudoud becuse is positive nd pprecible. Also V n1 n = α n 1n φ n 1 n V n φ n φ n 1 n = α n 1n φ n 1 Since 1 1 φ n 1n > 1 1 φ n 1 > 0, then V n 1 n V n1 = α n 1n 1 c with c 1 which implies V n 1 n V n1 + becuse α n 1n n 1 is odd, n is even. Then U n 1 n = αn1n 1 1 U n1 β n1 = 1 is pprecible nd strictly positive becuse 1 < α Since λ is positive infinitesiml nd β α = λ 1 λ + 1 then Now α n 1n 1 Replcing λ by which implies 1 U n1 n U n1 P D, we get = αn1n 1 1 λ n1 n 1 + λ α n 1n 1 α n 1n 1 becuse α < D. Since D > 3, Therefore 1 1 λ 1 + λ > αn 1n 1 > αn 1n 1 n1 n. 1 β n1 n α λ n 1n n 1 n λ 1 + λ n 1n. n 1 n λ 1 + λ n 1n = n n1 n 1n P D 1 α n 1. n 1 n n 1 n λ 1 + λ n 1n n n1 n 1n P D n 1 1. n 1n n 1 n P D n 1 n n 1 1 α n 1n 1. n n 1n P. n1n 3 3 1n n 1 n λ 1 + λ n 1n n 1n P. n1n β α, where n1 < 0. Journl of Logic & Anlysis 1:4 009
9 Decomposition of terms in Lucs sequences 9 nd then U n 1 n U n1 +. Concerning V n 1 n V n1, we hve V n 1 n V n1 = α n 1n φ n 1 n 1 1 φ n 1. The fcts tht φ n 1n is pprecible, 1 1 φ n 1 ]0, 1[ nd α n 1n 1 + led to V n 1 n +. C β α ±1. The fct tht U n1 n U n1 = α n 1n 1, V n1 n V n1 α n 1n 1 +, nd b re pprecible nd positive men tht nd V n 1 n V n1 + which finishes the proof. Lemm.5 For every i, U i < U i+1 & V i < V i+1. V n1 = α n 1n 1 b, where U n1 n U n1 + Proof Let i. We hve three cses. A β α = 1 φ. By Lemm., λ + nd then α +. Now U i+1 = U i α i+1 β i+1 1 r i+1 α i β i = α 1 r i, where r = β α = 1 φ. Since 1 ri+1 > 1 r i > 0 nd α +, then U i+1 > 1. Also we hve V i r i r i+1 = α U i V i 1 + r i. Since 1 + r i is pprecible nd α +, then V i+1 +. Tht is, V i+1 > 1. V i V i B β = 1 + φ. By Lemm., λ is positive infinitesiml. Since P is supposed α positive nd λ is positive infinitesiml, we directly get Q. Therefore, from Q < 0, we obtin 0 < U < U 3 < U 4 <.... Similrly, 0 < V < V 3 < V 4 <.... C β α ±1. According to Lemm., λ is pprecible nd strictly positive. U i+1 1 r i+1 = α U i 1 r i with r = β α = λ 1, where λ is different from 1 becuse λ + 1 otherwise Q = 0. We divide the rest of the proof into the following cses. P 1 λ ]0, 1[. Then 0 < < 1 nd consequently Q < 0. Hence, s P 4Q in the cse B of this lemm, we hve 0 < U < U 3 < U 4 <... nd 0 < V < V 3 < V 4 <.... Journl of Logic & Anlysis 1:4 009
10 10 Abdelmdjid Boudoud λ > 1. Then r is pprecible nd positive nd r < 1. Since U i+1 = U i 1 r i+1 α 1 r i nd 1 r i+1 > 1 r i > 0, then U i+1 > α > 1. Concerning V i+1, U i V i P the fct tht λ = > 1 implies Q > 0 nd consequently P 3 becuse P 4Q D = P 4Q > 0. Then α nd β > 0 becuse β = P P 4Q. Therefore, α i α 1 > β i 1 β. Indeed, if 0 < β 1, then β i 1 nd 0 1 β < 1 which implies 0 β i 1 β < 1. Moreover, α i α 1 1. Hence α i α 1 > β i 1 β which is evidently verified when β > 1. V i+1 > V i nd the proof is complete. Lemm.6 If P, Q is not stndrd then V V 1 +. Proof We hve two cses. Therefore α i+1 + β i+1 > α i + β i ; i.e. A P +. We divide the proof of A further into the following cses. 1 Q stndrd. Then V = P Q +. V 1 P Q. Then V = P Q +. V 1 P 3 Q +. Suppose tht V = l with l being limited. Then P Q = l > 0 V 1 P P Q > 0 becuse D > 0. Hence P Pl = Q which implies Q = 1 P Pl. Then D = P 4Q = P P Pl = P + Pl which mens tht D < 0 nd this is contrdiction. Then V V 1 +. B P stndrd. In this cse Q nd we show esily tht V finishes the proof. V 1 +. This Lemm.7 n my be written ccording to one nd only one of the following forms. Journl of Logic & Anlysis 1:4 009
