A linear recurrence sequence of composite numbers


 Solomon McCormick
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1 LMS J Comput Math 15 (2012) C 2012 Author do:101112/s A lnear recurrence sequence of composte numbers Jonas Šurys Abstract We prove that for each postve nteger k n the range 2 k 10 and for each postve nteger k 79 (mod 120) there s a kstep Fbonacclke sequence of composte numbers and gve some examples of such sequences Ths s a natural extenson of a result of Graham for the Fbonacclke sequence 1 Introducton For each nteger k 2 one can defne a kstep Fbonacclke sequence, that s, the sequence of ntegers (x n ) n=0 satsfyng the followng relaton x n = k =1 for n = k, k + 1, k + 2, Snce the values of x 0, x 1,, x determne the kstep Fbonacclke sequence we denote t by S k (x 0, x 1,, x ) The terms of the sequence S k (0, 0,, 0, 1) are wellknown Fbonacc kstep numbers Flores [4] developed the calculaton of Fbonacc kstep numbers wthout recurson Noe and Post [9] showed that Fbonacc kstep numbers are nearly devod of prmes n the frst terms for k 100 The am of ths paper s to prove the followng theorem Theorem 11 For each postve nteger k n the range 2 k 10 and for each postve nteger k 79 (mod 120) there are postve ntegers a 0, a 1,, a such that gcd(a 0, a 1,, a ) = 1 and the sequence S k (a 0, a 1,, a ) conssts of composte numbers only x n Graham [5] proved Theorem 11 for k = 2 n 1964 He showed that the sequence S 2 ( , ) contans no prme numbers Several authors (see [7, 8, 15]) made progress n fndng smaller ntal values Currently, the smallest known sequence (n the sense that the maxmum of the frst two elements s the smallest postve nteger) s due to Vsemrnov [14]: S 2 ( , ) The complete analyss of a bnary lnear recurrence sequence of composte numbers s gven n [12] and ndependently n [3] If (c 1, c 2 ) Z 2, where c 2 0 and (c 1, c 2 ) (±2, 1), then there exst two postve relatvely prme composte ntegers a 0, a 1 such that the sequence gven by a n+1 = c 1 a n + c 2 a n 1, n = 1, 2,, conssts of composte numbers only Alternatvely, t s easly seen that for (c 1, c 2 ) = (±2, 1) every nonperodc sequence a n+1 = c 1 a n + c 2 a n 1, n = 1, 2,, wth gcd(a 0, a 1 ) = 1 contans nfntely many prme numbers Somer [12] was usng deep results of Blu et al [1], Cho [2], and also the theorem of Parnam and Shorey [11] Receved 24 February 2012; revsed 27 July Mathematcs Subject Classfcaton 11B37 (prmary), 11B39, 11B50, 11Y55 (secondary)
2 A LINEAR RECURRENCE SEQUENCE OF COMPOSITE NUMBERS 361 n hs proof, whle [3] contans some explct calculatons, and examples of sequences for fxed c 1, c 2 and the proof do not use external results All proofs of Theorem 11 for k = 2 are based on the fact that the Fbonacc sequence s a regular dvsblty sequence, that s, F 0 = 0 and F n F m f n m However, by a result of Hall [6], there are no regular dvsblty sequences n the case S(0, a 1, a 2 ) for any a 1, a 2 Z These dffcultes have recently been overcome, and Theorem 11 was proved for k = 3 n [13], where we constructed the sequence S 3 ( , , ) of composte numbers Secton 3 of ths paper s devoted to the generalsaton of the proof developed n [13] We wll descrbe the set of postve nteger trples and show how to prove