A talk give at the Natioal Chiao Tug Uiversity (Hsichu, Taiwa; August 5, 2010 O Divisibility cocerig Biomial Coefficiets Zhi-Wei Su Najig Uiversity Najig 210093, P. R. Chia zwsu@ju.edu.c http://math.ju.edu.c/ zwsu August 5, 2010
Abstract Biomial coefficiets arise aturally i combiatorics. Recetly the speaker iitiated the study of certai divisibility properties of biomial coefficiets, ad products or sums of biomial coefficiets. I this talk we itroduce the speaker s related results ad various cojectures. The materials come from the author s two preprits: 1. Z. W. Su, Products ad sums divisible by cetral biomial coefficiets, arxiv:1004.4623. 2. Z. W. Su, O divisibility cocerig biomial coefficiets, arxiv:1005.1054.
Catala umbers For N = {0, 1, 2,...}, the th Catala umber is give by C = 1 ( ( ( 2 2 2 =. + 1 + 1 Recursio. C 0 = 1 ad C +1 = =0 C k C k ( = 0, 1, 2,.... k=0 Geeratig Fuctio. C x = 1 1 4x. 2x Combiatorial Iterpretatios. The Catala umbers arise i may eumeratio problems. For example, C is the umber of biary parethesizatios of a strig of + 1 letters, ad it is also the umber of ways to triagulate a covex ( + 2-go ito triagles by 1 diagoals that do ot itersect i their iteriors.
Geeralized Catala umbers Let h Z + = {1, 2, 3,...}. The first-kid Catala umbers of order h: C (h k = 1 ( ( ( (h + 1k (h + 1k (h + 1k = h (k N. hk + 1 k k k 1 (As usual, ( x = 0 for = 1, 2,.... The secod-kid Catala umbers of order h: C (h k = h ( ( ( (h + 1k (h + 1k (h + 1k = h k + 1 k k k + 1 (k N. Remark. Those C k = C (1 k = C (1 k are ordiary Catala umbers.
Whe l + 1 ( k+l l for all = 1, 2, 3,...? Problem For what values of k, l Z + we have ( l + 1 k + l for all = 1, 2, 3,...? l As ( l+ (l l = (l + 1C, ( l + 1 + l for all = 1, 2, 3,.... l Theorem. Let k ad l be positive itegers. The for ay N we have ( ( l + 1 k + l (k, l + 1, i.e., l + 1 k + l k k. l I particular, if all prime factors of k divides l the l + 1 ( k+l l for every = 0, 1, 2,....
The p-adic order of! Let p be a prime. The p-adic order of a iteger x is give by (I particular, ν p (0 = +. ν p (x = sup{a N : p a x}. A Kow Fact. For ay prime p ad Z + we have ν p (! = p i. Proof. Observe that ν p (! = = ν p (k = k=1 i=1 1 = k=1 p i k i=1 k=1 i=1 k 1 i=1 p i k p i = i=1 p i.
Legedre s Theorem Legedre s Theorem. For ay prime p ad Z + we have ν p (! = ρ p(, p 1 where ρ p ( is the sum of the digits of i the expasio of i base p. Proof. Write = k i=0 a ip i with a i {0, 1,..., p 1}. For i = 1,..., k we have k i 1 i 1 = p i a j p j i + r i with r i = a j p j (p 1p j = p i 1. Thus j=i ν p (! = = k i=1 k p i = a j j=1 i=1 k i=1 j p j i = j=0 k a j p j i j=i j=0 k p j 1 a j p 1 = ρ p(. p 1 j=1
A Lemma Lemma. Let m Z + ad k, l, Z. The k + l k l + 1 k k 1 + 0. m m m m m Proof. If m k, the k l + 1 k 1 l + 1 (k 1 + (l + 1 + = +. m m m m m If m (l + 1, the k l + 1 + = m m k + m l m k + l Now assume that m k ad m (l + 1. Clearly m is relatively prime to. Thus m k ad hece k + l k l + 1 k k 1 + m m m m m = k m + l + 1 m 1 k m l + 1 m + k ( k m m 1 = 0. m.
Proof of l + 1 k ( k+l l Let k, l, be ay positive itegers. With the help of the above lemma, for ay prime p we have ( ( k k+l ( k (k + l!k! ν p = ν p l + 1 (k!(l + 1!(k 1! ( k + l k l + 1 k k 1 = p j p j p j + p j p j 0. j=1 Therefore ( l + 1 k + l k. k
A cojecture o divisibility of biomial coefficiets If all prime factors of k Z + divides l Z +, the ( l + 1 k + l for every = 0, 1, 2,... l sice (l + 1, k = 1. Cojecture (Su, 2010. Let k ad l be positive itegers. If ( l + 1 k + l l for all sufficietly large positive itegers, the each prime factor of k divides l. I other words, if k has a prime factor ot dividig l the there are ifiitely may positive itegers such that l + 1 ( k+l l.