11 Decomposition of terms in Lucs sequences 11 i ii iii iv n = p + is prime. n = s p, where s 1 is limited nd p + is prime. n = n 1 n where one of n 1,n is odd greter thn or equl to 3, the other is unlimited. n = ω+1 with ω +. Proof It is well-known tht n must be written uniquely s t α 1 1 tα...tαr r, where t 1 < t <.. < t r re prime numbers, α i 1 i = 1,,..., r. Two cses rise. 1 r = 1. Either t 1 =, in which cse n is of the fourth form, or t 1 > which implies tht n is of the first form if α 1 = 1 becuse in this cse n = t 1, otherwise α 1 > 1, n = t α 1 1 = t 1 t α which shows tht n is of the third form becuse the quntities t 1 nd t α re both odd nd obviously t lest one of them is unlimited. r > 1. Here two subcses rise. t 1 =. If α 1 is limited, then we divide the proof into two further cses. 1 α = 1. If r =, then n = α 1t which shows tht t is unlimited nd consequently n is of the second form, otherwise i.e. r > the fct tht n = α 1t t α tαr r where t is odd nd α 1t α tαr r + becuse α 1t α tαr r > t nd the product t. α 1t α tαr r = n +, shows tht n is of the third form. α > 1. In this cse n is of the third form becuse n = α 1t α tα tαr r = t. α 1t α 1 t α tαr r where t is odd nd by the sme resoning s 1 bove, r > α 1t α 1 t α tαr r +. If α 1 is unlimited, then the fct tht n = α 1t α tα tαr r t is odd nd α 1t α 1 t α tαr r form. = t. α 1t α 1 t α tαr r, where + permits us to conclude tht n is of the third b t 1 >. Here lso, using the sme resoning s bove nd the fct tht n = t α 1 1 tα...tαr r = t 1 t α t α...tαr r +, we conclude tht n is of the third form. We now prove tht n cnnot be written simultneously ccording to two of the bove indicted forms. Indeed, we prove this for the second nd the third form the other cses re trivil. Suppose tht n = s p where s 1 is limited nd p + is prime nd lso n = n 1 n, where for exmple n 1 is odd greter thn or equl to 3 nd n is unlimited. Since the decomposition of n in prime fctors is unique, then n 1 = p, n = s which is contrdictory becuse n is unlimited nd cnnot be equl to s. Journl of Logic & Anlysis 1:4 009
12 1 Abdelmdjid Boudoud 3 Proof of Theorem Proof for U n. We consider the following two subcses. D I n is prime. By 1 5, U n = mod n. Then U n = u + kn, where n u { 1, 0, +1}. Since U n is, ccording to Lemm.3, of the form ωn with ω is n unlimited rel number, the integer k must be unlimited. Consequently the proof is finished for this cse. II n is composite i.e. n = n 1 n, where n 1 n > 1. By 1 6, U n = CU n1, where C is n integer which is, ccording to Lemm.4, unlimited. On the other hnd since n 1 +, then by Lemm.5, U n1 is unlimited. Thus the proof is finished for U n. 3. Proof forv n By Lemm.7, we need to consider the following four cses. I n = p + is prime. We hve two subcses to consider. P limited. By 1 5, V p = P modp; i.e.v p = P + kp. Since P is limited, k must be, ccording to Lemm.3, unlimited. b P unlimited. According to 1 6, V 1 V p ; i.e V p = V 1 N. By Lemm.5, we hve V < V 3 <... < V n <.... By Lemm.6, V V 1 + which implies from V V 1 < V p V 1 tht V p + nd V 1 then N is unlimited. Thus the proof is finished for this cse becuse V 1 = P nd P +. II n = s p, where s 1 is limited nd p + is prime. P nd Q re both limited. For s 1 define the formul A s "For n of the form s p, V n my be written s g 1 + g p where g 1 resp. g is limited resp. is n unlimited integer". We hve A 1. Indeed, let n = p; by 1 4 V n = V p = V p Q p. Journl of Logic & Anlysis 1:4 009
13 Decomposition of terms in Lucs sequences 13 Applying 1 5 nd 1 8 yields V p = P + kp Q + lp. Hence V n = V p = P + Pkp + k p Q lp = P Q + Pk + k p l p. If g 1 = P Q nd g = Pk + k p l, then g 1 is limited nd, ccording to Lemm.3, g is unlimited. Consequently, A 1. Suppose A s for s 1 limited integer nd prove A s + 1. Indeed, by 1 4 V s+1 p = V s p = V s p Q sp. From As resp. 1 8 we hve V s p = g 1 + g p, where g 1 is limited nd g is unlimited resp. Q sp = p Q s = Q s + fp with f n integer. Replcing by these vlues, we get V s+1 p = V s p Q s p = g 1 + g p = g 1 Q s + f p, Q s + fp where f = g 1 g + g p f. Since g 1, Q nd s re limited, then g 1 Qs is limited; the integer f, ccording to Lemm.3, must be unlimited. Consequently, As + 1. Then by 1 9, st s 1 A s. b P or Q is unlimited. By 1 6, V s V s p, i.e. V s p = V sc with c being n integer. By Lemm.4, c is unlimited. By Lemm.6, V + nd by Lemm.5, V < V 3 < V 4 <.... Hence V s is unlimited. This completes the proof for this cse. III n = n 1 n, where one of n 1, n is odd greter thn or equl to 3, the other is unlimited. Suppose n 1 3 is odd nd n +. Then V n1 n = V n C, where by 1 6 C is n integer which is, ccording to Lemm.4., unlimited. Since n +, then by Lemm.5 V n is unlimited. This finishes the proof for this cse. IV n = ω+1 with ω +. Q is even Q = t, t Z. By 1 4, we hve V n = V ω+1 = V. ω = V ω Qω. Journl of Logic & Anlysis 1:4 009
14 14 Abdelmdjid Boudoud Applying the fct tht ω =. ω 1 nd 1 4 yield V ω = V. ω 1 = V ω 1 Q ω 1 which, when substituted in V n = V ω Qω, gives V n = V ω+1 = V Q ω 1 Q 3 1 ω ω 1 = 4 V ω 1 mod Q ω 1. Similrly, the fct tht V ω 1 = V. ω nd 1 4 give 3 V n = V ω+1 = 8 V ω mod Q ω nd so on. Let f + be n integer such tht ω f +. The previous process permits to write 3 3 V n = V ω+1 = f +1 V ω f mod Q ω f, where V ω f is unlimited. Now, if V ω f is even then V ω+1 = γ t, where γ = min f +1, ω f + nd t is n integer. This shows tht V n = V ω+1 = γ 1 γ t, where γ 1 nd γ re two unlimited integers stisfying γ 1 + γ = γ. Otherwise i.e. V ω f is odd, we hve [ V ] f +1 V n 1 = ω f 1 + kq ω f. Since f +1 V ω f 1 is difference between squres, then [ V [ f V ] f V n 1 = ω f 1] ω f kq ω f. By the sme resoning bout the difference f V ω f 1 [ V ] [ f 1 V ] [ f 1 V ] f V n 1 = ω f 1 ω f + 1 ω f + 1 nd so on. Thus we cn write V n 1 s [ V [ f t V V n 1 = ω f 1] ω f f t [ V ] [ f 1 V f + ω f + 1 ω f + 1 [ V ] f t 1 + 1] ω f + 1 ] + kq ω f, + kq ω f +... Journl of Logic & Anlysis 1:4 009