Theorem 11 f ths set s gven In 4 we wll prove Theorem 11 for all k 79 (mod 120) and construct correspondng sequences for these cases Fnally, we wll gve an algorthm for the constructon of the set of postve nteger trples, and lst examples of kstep Fbonacclke sequences for k n the range 4 k 10 2 Auxlary lemmas We start wth the followng elementary property of the kstep Fbonacclke sequence Let a = (a 0, a 1,, a ) Z k Defne S k (a) = S k (a 0, a 1,, a ) We wll denote by F k the set of all kstep Fbonacclke sequences Lemma 21 We have that F k s a free abelan group of rank k, and the map s an somorphsm of abelan groups Z k F k, The proof of ths fact s straghtforward Defne a S k (a) (s () n ) n=0 = S k (δ 0, δ 1,, δ ) for = 1, 2,, k 1, where δj s Kronecker s delta symbol Let p be a prme number and let (y n ) n=0 S k (0, a 1, a 2,, a ) (mod p) for a 1, a 2,, a Z Lemma 21 mples y n a s n () (mod p) (21) =1 Lemma 22 Fx k 3 Let p be a prme number and let (y n ) n=0 S k (0, a 1, a 2,, a ) (mod p) wth some a Z for n the range 1 k 1 Suppose that m 2 s an nteger If y m 0 (mod p) for satsfyng 1 k 1, then y lm 0 (mod p) for l = 0, 1, 2, Proof Let A =
3 362 J ŠIURYS be a k k matrx and Y n = (y n+, y n+k 2,, y n+1, y n ) Then the recurrence relaton y n+k = y n+ + y n+k y n+1 + y n can be rewrtten n the matrx form Y n+1 = Y n A, for n = 0, 1, 2 In partcular, Y n = Y 0 A n and Y lm = (y lm+, y lm+k 2,, y lm+1, y lm ) = (y, y k 2,, y 1, y 0 )(A m ) l (22) Let B = A m Ths s a k k matrx wth nteger coeffcents By the Cayley Hamlton theorem, B k = b 0 I + b 1 B + b 2 B b B, for some ntegers b 0, b 1,, b Snce Y lm = Y 0 B l we fnd that Y lm = b 0 Y (l k)m + b 1 Y (l k+1)m + + b Y (l 1)m for l k Consderng the last entres for these vectors, we have that The lemma follows by nducton Defne the matrx y lm = b 0 y (l k)m + b 1 y (l k+1)m + + b y (l 1)m B k,m = s (1) m s (1) 2m s (1) ()m s (2) m s (2) 2m s (2) ()m s () m s () 2m s () ()m for each postve nteger m Let B k,m be the determnant of the matrx (23) (23) Lemma 23 Let m 2 be an nteger If p s prme number and B k,m 0 (mod p) then there exst a 1, a 2,, a Z such that a s not dvsble by p for at least one = 1, 2, k 1, and a s () lm 0 (mod p) for l = 0, 1, 2, =1 Proof Set y n = =1 a s () n Snce y 0 = =1 a s () 0 = 0, by Lemma 22, t suffces to show that there exst sutable a 1, a 2,, a Z such that y lm 0 (mod p) for l = 1, 2,, k 1 Our am s to solve the followng system of lnear equatons: (a 1, a 2,, a )B k,m (0, 0,, 0) (mod p) (24) Let us consder system (24) as a homogeneous lnear system over the fnte feld Z/pZ The assumpton B k,m 0 (mod p) mples that the rank of the system (24) s at most k 2 Therefore, the system has a nontrval soluton n Z/pZ In other words, there exst a 1, a 2,, a Z such that a s not dvsble by p for at least one = 1, 2, k 1 3 General case Let I be a postve nteger (to be defned later) Our goal s to fnd a fnte set S k (N) of postve nteger trples (p, m, r ) ( = 1, 2,, I) wth the followng propertes
