Factors of certain sums involving central q-binomial coefficients
aa r X i v : . [ m a t h . N T ] F e b Factors of certain sums involving central q -binomialcoefficients Victor J. W. Guo and Su-Dan Wang School of Mathematics and Statistics, Huaiyin Normal University, Huai’an 223300, Jiangsu,People’s Republic of China [email protected] College of Mathematics Science, Inner Mongolia Normal University, Huhhot 010022, Inner Mongolia,People’s Republic of China [email protected]
Abstract.
Recently, Ni and Pan proved a q -congruence on certain sums involving central q -binomial coefficients, which was conjectured by Guo. In this paper, we give a general-ization of this q -congruence and confirm another q -congruence, also conjectured by Guo.Our proof uses Ni and Pan’s technique and a simple q -congruence observed by Guo andSchlosser. Keywords : congruences; q -binomial coefficients; cyclotomic polynomials. : 11B65, 05A10, 05A30
1. Introduction
In 1914, Ramanujan [24] obtained a number of representations for 1 /π . One such instance,though not listed in [24], is the identity ∞ X k =0 k + 1( − k (cid:18) kk (cid:19) = 2 π , (1.1)which was first proved by Bauer [1] in 1859. Such formulas for 1 /π gained popularity in1980’s for the reason that they can be utilized to provide fast algorithms for computingdecimal digits of π . See, for example, the Borwein brothers’ monograph [2]. Recently,Guillera [4] gave a general method to prove Ramanujan-type series. For a q -analogue of(1.1), see [14].In 1997, Van Hamme [27] conjectured that 13 Ramanujan-type series possess nice p -adic analogues, such as ( p − / X k =0 k + 1( − k (cid:18) kk (cid:19) ≡ p ( − ( p − / (mod p ) , (1.2) * Corresponding author. p is an odd prime. The congruence (1.2) was confirmed by Mortenson [20] using a F transformation, and was reproved by Zudilin [30] employing the WZ (Wilf–Zeilberger)method. In 2013, also via the WZ method, Sun [26] gave the following generalization of(1.2): for any integer n > n − X k =0 (4 k + 1) (cid:18) kk (cid:19) ( − n − k − ≡ n (cid:18) nn (cid:19) ) . (1.3)Recently, among other things, Ni and Pan [21] proved the following extension of (1.3):for each n > r > n − X k =0 (4 k + 1) (cid:18) kk (cid:19) r ( − r ( n − k − ≡ r − n (cid:18) nn (cid:19) ) . (1.4)In fact, Ni and Pan also gave a q -analogue of (1.4): for each n > r >
2, modulo(1 + q n − ) r − [ n ] (cid:2) n − n − (cid:3) , n − X k =0 ( − k q k +( r − k [4 k + 1] (cid:20) kk (cid:21) r − ( − q k +1 ; q ) r − n − k − ≡ , (1.5)11 + q n − n − X k =0 q ( r − k [4 k + 1] (cid:20) kk (cid:21) r ( − q k +1 ; q ) rn − k − ≡ . (1.6)Here ( a ; q ) n = (1 − a )(1 − aq ) · · · (1 − aq n − ) stands for the q - shifted factorial , and [ n ] =[ n ] q = 1 + q + · · · + q n − is the q -integer . The q -congruences (1.5) and (1.6) were originallyconjectured by the first author [9, Conjecture 5.4]. They are obviously true for r = 1 (theleft-hand sides have closed forms; see [9]). The r = 2 case of (1.5) is a q -analogue of (1.4)and was first proved by the first author himself. The r = 2 case of (1.6) was first obtainedby the present authors [12].The first aim of this paper is to prove the following generalizations of (1.5) and (1.6). Theorem 1.1.
For each n > , r > and s > , modulo (1 + q n − ) r − [ n ] q (cid:2) n − n − (cid:3) q , n − X k =0 ( − k q k +(2 r − s − k [4 k + 1] s [4 k + 1] q (cid:20) kk (cid:21) r − q ( − q k +2 ; q ) r − n − k − ≡ , (1.7)11 + q n − n − X k =0 q (2 r − s − k [4 k + 1] s [4 k + 1] q (cid:20) kk (cid:21) rq ( − q k +2 ; q ) rn − k − ≡ . (1.8)In particular, letting q → Corollary 1.2.
For each n > , r > and s > , n − X k =0 (4 k + 1) s +1 (cid:18) kk (cid:19) r ( − r ( n − k − ≡ r − n (cid:18) nn (cid:19) ) . n , n − X k =0 [3 k + 1] ( q ; q ) k q − ( k +12 )( q ; q ) k ( q ; q ) k ≡ [ n ] q (1 − n ) / (mod [ n ]Φ n ( q ) ) , (1.9)where Φ n ( q ) is the n -th cyclotomic polynomial in q , i.e.,Φ n ( q ) = n Y k n gcd( n,k )=1 ( q − ζ k )with ζ being an n -th primitive root of unity.The second aim of this paper is to prove the following q -congruence, which was origi-nally conjectured by the first author [8, Conjecture 1.7]. Theorem 1.3.