15 Decomposition of terms in Lucs sequences 15 where t is n integer stisfying 1 t < f. Now choose t 0 + such tht t 0 < f nd t 0 + < ω f. This is possible becuse, from the fct tht min f, ω f +, we cn choose n integer s + such tht s min f, ω f nd tke therefter t 0 = s 3. Since Q ω f contins the fctor ω f nd the product [ V f t 0 t0 [ V ] f i ω f 1] ω f + 1 contins k with k t 0 +, then i=0 where N is n integer. Therefore V n 1 = t 0+ N, V n 1 = V ω+1 1 = t 1 t N, where t 1 nd t re two unlimited integers stisfying t 1 + t = t 0 +. b Q is odd Q = t +1, t Z. We put n o = ω. If Q = ±1, then by 1 4 V n = V n0 = V n0 Q n 0 = V n0 becuse n 0 is even. V n0 is, by Lemm.3, unlimited. Otherwise i.e. Q ±1 we divide the proof into the following cses. 1 P is even. By 1 3 nd the induction, we show esily tht ech V l l 0 is even. Moreover V, becuse otherwise P Q = which implies D = P 4Q = Q. The fct tht D > 0 implies tht Q > 0 i.e. Q < 0 becuse Q Z nd this contrdicts P Q =. In the sme wy V. Now, we prove tht V n is the product of two unlimited integers. Indeed, by 1 4 Then which implies V n = V ω+1 = V n0 = V n 0 Q n 0. V n = V n 0 4 Q n 0 + = V n0 V n0 + Q n V n = V n0 V n0 + Q n0/ 1 Q n0/ + 1. Becuse n 0 is divisible by, then pplying 1 4 shows tht V n0 cn be written s V n0 = V n0 / = V n 0 / Q 4 n0/ 1 which, when substituted in 3 4, gives V n = [V n0/ 4 Q n0/ Vn0 1] + Q n0/ 1 Q n0/ + 1. Journl of Logic & Anlysis 1:4 009
16 16 Abdelmdjid Boudoud Hence, 3 5 V n = V n0 / V n0 / + Vn0 + Q n 0/4 1 Q n 0/4 + 1 V n0 + Q n 0/4 1 Q n 0/4 + 1 Q n 0/ + 1. Becuse n 0 / is divisible by, then pplying 1 4 shows tht V n0 / cn be written s V n0 / = V n 0 /4 Qn 0/4 4 + = V n0/4 4 Q n0/4 1 which, when substituted in 3 5, gives V n0 = V n0 /4 V n0 /4 V + n0 / + Vn Q n0/8 1 Q n0/8 + 1 V n0 / + Vn0 + Q n0/8 1 Q n0/8 + 1 Q n0/4 + 1 V n0 + Q n0/8 1 Q n0/8 + 1 Q n0/4 + 1 Q n0/ + 1 nd so on. So the process of pplying 1 4 nd putting the difference between squres s product of two fctors, yields by induction V n = V n0 = V n0 / i 1 V n0 / i V n0 / + V n Q n 0/ i 1 Q n 0/ i + 1 V n0 / i +... V n0 / + V n0 + Q n 0/ i 1 Q n 0/ i + 1 Q n 0/ i 1 Vn0 + 1 / i V n0 + Q n 0/ i 1 Q n 0/ i + 1 Q n 0/ i Q n 0/ i Vn0 + 1 / i V n0 + Q n 0/ i 1 Q n 0/ i + 1 Q n 0/ i Q n 0/ Vn Q n 0/ i 1 Q n 0/ i + 1 Q Q n 0/ i Q n 0/ + 1 n 0 / + 1, where 1 i ω. In this formul, if we replce i by 1 we recover 3 4, by we recover 3 5, etc. Journl of Logic & Anlysis 1:4 009
17 Decomposition of terms in Lucs sequences 17 We tke i 0 + such tht n 0 i 0 1.We show tht V n is of the form i 0+1 t, where t is n integer. Indeed, ech element V n0 / j 0 j i 0 1 is even nd, ccording to Lemm.5, different from ± giving the fct tht V is different from these vlues. On the other hnd Q is odd nd different from ±1. Hence the formul 3 7 is the sum of i terms, where ech term is the product of i non-zero even integers. Therefore V n = i0+1 t = t 1 t t, where t 1 nd t re two unlimited integers stisfying t 1 +t = i 0 +1 nd t is n integer. P is odd. We prove by induction tht V l l 1 is odd. Indeed, V 1 = P Q this shows tht V is odd. Now suppose thtv l is odd with l 1. Then