4 A LINEAR RECURRENCE SEQUENCE OF COMPOSITE NUMBERS 363 (1) Each p s a prme number and p p j f j (2) The postve nteger p dvdes the determnant B k,m, where B k,m s the matrx (23) (3) The congruences x r (mod m ) (31) cover the ntegers; that s, for any nteger x there s some ndex, 1 I, such that x r (mod m ) Now, suppose that we already found the set S k (N) and that I s a fxed postve nteger Choose, where 1 I Snce B k,m 0 (mod p ), by Lemma 23, there exst a,1, a,2,, a, Z such that a,j s not dvsble by p for at least one j = 1, 2, k 1, and j=1 a,j s (j) lm 0 (mod p ) (32) for l = 0, 1, 2, We shall construct the sequence (x n ) n=0 = S k (x 0, x 1,, x ) satsfyng for n = 0, 1, 2, Set x n s (j) m r +n a,j (mod p ) = 1, 2, I (33) j=1 A,0 = s (j) m r a,j, j=1 A,1 = s (j) m a r +1,j, j=1 (34) A, = s (j) m a r +,j j=1 for = 1, 2,, I Snce the sequence (x n ) n=0 s defned by ts frst k terms, t suffces to solve the followng equatons: x 0 A,0 (mod p ), x 1 A,1 (mod p ), x A, (mod p ) for = 1, 2,, I By the Chnese remander theorem, the system of congruences (35) has the postve nteger soluton x 0 = X 0, x 1 = X 1,, x = X It s assumed that gcd(x 0, X 1,, X ) = 1 By (32) and (33), p dvdes x n f n r (mod m ), where {1, 2,, I} Snce congruences (31) cover the ntegers, we see that for every nonnegatve nteger n there s some, 1 I, such that p dvdes x n The sequence (x n ) n=i s strctly ncreasng, so x n must be composte for n I In ths way, we can construct the kstep Fbonacclke sequence of composte numbers (x n ) n=i f the set S k(n) s gven Note that the assumpton gcd(x 0, X 1,, X ) = 1 s unnecessarly restrctve We can always construct the soluton of (35) wth ths property Indeed, let gcd(x 1,, X ) = d 1, gcd(x 0, d 1 ) = d 0 > 1, and P = I =1 p Suppose that p s a prme number and p d 0 If p P, (35)
5 364 J ŠIURYS then, by (35), Let C = A,0 0 A,1 0 A, 0 (mod p), (mod p), (mod p) s (1) m r s (1) m r +1 s (1) m r + s (2) m r s (2) m r +1 s (2) m r + s () m r s () m r +1 s () m r + be a (k 1) k matrx over the fnte feld Z/pZ By (34) and (36), we get (36) (a,1, a,2,, a, )C (0, 0,, 0) (mod p) (37) The system of equatons (37) has a nontrval soluton f rank(c) k 2 But s (1) m r 1 s (1) m r s (1) m r +k 2 s (2) m rank(c) = rank r 1 s (2) m r s (2) m r +k 2 s () m r 1 s () m r s () m r +k 2 s (1) 0 s (1) 1 s (1) s (2) = rank 0 s (2) 1 s (2) s () 0 s () 1 s () = rank whch s a contradcton From ths f follows that = k 1, gcd(x 0, d 1, P ) = 1 (38) It s easy to check that f (X 0, X 1,, X ) s a soluton of (35), then (X 0 + lp, X 1,, X ) s also a soluton for all ntegers l Let gcd(x 0, P ) = d; then, by Drchlet s theorem on prme numbers n arthmetc progresson, we conclude that X 0 /d + lp/d s a prme number for nfntely many ntegers l So, gcd(x 0 /d + lp/d, d 1 ) = 1 for some l It follows from (38) that gcd(x 0 + lp, d 1 ) = 1 for some l, whch s the desred concluson 4 Proof of Theorem 11 for k 79 (mod 120) In ths secton we wll show that f k 79 (mod 120), then there exst a kstep Fbonacclke sequence of composte numbers We wll need the followng lemma
