Two Properties of Catalan-Larcombe-French Numbers

Transcription

1 Journal of Ineger Sequences, Vol. 9 06, Aricle 6.3. Two Properies of Caalan-Larcombe-French Numbers Xiao-Juan Ji School of Mahemaical Sciences Soochow Universiy Suzhou Jiangsu 5006 P. R. China xiaojuanji05@yahoo.com Zhi-Hong Sun School of Mahemaical Sciences Huaiyin Normal Universiy Huaian Jiangsu 300 P. R. China zhihongsun@yahoo.com Absrac Le P n be he Caalan-Larcombe-French numbers. The numbers P n occur in he heory of ellipic inegrals, and are relaed o he arihmeic-geomeric-mean. In his paper we invesigae he properies of he relaed sequence S n P n / n insead, since S n is an Apéry-lie sequence. We prove a congruence and an inequaliy for P n. Inroducion Le P n be he sequence given by P 0, P 8 and n P n 83n 3nP n 8n P n n.

2 The numbers P n are called Caalan-Larcombe-French numbers since Caalan firs defined P n in [], and Larcombe and French [6] proved ha n/ n n P n n n n n n n, where x is he greaes ineger no exceeding x. The numbers P n are relaed o he arihmeic-geomeric-mean. See [6] and A05375 in Sloane s On-Line Encyclopedia of Ineger Sequences. Le S n be defined by S 0, S and n S n 3n 3nS n 3n S n n. Comparing wih, we see ha Zagier noed ha S n As observed by Jovović [7] in 003, S n n n n/ S n P n n. n n. n n Recenly Z. W. Sun saed ha n S n n n n The firs few values of S n are shown below: n n 0,,,... n n n. S 0, S, S 0, S 3, S 676, S 5 30, S 6 896, S 7 90, S , S , S , S , S , S Le p be an odd prime. Jarvis, Larcombe, and French [3] proved ha if n a r p r a pa 0 wih a 0,a,...,a r {0,,...,}, hen P n P ar P a P a0 mod p. Jarvis and Verrill [5] showed ha P n 8 n P n mod p for n 0,,...,

3 and P mp r P mp r mod p r for m,r Z, where Z is he se of posiive inegers. For a prime p le Z p denoe he se of hose raional numbers whose denominaor is no divisible by p. Le p be an odd prime, n Z p and n 0,6 mod p. The second auhor [] proved ha S nn6 n6 p n where a is he Legendre symbol. p In 89 Franel [] inroduced he following Franel numbers f n : mod p, f n n 3 n n 0,,,... The firs few Franel numbers are as below: f 0, f, f 3 0, f 56, f 5 36, f 6 5, f Franel [] noed ha he sequence f n saisfies he recurrence relaion: n f n 7n 7nf n 8n f n n. Le r Z and p be a prime wih p 5,7 mod 8. The second auhor [] conjecured ha Sp r 0 mod p r and fp r 0 mod p r. 3 In his paper we prove 3 in he case r. We also prove he second auhor s conjecure []: Basic lemmas S mm m > S m S m for m,3,... Lemma Lucas heorem [8]. Le p be an odd prime. Suppose a a r p r a pa 0 and b b r p r b pb 0, where a r,...,a 0,b r,...,b 0 {0,,...,}. Then a ar a0 mod p. b Lucas heorem is ofen formulaed as follows. b r 3 b 0

4 Lemma [8]. Le p be an odd prime and a,b Z. Suppose a 0,b 0 {0,,...,p }. Then apa0 a a0 mod p. bpb 0 b b 0 Lemma 3 [, Lemma.7]. For any posiive ineger n we have S n n n n n Lemma [9]. Le p be an odd prime. Suppose n n p n 0 and p 0 wih,n Z and 0,n 0 {0,,...,}. Then n n n0 n0 p n0 p n n mod p. 0 p Lemma 5. Le p be an odd prime. Then / 0 p 0 mod p. p Proof. For 0 /, from Lemma we have p p and so p p p p p p p mod p 0. p p We firs assume p mod. Applying Lemma we ge 3 3 p 3 p mod p p 0 mod p. 3 p

5 and p p p p 0 mod p. p Also, By he above four congruences, we have 0 mod p. / p p p5/ p p 3 p 3 p p5/ p p p5/ p p 3 p 3 p 0 mod p. Thus he resul is rue for p mod. Now we assume p 3 mod. By Lemma, p p mod p. 5

6 As and we obain / p p / p3/ 0 mod p. 0 mod p 0 mod p, Hence he resul is also rue in his case. The proof is now complee. Lemma 6 [0, Theorem 3.3]. Le p be a prime wih p 5,7 mod 8. Then mod p. Lemma 7 [, Lemma.8]. Le m Z and,p Z. Then mp r p mp r mpr. /p i i, p i 3 Congruences for S p and f p Theorem 8. Le p be a prime wih p 5,7 mod 8. Then Moreover, S p S p f p f p 0 mod p. 6 mod p 3. mod p

7 Proof. For < < p and 0 s, from Lemma we see ha p p mod p, and sp sp p 0 mod p, sp s spp s p 0 mod p sp s p sp pspp p sp spp ps p s p 0 mod p. Now we asser ha / p sp 3 0 mod p for s 0,,,... We prove he resul by inducion on s. For 0 / we see ha p mod p. From Lemma 6 we now ha he resul is rue for s 0. Suppose ha holds for s. For s, applying Lemma we have / p p 3 p p Hence / p p hypohesis and Lemma we have / p 3 p p p p 3 0 mod p for p 7 p 3 mod p.. For <, by he inducive p p 3 0 mod p.