Let n > be an integer. Then n − X k =0 [3 k + 1] (cid:20) kk (cid:21) ( − q k +1 ; q ) n − k − q − ( k +12 ) ≡ q n − ) [ n ] (cid:20) n − n − (cid:21) ) . (1.10)Letting q → Corollary 1.4.
For each n > , n − X k =0 (3 k + 1) (cid:18) kk (cid:19) n − k − ≡ n (cid:18) nn (cid:19) ) . We point out that some other interesting congruences and q -congruences can be foundin [5–7, 10, 11, 13, 15–18, 22, 28, 29, 31].
2. Proof of Theorem 1.1
We need the following lemma, which is a special case of [21, Lemma 3.2] and can also bededuced from the q -Lucas theorem (see [23]). Lemma 2.1.
Let s and t be non-negative integers with t d − . Then ( q ; q ) sd + t ( q , q ) sd + t ≡ s (cid:18) ss (cid:19) ( q ; q ) t ( q, q ) t (mod Φ d ( q )) . We adopt some notation similar to that used by Ni and Pan [21]. For any positiveinteger n , let S ( n ) = ( d > d is odd and $ n − d +12 d % = j nd k) , ⌊ x ⌋ denotes the greatest integer not exceeding x . It is easy to see that, for anyinteger d > n −
1, the number ( d + 1) / n , and so d / ∈ S ( n ). This meansthat S ( n ) only contains finite elements. Let A n ( q ) = Y d ∈S ( n ) Φ d ( q ) ,C n ( q ) = Y d | n, d> d is odd Φ d ( q ) , It is clear that, if d | n , then d / ∈ S ( n ). Therefore, the polynomials A n ( q ) and C n ( q ) arerelatively prime.We now give the key lemma, which is similar to [21, Theorem 2.1]. Lemma 2.2.
Let ν ( q ) , ν ( q ) , . . . be a sequence of rational functions in q such that, forany positive odd integer d > , (i) ν k ( q ) is Φ d ( q ) -integral for each k > , i.e., the denominator of ν k ( q ) is relativelyprime to Φ d ( q ) ; (ii) for any non-negative integers s and t with t d − , ν sd + t ( q ) ≡ µ s ( q ) ν t ( q ) (mod Φ d ( q )) , where µ s ( q ) is a Φ d ( q ) -integral rational function only dependent on s ; (iii) ( d − / X k =0 ( q ; q ) k ( q ; q ) k ν k ( q ) ≡ d ( q )) . Then, for all positive integers n , n − X k =0 ( q ; q ) k ( q ; q ) k ν k ( q ) ≡ A n ( q ) C n ( q )) . (2.1) Proof.
Our proof is very similar to that of [21, Theorem 2.1]. In order to make thepaper more readable, we provide it here. For d ∈ S ( n ), we can write n = ud + v with( d + 1) / v d −
1. Thus, for any v t d −
1, the expression ( q ; q ) t / ( q ; q ) t iscongruent to 0 modulo Φ d ( q ). In view of Lemma 2.1, we obtain( q ; q ) ud + t ( q ; q ) ud + t ≡ d ( q )) . n − X k =0 ( q ; q ) k ( q ; q ) k ν k ( q ) ≡ u X s =0 d − X t =0 ( q ; q ) sd + t ( q ; q ) sd + t ν sd + t ( q ) ≡ u X s =0 s (cid:18) ss (cid:19) µ s ( q ) d − X t =0 ( q ; q ) t ( q ; q ) t ν t ( q ) ≡ d ( q )) , where we have used the condition (iii) and the fact that ( q ; q ) t ν t ( q ) / ( q ; q ) t is congruentto 0 modulo Φ d ( q ) for ( d +1) / t d −
1. This proves that (3.1) is true modulo A n ( q ).On the other hand, for d | n , we assume that u = n/d . Still by Lemma 2.1, we have n − X k =0 ( q ; q ) k ( q ; q ) k ν k ( q ) = u − X s =0 d − X t =0 ( q ; q ) sd + t ( q ; q ) sd + t ν sd + t ( q ) ≡ u − X s =0 s (cid:18) ss (cid:19) µ s ( q ) d − X t =0 ( q ; q ) t ( q ; q ) t ν t ( q ) ≡ d ( q )) . This proves that (3.1) is also true modulo C n ( q ). ✷ We also require the following easily proved result, which is due to Guo and Schlosser[11, Lemma 3.1].
Lemma 2.3.