6 A LINEAR RECURRENCE SEQUENCE OF COMPOSITE NUMBERS 365 Lemma 41 Suppose that the numbers k, p and the sequence (y n ) n=0 are defned as n Lemma 22 If there s a postve nteger l such that and l 1 n=0 y n 0 (mod p) (41) y n y n l (mod p) for n = l, l + 1, l + 2,, (42) then for every nonnegatve nteger t the sequence (y (t) n ) n=0 S tl+k (y 0, y 1,, y tl+ ) (mod p) has the followng property: for n = 0, 1, 2, y (t) n y n (mod p) (43) Proof If t = 0, then the statement of the lemma s trval Let t 1 By the defnton of the sequence y n (t), n = 0, 1, 2,, (43) s true for n = 0, 1,, tl + k 1 Let r k be an nteger and suppose (43) s true for n = 0, 1, tl + r 1 By (41) and (42), r+l 1 n=r y n 0 (mod p) for any postve nteger r Thus we have tl+r 1 y (t) tl+r =r k y r 1 =r k By nducton, (43) s true for n = 0, 1, 2, y y r y tl+r (mod p) Assume that k = 4 and B 4,3 s the matrx defned n (23) It s easy to check that B 4,3 = = 11, and (1, 2, 0)B 4,3 = (0, 0, 0) (mod 11) By Lemma 23, the sequence (y n ) n=0 S 4 (0, 1, 2, 0) (mod 11) has the followng property: 11 y 3n (44) for n = 0, 1, 2, We calculate the frst elements of sequence (y n ) n=0 (mod 11): 0, 1, 2, 0, 3, 6, 0, 9, 7, 0, 5, 10, 0, 4, 8, 0, 1, 2, 0, By a smple nducton, one can prove that the sequence (y n ) n=0 (mod 11) s perodc The length of the perod s 15 and 14 =0 y 0 (mod 11) By Lemma 41 appled to k = 4, l = 15, p = 11 and to the sequence (y n ) n=0, we conclude that the sequence (y (t) n ) n=0 S 15t+4 (y 0, y 1,, y 15t+3 ) (mod 11) satsfes the property (43) for t = 0, 1, 2, It follows that (y (t) n ) n=0 satsfes the property (44) for t = 0, 1, 2,
7 366 J ŠIURYS Now, let k = 7 It s easy to check that B 7,3 = = and (1, 2, 0, 2, 4, 0)B 7,3 = (0, 0, 0, 0, 0, 0) (mod 5), (1, 2, 0, 9, 1, 0)B 7,3 = (0, 0, 0, 0, 0, 0) (mod 17) Lemma 23 mples that the sequence (u n ) n=0 S 7 (0, 1, 2, 0, 2, 4, 0) (mod 5) has the property 5 u 3n (45) for n = 0, 1, 2, and the sequence (v n ) n=0 S 7 (0, 1, 2, 0, 9, 1, 0) (mod 17) has the property 17 v 3n (46) for n = 0, 1, 2, The frst members of the sequence (u n ) n=0 (mod 5) are 0, 1, 2, 0, 2, 4, 0, 4, 3, 0, 3, 1, 0, 1, 2, 0, 2, 4, 0,, and those of the sequence (v n ) n=0 (mod 17) are 0, 1, 2, 0, 9, 1, 0, 13, 9, 0, 15, 13, 0, 16, 15, 0, 8, 16, 0, 4, 8, 0, 2, 4, 0, 1, 2, 0, 9, 1, 0 By nducton, one can prove that the sequences (u n ) n=0 and (v n ) n=0 are perodc wth the length of the perod 12 and 24, respectvely Snce 11 =0 u 0 (mod 5) and 23 =0 v 0 (mod 17), by Lemma 41 appled to (u n ) n=0 and (v n ) n=0, we derve that the sequences and (u (t) n ) n=0 S 12t+7 (u 0, u 1,, u 12t+6 ) (v (t) n ) n=0 S 24t+7 (v 0, v 1,, v 24t+6 ) satsfy the property (43) for t = 0, 1, 2, Hence, the sequence u (t) n for n = 0, 1, 2, satsfes the property (45), and the sequence v n (t) for n = 0, 1, 2, satsfes the property (46) for t = 0, 1, 2, Set t 1 = 8t + 5, t 2 = 10t + 6, t 3 = 5t + 3 for some postve nteger t Our goal s to fnd a sequence x (t) n for n = 0, 1, 2, satsfyng the followng condtons for every postve nteger n: x (t) n x (t) n x (t) n y (t1) n (mod 11), u (t2) n+1 (mod 5), v (t3) n+2 (mod 17) (47)
8 A LINEAR RECURRENCE SEQUENCE OF COMPOSITE NUMBERS 367 Usng the defnton of the sequences y (t) n, u (t) n, and v (t) n for n = 0, 1, 2, we can rewrte (47) as (x (t) n ) n=0 S 120t+79 (y 0, y 1,, y 120t+78 ) (mod 11), (x (t) n ) n=0 S 120t+79 (u 1, u 2,, u 120t+78, u 7 ) (mod 5), (x (t) n ) n=0 S 120t+79 (v 2, v 3,, v 120t+78, v 7, v 8 ) (mod 17) By the Chnese remander theorem, the system of equatons (47) has a soluton for every nonnegatve nteger t For t = 0 we fnd that (x (0) n ) n=0 = S 79 (121, 782, 145, 902, 289, 710, 264, 493, 