Some umerical examples Let k ad l be positive itegers such that ot all prime factors of k divides l. Defie f (k, l as the smallest positive iteger such that l + 1 ( k+l k. Via Mathematica we obtaied the followig data: f (7, 36 = 279, f (10, 192 = 362, f (11, 100 = 1187, f (13, 144 = 2001, f (22, 200 = 6462, f (31, 171 = 1765; f (43, 26 = 640, f (53, 32 = 790, f (67, 56 = 2004, f (73, 61 = 2184, f (74, 62 = 885, f (97, 81 = 2904, f (179, 199 = 28989, f (223, 93 = 13368, f (307, 189 = 31915.
Bober s Recet Work I 2009 J. W. Bober [J. Lodo Math. Soc.] determied all those a 1,..., a r, b 1,..., b r+1 Z + = {1, 2, 3,...} with a 1 + + a r = b 1 + + b r+1 such that (a 1! (a r! (b 1! (b r+1! Z for all Z+. I particular, if k ad l are positive itegers the ( l ( kl l (kl!((k 1! ( k = (k!((l 1!((k 1l! Z for all Z+, k = l, or {k, l} {1, 2}, or {k, l} = {3, 5}.
More Results o Products of Two Biomial Coefficiets Theorem (Su, 2010. Let k, Z +. (i we have ( ( k 2 2 C (k 1 2. Moreover, ( 2 (k 1 C 2 two. (ii We have /(2 ( k is odd if ad oly if is a power of ( k (2k 1C ( 2k 2 ( ad (2k 1C 2k ( 2 / k is odd if ad oly if + 1 is a power of two. (iii Let (k + 1 be the odd part of k + 1. The ( 2 (k + 1 C (k 1, ( 2k k ad (k + 1 C (k 1 ( 2k ( k / 2 is odd if ad oly if (k 1 + 1 is a power of two.,
Some Lemmas Lemma 1. Let m Z + ad k, Z. The we have 2k k (k 1 2(k 1 + m m m m + 1 2k 1 2k 2 +. m m m Lemma 2. Let m > 2 be a iteger. For ay k, Z we have 2k k + 1 k 2 k (k 1 + 1 + + + + +. m m m m m m m
Aother Theorem Theorem (Su, 2010. For ay positive itegers k ad, we have 2 k 1 ( 2 ( 2(2 k 1 (2 k C (2k 2. 1 I particular, S = ( 6 3 3( 2(2 + 1 ( Z for all = 1, 2, 3,.... 2 A Key Lemma. Let p be a prime ad let k N ad Z +. The ρ p ((p k 1 { (p k } 1 = p 1 p j k ad hece the expasio of (p k 1 i base p has at least k ozero digits. j=1
Proof of the Lemma For ay m Z +, by Legedre s theorem we have ρ p (m p 1 = m p 1 ν m m p(m! = p j p j = Observe that ( p p k k 1 1 ad hece j=1 ( p k = = j=1 (p k!!((p k 1! k ν p ((p k! ν p (! ν p (((p k 1! p k (p k 1 = p j p j p j = j=1 j=1 j=1 j=1 k (p p k j k 1 = j=1 p j j=1 p j j=1 { } m p j. { (p k } 1.
A Geeral Cojecture o Expasios i Base m Cojecture. Let m > 1 be a iteger ad k ad be positive itegers. The the sum of all digits i the expasio of (m k 1 i base m is at least k(m 1. Also, the expasio of mk 1 m 1 i base m has at least k ozero digits. Remark. Hao Pa has proved the cojecture fully.
Properties of S It is iterestig to ivestigate the iteger sequece ( 6 3 S = 3( 2(2 + 1 ( ( = 1, 2, 3,.... 2 S 1 = 5, S 2 = 231, S 3 = 14568, S 4 = 1062347, S 5 = 84021990. By Stirlig s formula, S 108 /(8 π as +. Set S 0 = 1/2. Usig Mathematica we fid that S k x k = si( 2 3 arcsi(6 ( 3x 8 0 < x < 1 3x 108 k=0 ad i particular S k 108 k = 3 k=0 3 8. Mathematica also yields that S k (2k + 3108 k = 27 3 256. k=0
A Cojecture o the Sequece {S } 1 Oe ca easily show that for ay odd prime p. S p 15 30p + 60p 2 (mod p 3 Cojecture (Su, 2010 (i Let Z + = {1, 2, 3,...}. The S is odd if ad oly if is a power of two. Also, 2 + 3 3S. (ii For ay prime p > 3 we have p 1 k=1 S k 108 k { 0 (mod p if p ±1 (mod 12, 1 (mod p if p ±5 (mod 12.