Let d be a positive odd integer. Then, for k ( d − / , we have ( q ; q ) ( d − / − k ( q ; q ) ( d − / − k ≡ ( − ( d − / − k ( q ; q ) k ( q ; q ) k q ( d − / k (mod Φ d ( q )) . In order to simplify the proof of Theorem 1.1, we need to present the following result.
Theorem 2.4.
Let n > , r > and s > be integers. Then n − X k =0 ( − k q k +(2 r − s − k [4 k + 1] s [4 k + 1] q ( q ; q ) r − k ( q ; q ) r − k ≡ A n ( q ) C n ( q )) , (2.2) n − X k =0 q (2 r − s − k [4 k + 1] s [4 k + 1] q ( q ; q ) rk ( q ; q ) rk ≡ A n ( q ) C n ( q )) . (2.3) Proof.
We only give a proof of (2.2), since the proof of (2.3) is exactly the same as thatof (2.2). For any non-negative integer k , let ν k ( q ) = ( − k q k +(2 r − s − k [4 k + 1] s [4 k + 1] q ( q ; q ) r − k ( q ; q ) r − k . d , the rational function ν k ( q ) is Φ d ( q )-integral, since( q ; q ) k ( q ; q ) k = (cid:20) kk (cid:21) q − q ; q ) k , and ( − q ; q ) k is relatively prime to Φ d ( q ). Using q d ≡ d ( q )) and applyingLemma 2.1 with q q , we have ν sd + t ( q ) = ( − sd + t q sd + t ) +(2 r − s − sd + t ) [4( sd + t ) + 1] s [4( sd + t ) + 1] q ( q ; q ) r − sd + t ( q ; q ) r − sd + t ≡ ( − s (2 r − s (cid:18) ss (cid:19) r − ν t ( q ) (mod Φ d ( q )) , for non-negative integers s and t with 0 t d −
1. Thus, the sequence ν ( q ) , ν ( q ) , . . . satisfies the requirements (i) and (ii) of Lemma 2.2.We now verify the requirement (iii) of Lemma 2.2, i.e., ( d − / X k =0 ( − k q k +(2 r − s − k [4 k + 1] s [4 k + 1] q ( q ; q ) r − k ( q ; q ) r − k ≡ d ( q )) . (2.4)In fact, by Lemma 2.3 with q q , it is easy to verify that, for 0 k ( d − /
2, the k -th and (( d − / − k )-th terms on the left-hand side of (2.4) cancel each other moduloΦ d ( q ), i.e.,( − ( d − / − k q d − / − k ) +(2 r − s − d − / − k ) × [2 d − k − s [2 d − k − q ( q ; q ) r − d − / − k ( q ; q ) r − d − / − k ≡ − ( − k q k +(2 r − s − k [4 k + 1] s [4 k + 1] q ( q ; q ) r − k ( q ; q ) r − k (mod Φ d ( q )) . This proves (2.4). By Lemma 2.2, we conclude that (2.2) is true modulo A n ( q ) C n ( q ). ✷ We collected adequate ingredients and are able to prove Theorem 1.1.
Proof of Theorem . For 0 k n −
2, the q -factorial ( − q k +2 ; q ) n − k − contains thefactor 1 + q n − , and for k = n −
1, we have (cid:20) kk (cid:21) q = (cid:20) n − n − (cid:21) q = (1 + q n − ) (cid:20) n − n − (cid:21) q . Therefore, the left-hand sides of (1.7) and (1.8) are both divisible by (1 + q n − ) r − .6n what follows, we shall prove that the left-hand sides of (1.7) and (1.8) are bothdivisible by [ n ] q (cid:2) n − n − (cid:3) q . It is easy to see that the q -binomial coefficient (cid:2) nk (cid:3) q has the thefollowing factorization (see [3, Lemma 1]): (cid:20) nk (cid:21) q = Y d ∈D n,k Φ d ( q ) , where D n,k := (cid:26) d > (cid:22) kd (cid:23) + (cid:22) n − kd (cid:23) < j nd k(cid:27) . It is easily seen that 1 < d ∈ D n − ,n − is odd if and only d ∈ S ( n ), and so[ n ] q (cid:20) n − n − (cid:21) q = A n ( q ) C n ( q ) Y d | nd > Φ d ( q ) · Y d ∈D n − ,n − d is even Φ d ( q ) . (2.5)Note that (cid:20) kk (cid:21) q ( − q k +2 ; q ) n − k − = ( q ; q ) k ( q ; q ) k ( − q ; q ) n − . By Theorem 2.4, the left-hand sides of (1.7) and (1.8) are both divisible by A n ( q ) C n ( q ).It remains to show that (1.7) and (1.8) also hold modulo Y d | nd > Φ d ( q ) · Y d ∈D n − ,n − d is even Φ d ( q ) . Firstly, let d | n be an even integer. Then1 + q d = 1 − q d − q d ≡ d ( q )) . It follows that, for 0 k < n − d/