V l+1 = V l Q l so V l+1 is lso odd. On the other hnd V 1, otherwise P Q = 1 then the fct tht D = P 4Q = 1 Q > 0 implies tht Q < 0 nd this contrdicts P Q = 1. In the sme wy V 1. By 1 4 V n = V ω+1 = V n0 = V n 0 Q n 0 Then V n + 1 = V n Q n 0 = V n0 1 V n Q n 0 So 3 8 V n + 1 = V n0 + 1 V n0 1 + By Q n 0/ 1 + Q n 0/. V n0 + 1 = V n0 / + 1 = [V n0/ 1 + Qn 0/ ] = [V n0 / V 1 n0 / Q n 0/ ] which, when substituted in 3 8, gives V n + 1 = [V n0 / V 1 n0 / Q n 0/ ] Vn Q n 0/ 1 + Q n 0/ = V n0 / 1 V n0 / + 1 Vn Q n 0/ Vn Q n 0/ 1 + Q n 0/ Journl of Logic & Anlysis 1:4 009
18 18 Abdelmdjid Boudoud Then 3 9 V n + 1 = V n0 / 1 V n0 / + 1 Vn Q n 0/4 1 + Q n 0/4 Vn Q n 0/4 1 + Q n 0/4 1 + Q n 0/ Agin, setting n 0 =.n 0 4 nd clculting by 1 4 n expression for V n 0 / + 1 nd substituting in 3 9, we get nother formul for V n + 1, nd so on. So this process yields by induction V n + 1 = V ω = V n0 / i + 1 V n0 / i 1 V n0 / i V n0 / 1 V n Q n 0/ i Q n 0/ i+1 Vn0 / i 1 1 V n0 / i 1... V n Q n 0/ i Q n 0/ i Q n 0 / i Vn0 / i 1... V n Q n 0/ i Q n 0/ i Q n 0 / i 1 + Q n 0/ i 1 Vn0 / i V n Q n 0/ i Q n 0/ i Q n 0 / i 1 + Q n 0/ i Q n 0/ Vn Q n 0/ i Q n 0/ i Q n 0 / i 1 + Q n 0/ i Q n 0/, 3 10 where 0 i ω 1. In this formul, if we replce i by 0 we recover 3 8, by 1 we recover 3 9 etc.... We tke i 0 + such tht n 0 i 0. We show tht V n + 1 is of the form i 0+ t, where t is n integer. Indeed, ech element V n0 / j 0 j i 0 is odd nd, ccording to Lemm.5, different from ±1 giving the fct tht V is different from these vlues. On the other hnd Q is odd nd different from ±1. Hence the formul 3 10 is the sum of i 0 + terms, where ech term is the product of i 0 + non-zero even integers. From this V n + 1 = V n0 + 1 = t 1 t t, where t 1 nd t re two unlimited integers stisfying t 1 + t = i 0 + nd t is n integer. This finishes the proof of this cse nd therefore the theorem. Journl of Logic & Anlysis 1:4 009
19 Decomposition of terms in Lucs sequences 19 4 In clssicl terms In order to find the clssicl equivlence of our result we use the reduction lgorithm of externl formuls of Nelson [5]. We denote the stndrd set { P, Q Z } : P 4Q > 0 by H. Theorem.1 my be written s n P, Q st t : n t st U, V l 1, l, l 1, l st i N U n = U + l 1 l, V n = V + l 1 l & min l 1, l, l 1, l > i, where n, t, i rnge over N, P, Q rnge over H nd U, V, l 1, l, l 1, l Using ideliztion, this formul is equivlent to rnge over Z. n P, Q st t : n t st U, V stfini I l 1, l, l 1, l i I U n = U + l 1 l, V n = V + l 1 l & min l 1, l, l 1, l > i where I belongs to P f N the set of finite subsets of N. This formul is equivlent to n, P, Q st t, U, V stfini I n t l 1, l, l 1, l i I U n = U + l 1 l, V n = V + l 1 l & min l 1, l, l 1, l > i If we refer to st st in the lexicon cited in [5], the lst formul is equivlent to Ĩ fini R n, P, Q t, U, V R n t l 1, l, l 1, l i Ĩ t, U, V U n = U + l 1 l, V n = V + l 1 l & min l 1, l, l 1, l > i,,. 4 1 where Ĩ is mpping tht ssocites with ech t, U, V N Z finite subset Ĩ t, U, V N. Now we prove tht 4 1 is equivlent to Journl of Logic & Anlysis 1:4 009