865, 693, 731, 560, 66, 697, 195, 407, 34, 310, 484, 663, 325, 803, 306, 205, 121, 357, 230, 902, 884, 30, 264, and for t > 0 we defne 408, 695, 693, 476, 50, 66, 867, 535, 407, 544, 395, 484, 323, 580, 803, 221, 35, 121, 102, 655, 902, 119, 370, 264, 918, 780, 693, 136, 305, 66, 782, 365, 407, 289, 820, 484, 493, 920, 803, 731, 120, 121, 697, 910, 902, 34, 200, 264), (x (t) n ) n=0 = S 120t+79 (x (0) 0, x(0) 1,, x(0) 120t+78 ) By (43), (47) and by the propertes (44) (46), t follows mmedately that the followng hold If n 0 (mod 3), then x (t) n 0 (mod 11) If n 1 (mod 3) then x (t) n 0 (mod 17) If n 2 (mod 3) then x (t) n 0 (mod 5) Snce x (0) n > 17 for n = 0, 1, 2,, we conclude that x (t) n for n = 0, 1, 2, s a kstep Fbonacclke sequence of composte numbers for k = 120t + 79 and t = 0, 1, 2, 5 An algorthm for the constructon of the set S k (N) The constructon of the set S k (N) splts nto two parts We frst generate the fnte set s k (N) = {(p 1, m 1 ), (p 2, m 2 ),, (p I, m I ), }, where p s a prme number and m s a postve nteger (Algorthm 1) Then we try to construct the coverng system {r 1 (mod m 1 ), r 2 (mod m 2 ),, r I (mod m I )} for I I Algorthm 2 gves the answer I can t construct a coverng system or returns a coverng system In the second case, we construct the set S k (N) = {(p, m, r )} These algorthms were mplemented usng a computer algebra system PARI/GP [10] The only thng we can control n the constructon of the set S k (N) s the nteger N If Algorthm 2 gves an answer I can t construct a coverng system, then we can try to choose a dfferent N and try agan We can have dfferent sets S k (N) for dfferent values of N The mplementaton of these algorthms takes less than one mnute to gve an answer on a modestly powered computer (Athlon XP 2100+) for 3 k 10 and for good choce of N Defne A N = {1, 2,, N} for some postve nteger N and let A N (m, r) = {a a A N, a r (mod m)} Emprcal results suggest that we can choose sutable a N for any postve nteger k 2, so we state a followng conjecture Conjecture 1 Let k 2 be some fxed postve nteger Then there exst postve ntegers a 0, a 1,, a such that gcd(a 0, a 1,, a ) = 1 and the sequence S k (a 0, a 1,, a ) contans no prme numbers
9 368 J ŠIURYS Algorthm 1 Construct the set s k (N) Requre: k 2, N 2 Ensure: The set s k (N) 1: prmes lst {} 2: s k (N) {} 3: dvsors lst lst of N dvsors 4: for d dvsors lst do 5: Construct the matrx B k,d {see 2} 6: determnant B k,d 7: factors lst prme factors of determnant 8: for factor factors lst do 9: f factor prmes lst then 10: Put factor n prmes lst 11: Put (factor, dvsor) n s k (N) 12: end f 13: end for 14: end for 15: return s k (N) Algorthm 2 Construct a coverng system Requre: A fnte set of postve ntegers {m 1, m 2,, m I } Ensure: The coverng system {r 1 (mod m 1 ), r 2 (mod m 2 ),, r I (mod m I )} 1: N lcm(m 1, m 2,, m I ) 2: Coverng set {} 3: B A N 4: for from 1 to I do 5: MAX 0 6: for r from 0 to m 1 do 7: f MAX < A N (m, r) B then 8: r r 9: end f 10: Put r (mod m ) n Coverng set 11: B B\A N (m, r ) 12: f B = {} then 13: return Coverng set 14: end f 15: end for 16: end for 17: prnt I can t construct a coverng system 6 Examples of sequences for k = 4, 5,, 10 Snce the case k = 2 s proved n [5] and the case k = 3 n [13], n ths secton we wll prove Theorem 11 for k = 4, 5,, 10 As was notced n 3, we only need to construct the