8 Also, p mod pand 0 mod p for {0,,..., Hence / p 3 / / / / 3 p }. By Lemma 5, p p p p p p p p p p p p p p p p p 3 p 3 0 mod p. 3 / 3 / f p / s0 / p p sp p p p p p p p p 3 0 mod p. p p p p 3 Se H 0 H 0, 0, H i and H, i<j for ij Z. For 0 s /, i is easily seen ha H 0 mod p, s phs p H s, mod p 3 and so s ph sp HsH s, mod p 3. By Lemma 7, for 0 / we see ha p sp s sp i,p i p i 8 sp p s i,p i i mod p 3.

9 Applying, Lemma 6 and he ideniy ab b cd d we derive ha S p / s0 p p p p p p / / p sp s0 / s0 / p p p p p p / s0 f p / s0 / s s0 s / s0 / / s0 a c c d ac / a bd b sp p sp p sp sp p /3 sp p sp / / s 3 s 6 s / / p sp p sp p sp 3 3 sp p 3 6 H 3 Hs phs H s, / mod p 3. p sp Summarizing he above proves he heorem. An inequaliy involving S m Theorem 9. For m,3,,... we have S mm m > S m S m. 3 i,p i p sp Proof. I is easily seen ha S mm m > S m S m for m 3,,...,3. 9 i 3

10 Now suppose m and mm S m > S m S m. By, Lemma 3 and he inducive hypohesis we have S mm m S m S m S mm m 3m 3m S m m S m 3m m S m m m > mm S m 3m 3m S m m 3m mm m m 3m3 S m 0m 5 60m 5m 3 8m 36mS m 8m 5 30m 56m 3 3m 9m96S m S m m m 3m3m 3 m 6S m m m 3m3m 3 m m m m m S m Fm,, where Fm, 5m 5 5m 3m 3 7m 9m3 8m 5 0m 6m 3 m m6. Form iiseasilyseenha3 < m7m5 m3m <, 5m 5 5m 3m 3 7m 9m3 > 0, 8m 5 0m 6m 3 m m6 < 0,6m 7 75m 6 3m 5 83m 6m 3 7m 87m > 0, and Fm, > Fm, for 0,,...,m3. Thus, Fm,m3Fm, > Fm,5Fm, > 0 and Fm, Fm, 5 3 5m5 5m 3m 3 7m 9m3 8m 5 0m 6m 3 m m6 > 0 for. 0

11 From he above we derive ha S mm m S m S m 6S m m m > Fm, m m 3m3m 3 m m m m m Fm, m m 6S m m Fm,mFm,0 m m 3m3m 3 m m m8 Fm,mFm, 3mm3 m m6 Fm,Fm,m3 m m3 > 3m 5m6Fm,m m7m5 Fm,0 m3m 6S m8 m m m m 3m3m 3 m S m8 m m 6mm7 8m7m5 m m 3m3m 3 m m3m 0m 5 0m 0m 3 56m 7m 8m 5 30m 56m 3 3m 9m96 3mm3 m7m5 m3m 6mm7 > 0m 5 0m 0m 3 56m 7m 3mm36 8m 5 30m 56m 3 3m 9m96 S m8 m m m m 3m3m 3 m 6m 7 75m 6 3m 5 83m 6m 3 7m 87m 6S m8 m m m m 3m3m 3 m > 0. Hence he inequaliy is proved by inducion.

12 Corollary 0. For m,3,,... we have P mm m > P m P m. Proof. Since P m m S m, he resul follows from Theorem 9. References [] E. Caalan, Sur les nombres de Segner, Rend. Circ. Ma. Palermo 887, [] J. Franel, On a quesion of Laisan, L Inermédiaire des Mahémaiciens 89, 5 7. [3] A. F. Jarvis, P. J. Larcombe, and D. R. French, On small prime divisibiliy of he Caalan-Larcombe-French sequence, Indian J. Mah , [] X. J. Ji and Z. H. Sun, Congruences for Caalan-Larcombe-French numbers, preprin, 05, hp://arxiv.org/abs/ v. [5] F. Jarvis and H. A. Verrill, Supercongruences for he Caalan-Larcombe-French numbers, Ramanujan J. 00, [6] P. Larcombe and D. French, On he oher Caalan numbers: a hisorical formulaion re-examined, Congr. Numer , [7] P. J. Larcombe and D. R. French, A new generaing funcion for he Caalan-Larcombe- French sequence: proof of a resul by Jovovic, Congr. Numer , 6 7. [8] R. Mešrović, Lucas heorem: is generalizaions, exensions and applicaions 878 0, preprin, 0, hp://arxiv.org/abs/ [9] B. Sagan, Congruences via Abelian groups, J. Number Theory , [0] Z. H. Sun, Congruences concerning Legendre polynomials II, J. Number Theory , [] Z. H. Sun, Congruences involving Franel and Caalan-Larcombe-French numbers, preprin, 05, hp://arxiv.org/abs/ Mahemaics Subjec Classificaion: Primary A07; Secondary 05A0, 05A9, 05A0. Keywords: congruence, inequaliy, Caalan-Larcombe-French number. Concerned wih sequence A05375.

13 Received November 05; revised versions received November 05; February Published in Journal of Ineger Sequences, April Reurn o Journal of Ineger Sequences home page. 3