20 0 Abdelmdjid Boudoud 4 Ĩ T, W N n, P, Q t, U, V [0, T] [ W, W] n t l 1, l, l 1, l i Ĩ t, U, V U n = U + l 1 l, V n = V + l 1 l & min l 1, l, l 1, l. > i Indeed, let Ĩ : N Z P f N be set-vlued mpping. Then ccording to 4 1 fini R N Z nd therefore there exist smller integers T nd W such tht R [0, T] [ W, W]. Let n, P, Q N H, by 4 1 t, U, V R [0, T] [ W, W] ; tht is, t, U, V [0, T] [ W, W]. Now if n t, then l 1, l, l 1, l i Ĩ t, U, V U n = U + l 1 l, V n = V + l 1 l l & min 1, l, l 1, l > i which shows tht For the converse one tkes, R = [0, T] [ W, W]. Hence we hve the following internl formultion of Theorem.1. Proposition 4.1 For ny set-vlued mpping Ĩ : N Z P f N there exists T, W N such tht for ll n, P, Q N H there exists t, U, V [0, T] [ W, W] such tht if n t, then U n resp. V n differs by U resp. V from product of two integers whose bsolute vlue is greter thn or equl to i for ll i Ĩ t, U, V. Below we formulte, from the proof of Theorem.1, two prticulr cses of tht result, nd give their reductions seprtely. 1 U n differs from product of two unlimited integers by n integer U with 1 U 1. If n is not of the form s p with s 0 being limited integer nd p + being prime, then V n differs from product of two unlimited integers by n integer V with V. The reductions of these prticulr cses re s follows. 1 The first is equivlent to [ st t : n t = U, l n P, Q 1, l st ] i Un P, Q = U + l 1 l : U 1 & min l 1, l i, where n, t, i N, U, l 1, l Z 3. By ideliztion nd trnsfer the previous formul trnsforms, while remining equivlent, to Journl of Logic & Anlysis 1:4 009
21 Decomposition of terms in Lucs sequences 1 fini I fini T n, P, Q t T [ ] n t = U, l 4 3 1, l i I Un P, Q = U + l 1 l : U 1 & min l 1, l i, where I nd T belong to the power set of N. 4 3 is equivlent to i N T N n, P, Q N H 4 4 [ n T = U, l 1, l U n P, Q = U + l 1 l with U 1 & min l 1, l i Indeed, let i N. Then, ccording to 4 3, fini T N. Now put T = mx t T t. Now if n, P, Q N H such tht n T, then n t, t T. Hence from 4 3 U, l 1, l with U n P, Q = U+l 1 l, U 1 nd min l 1, l i. Consequently 4 3 = 4 4. Conversely, let I be finite subset of N nd put i = mx i I i. For i there is, { T} ccording to 4 4, T N. Consider T = s finite subset of N. Now if n, P, Q N H such tht n T, then U, l 1, l with U n P, Q = U + l 1 l with U 1 nd min l 1, l i ; tht is, min l 1, l i i I. Hence 4 4 = 4 3 nd consequently we hve the following proposition. Proposition 4. For ny integer i N there exists n integer T N such tht for ll n, P, Q N H stisfying n T, the term U n P, Q differs from product of two integers whose bsolute vlue is greter thn or equl to i by n integer U with 1 U 1. The second prticulr cse is equivlent to n P, Q st t 1, t n t 1 & n / P = t V, l 1, l st i V n P, Q = V + l l 1 l with V {0, ±1, ±} nd min 1, l where n, i N, t 1, t N, V, l 1, l {0, ±1, ±} Z Z. Using ideliztion nd trnsfer the previous formul trnsforms, while remining equivlent, to fini I fini T n, P, Q t 1, t T n t 1 & n 4 5 t / P = V, l 1, l i I V n P, Q = V + l l 1 l : V {0, ±1, ±} nd min 1, l i ]., i, Journl of Logic & Anlysis 1:4 009