O the Compaio Sequece {T } 1 Here is a cojecture o a compaio sequece of {S } 0. Cojecture (Su, 2010. There are positive itegers T 1, T 2, T 3,... such that S k x 2k+1 + 1 24 T k x 2k = cos( 2 3 arccos(6 3x 12 k=0 k=1 for all real x with x 1/(6 3. Also, T p 2 (mod p for ay prime p. Values of T 1,..., T 8 : 1, 32, 1792, 122880, 9371648, 763363328, 65028489216, 5722507051008.
More Results Theorem (Su, 2010. For ay Z + we have ( ( ( 5 3 1 (6 + 1 C (4 3 3 1, A Lemma. (i For ay real umber x we have ( 5 1 1 {12x} + {5x} + {2x} {4x} + {15x}. C (2 5. (ii Let x be a real umber with {5x} {2x} 1/2. The {5x} 2/3. A Auxiliary Theorem. Let m > 1 ad be itegers. (i If 3 m, the 15 1 2 4 12 + 2 2 5 1 + + + +. m m m m m m (ii If 5 m, the 15 1 m + 2 m 10 + 1 m + 4 + m 3 1 m.
More Cojectures Defie the ew sequece {t } 1 of itegers by ( 5 1 (2 ( 1 C 5 1 ( 15 5 1 5 t = ( 3 = (10 + 1 (. 3 Cojecture (Su, 2010. Let be ay positive iteger. We have 21t 0 (mod 10 + 3. If 3, the (10 + 3 7t. If 7 + 1, the (10 + 3 3t. Cojecture (Su, 2010. Let k ad l be itegers greater tha oe. (i If ( k ( ( l kl l 1 for all Z +, the k = l, or l = 2, or {k, l} = {3, 5}. (ii If ( ( ( k l kl 1 l for all Z +, the k = 2, ad l + 1 is a power of two.
O Products of Three Biomial Coefficiets Theorem 1. For ay oegative itegers k ad we have ( ( ( ( 2k 4 + 2k + 2 2 + k + 1 2 k + 1 k 2 + k + 1 2k ad ( 2k (2 + 1 k ( 2 C +k ( + k + 1 2k. Lemma 1. Let m Z + ad k, Z. The we have 4 + 2k + 2 2 + k + 1 k 2k + 2 2 m m m m k + 1 +, m m uless 2 m ad k + 1 m/2 (mod m i which case the right-had side of the iequality equals the left-had side plus oe.
Aother Lemma Lemma 2. Let m Z + ad k, Z. The we have 2 + 2k + k k 2k + 2 2 m m m m 2 + 1 k + 1 2 +, m m m uless 2 m ad k + 1 m/2 (mod m i which case the right-had side of the iequality equals the left-had side plus oe.
A Theorem o Sums of Biomial Coefficiets I 1914 Ramauja got that ad k=0 4k + 1 ( 64 k ( 2k 3 = 2 k π 2 ( 4k 2k (20k + 3 ( 2 10 k = 8 π. k=0 ( 2k k Via Theorem 1 ad the WZ method, we obtai Theorem 2. For ay positive iteger we have ( 2 ( 2k 3 4(2 + 1 (4k + 1 ( 64 k k ad k=0 k=0 ( 2 ( 2k 2 ( 4k 4(2 + 1 (20k + 3 ( 2 10 k. k 2k
A Further Cojecture Cojecture (Su, 2009 (i For Z + set 1 a := 2(2 + 1 ( 1 ( 2k 2 ( 4k (20k + 3 ( 2 10 1 k. 2 k 2k k=0 The ( 1 1 a Z + for all = 2, 3, 4,.... (ii Let p be a odd prime. The p 1 k=0 20k + 3 ( 2 10 k ( 2k k 2 ( 4k 3p 2k where E 0, E 1, E 2,... are Euler umbers, ad p (p 1/2 k=0 ( 1 p provided p > 3. ( 1 + 3p 3 E p 3 (mod p 4 p ( 20k + 3 2k 2 ( 4k ( 2 10 k k 2k (2 p 1 + 2 (2 p 1 1 2 (mod p 4
A Further Cojecture (Cotiued (iii For ay prime p 2, 5, we have We also have p 1 k=0 ( 2k k 2 ( 4k 2k ( 2 10 k 4x 2 2p (mod p 2 if p 1, 9 (mod 20 & p = x 2 + 5y 2, 2(p x 2 (mod p 2 if p 3, 7 (mod 20 & 2p = x 2 + 5y 2, 0 (mod p 2 if p 11, 13, 17, 19 (mod 20. Remark. Let p 2, 5 be a prime. By the theory of biary quadratic forms, if p 1, 9 (mod 20 the p = x 2 + 5y 2 for some x, y Z; if p 3, 7 (mod 20 the 2p = x 2 + 5y 2 for some x, y Z. The speaker has made lots of other cojectures similar to the above oe.
Thak you!