10 A LINEAR RECURRENCE SEQUENCE OF COMPOSITE NUMBERS 369 set S k (N) Below we lst some examples of sequences (x n ) n=0 for each k n the range 4 k 10 (x n ) n=0 = S 4 ( , , , ) (x n ) n=0 = S 5 ( , , , , ) (x n ) n=0 = S 6 ( , , , , , ) (x n ) n=0 = S 7 (49 540, , , , 9962, , ) (x n ) n=0 = S 8 ( , , , , , , , ) (x n ) n=0 = S 9 ( , , , , , , , , ) (x n ) n=0 = S 10 ( , , , , , , , , , ) Snce the set S k (N) s essental n the constructon of a kstep Fbonacclke sequence S k (x 0, x 1,, x ), we gve ths set for each k n the range 4 k 10 (see Tables 1 7) Table 1 The set S 4(360) p m r B 4,m
11 370 J ŠIURYS Table 2 The set S 5(16) p m r B 5,m Table 3 The set S 6(32) p m r B 6,m Table 4 The set S 7(6) p m r B 7,m Table 5 The set S 8(30) p m r B 8,m Table 6 The set S 9(12) p m r B 9,m Table 7 The set S 10(8) p m r B 10,m
12 A LINEAR RECURRENCE SEQUENCE OF COMPOSITE NUMBERS 371 Fnally, we gve the coeffcents of the system of equatons (35) (see Tables 8 15) It s necessary because n Lemma 23 we prove only the exstence of these coeffcents; that s, wth the same set S k (N) we can fnd the dfferent kstep Fbonacclke sequence S k (x 0, x 1,, x ) Table 8 Coeffcents of (35) for k = A, A, A, A, Table 9 Coeffcents of (35) for k = A, A, A, A, Table 10 Coeffcents of (35) for k = A, A, A, A, A, Table 11 Coeffcents of (35) for k = A, A, A, A, A, A, Table 12 Coeffcents of (35) for k = A, A, A, A, A, A, A,
13 372 J ŠIURYS Table 13 Coeffcents of (35) for k = A, A, A, A, A, A, A, A, Table 14 Coeffcents of (35) for k = A, A, A, A, A, A, A, A, A, Table 15 Coeffcents of (35) for k = A, A, A, A, A, A, A, A, A, A, References 1 Y Blu, G Hanrot, P M Vouter and M Mgnotte, Exstence of prmtve dvsors of Lucas and Lehmer numbers, J rene angew Math 539 (2001) S L G Cho, Coverng the set of ntegers by congruence classes of dstnct modul, Math Comp 25 (1971) A Dubckas, A Novkas and J Šurys, A bnary lnear recurrence sequence of composte numbers, J Number Theory 130 (2010) I Flores, Drect calculaton of kgeneralzed Fbonacc numbers, Fbonacc Quart 5 (1967) R L Graham, A Fbonacclke sequence of composte numbers, Math Mag 37 (1964) M Hall, Dvsblty sequences of thrd order, Amer J Math 58 (1936) D E Knuth, A Fbonacclke sequence of composte numbers, Math Mag 63 (1990) J W Ncol, A Fbonacclke sequence of composte numbers, Electron J Combn 6 (1999) research paper R44
14 A LINEAR RECURRENCE SEQUENCE OF COMPOSITE NUMBERS T D Noe and J V Post, Prmes n Fbonacc nstep and Lucas nstep sequences, J Integer Seq 8 (2005) artcle The PARI Group, Bordeaux PARI/GP, verson, 235, 2006, 11 J C Parnam and T N Shorey, Subsequences of bnary recursve sequences, Acta Arth 40 (1982) L Somer, Secondorder lnear recurrences of composte numbers, Fbonacc Quart 44 (2006) J Šurys, A trbonacclke sequence of composte numbers, Fbonacc Quart 49 (2011) M Vsemrnov, A new Fbonacclke sequence of composte numbers, J Integer Seq 7 (2004) artcle H S Wlf, Letters to the edtor, Math Mag 63 (1990) 284 Jonas Šurys Department of Mathematcs and Informatcs Vlnus Unversty Naugarduko 24, Vlnus LT Lthuana
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