22 Abdelmdjid Boudoud where I resp. T belongs to the power set of N resp. N. Then 4 5 is equivlent to i N T 1, T N n, P, Q n T 1, n 4 6 i / P i = 0, 1,.., T = V, l 1, l V n P, Q = V + l l 1 l : V {0, ±1, ±} nd min 1, l Indeed, let i N. According to 4 5, fini T N. Suppose i T 1 = mx {t 1 : t 1, t T} nd T = mx {t : t 1, t T}. Let n, P, Q N H. Suppose tht n T 1 nd n i / P for i = 0, 1,.., T. Then for ll t 1, t T, n t 1 nd n t / P. Hence, ccording to 4 5, V, l 1, l V n P, Q = V + l 1 l with l V {0, ±1, ±} nd min 1, l i. Consequently Conversely, let I be finite subset of N. Put i = mx i I. i. Then for i there is, ccording to 4 6, T 1, T N. Set T = {0, 1,..., T 1 } {0, 1,..., T }, then T is finite. Let n, P, Q N H nd suppose tht for ll t 1, t T, n t 1 & n t / P. Then n T 1 nd n i / P i = 0, 1,.., T. Hence, ccording to 4 6, V, l 1, l such tht V n P, Q = V + l l 1 l with V {0, ±1, ±} nd min 1, l i ; tht l is i I min 1, l i. Hence 4 6= 4 5 nd consequently we hve the following proposition. Proposition 4.3 For ny integer i N there exists two integers T 1, T N such tht for ll n, P, Q N H stisfying n T 1 nd n i / P for i = 0, 1,.., T the term V n P, Q differs from product of two integers whose bsolute vlue is greter thn or equl to i, by n integer V with V. Generl remrk. In the clssicl literture concerning Lucs sequences, generlly the studies re concerned with the terms U k nd V k for k belonging to prticulr fmily of integers see for exmple [6, 7]. In this work we note from the previous propositions tht the min result expresses property of uniformity becuse the conclusion is vlid for ll k beyond certin rnk. Moreover, the ides used in proofs here cn lso be used to deduce stndrd results. For exmple one sees without pin tht Lemm.3 gives the size of U k nd V k wheres Lemm.5 gives the growth of U k nd V k. In ddition, the trnsltion Journl of Logic & Anlysis 1:4 009
23 Decomposition of terms in Lucs sequences 3 by the reduction lgorithm of lemms used previously, which is n opertion tht is not difficult, gives more clssicl results. But giving further detils would increse the length of this pper. Acknowledgment The uthor wishes to thnk the referees nd editor for their comments nd suggestions for improving n erlier version of this pper. References [1] A. Blog, p + without lrge prime fctors, Séminire de théorie des nombres de Bordeux, Exposé [] A. Boudoud, L conjecture de Dickson et clsses prticulières d entiers, Annles Mthémtiques Blise Pscl 13006, [3] Chen, M.J.R, On the representtion of lrge even integer s the sum of prime nd the product of t most two primes, Scienti Sinic , [4] F. Diener et G. Reeb, Anlyse Non-Stndrd, Hermnn, Éditeurs des Sciences et des Arts, [5] E. Nelson, Internl set theory: A new pproch to nonstndrd nlysis, Bull. Amer. Mth. Soc , ; doi: /s x. [6] Pulo Ribenboim, The Little Book of Big Primes, Springer-Verlg, [7] Pulo Ribenboim, My numbers my friends, Springer-Verlg 000. [8] S. Slerno - A. Vitolo, p + with few nd bounded prime fctors, Anlysis , Deprtment of Mthemtics, University of Msil, Ichbili BP Msil, Algeri boudoudb@yhoo.fr Received: 5 Mrch 008 Revised: 5 November 008 Journl of Logic & Anlysis 1:4 009
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