2, the q -shifted factorial ( − q k +2 ; q ) n − k − contains thefactor 1 + q n − d and is therefore congruent to 0 modulo Φ d ( q ). On the other hand, for n − d/ k n −
1, we have d ∈ D k,k , i.e., (cid:20) kk (cid:21) q ≡ d ( q )) . (2.6)Hence, for 0 k n −
1, we always have (cid:20) kk (cid:21) q ( − q k +2 ; q ) n − k − ≡ d ( q )) . (2.7)7his proves that (1.7) and (1.8) are true modulo Q d | nd > Φ d ( q ). Secondly, we considerthe case where d ∈ D n − ,n − is even. Write n = ud + v with 0 v d −
1. Then v > d/ k < ud + d/
2, the polynomial ( − q k +2 ; q ) n − k − has the factor 1 + q ud + d and is divisible by Φ d ( q ). Moreover, for ud + d/ k n −
1, we have d ∈ D k,k , and so(2.6) holds. Therefore, for 0 k n −
1, the q -congruence (2.7) also holds in this case.This proves that (1.7) and (1.8) are true modulo Q d ∈D n − ,n − d is even Φ d ( q ).Noticing that [ n ] q (cid:20) n − n − (cid:21) q = (1 + q n − )[2 n − q (cid:20) n − n − (cid:21) q , and (1 + q n − ) is relatively prime to [2 n − q (cid:2) n − n − (cid:3) q , the least common multiple of(1 + q n − ) r − and [ n ] q (cid:2) n − n − (cid:3) q is just (1 + q n − ) r − [ n ] q (cid:2) n − n − (cid:3) q . This completes theproof of the theorem. ✷
3. Proof of Theorem 1.3
We first give the following result.
Lemma 3.1.
Let d be a positive odd integer. Let s and t be non-negative integers with t d − . Then ( − q ; q ) sd + t ≡ s ( − q ; q ) t (mod Φ d ( q )) . Proof.
It is easy to see that ( − q ; q ) d − ≡ d ( q )) (see, for example, [12, Lemma3.2]). Hence, for 0 t d −
1, we have( − q ; q ) sd + t = ( − q sd +1 ; q ) t s − Y j =0 ( − q jd +1 ; q ) d ≡ ( − q ; q ) t ( − q ; q ) sd ≡ s ( − q ; q ) t (mod Φ d ( q )) , where we have used q d ≡ d ( q )). ✷ We also need the following result, which is similar to Lemma 2.2 and is a special caseof [21, Theorem 2.1].
Lemma 3.2.
Let ν ( q ) , ν ( q ) , . . . be a sequence of rational functions in q such that, forany positive odd integer d > , (i) ν k ( q ) is Φ d ( q ) -integral for each k > , i.e., the denominator of ν k ( q ) is relativelyprime to Φ d ( q ) ; for any non-negative integers s and t with t d − , ν sd + t ( q ) ≡ µ s ( q ) ν t ( q ) (mod Φ d ( q )) , where µ s ( q ) is a Φ d ( q ) -integral rational function only dependent on s ; (iii) d − X k =0 ( q ; q ) k ( q ; q ) k ν k ( q ) ≡ d ( q )) . Then, for all positive integers n , n − X k =0 ( q ; q ) k ( q ; q ) k ν k ( q ) ≡ A n ( q ) C n ( q )) . (3.1)We can now prove Theorem 1.3. Proof of Theorem . Similarly as before, for 0 k n −
2, we have( − q k +1 ; q ) n − k − ≡ q n − ) , and for k = n −
1, there holds (cid:2) kk (cid:3) q ≡ q n − ) . This means that the left-handside of (1.10) is divisible by (1 + q n − ) .In what follows, we shall prove that the left-hand side of (1.10) is divisible by [ n ] (cid:2) n − n − (cid:3) .Letting q q in (2.5), we have[ n ] (cid:20) n − n − (cid:21) = A n ( q ) C n ( q ) Y d | nd > Φ d ( q ) · Y d ∈D n − ,n − d is even Φ d ( q ) . For any non-negative integer k , let ν k ( q ) = q − ( k +12 )[3 k + 1] ( q ; q ) k ( − q ; q ) k ( q ; q ) k . Then, applying Lemmas 2.1 and 3.1, for any positive odd integer d and non-negativeintegers s and t with 0 t d −
1, we have ν sd + t ( q ) = q − ( sd + t +12 )[3( sd + t ) + 1] ( q ; q ) sd + t ( − q ; q ) sd + t ( q ; q ) sd + t ≡ s (cid:18) ss (cid:19) ν t ( q ) (mod Φ d ( q )) . d − X k =0 q − ( k +12 )[3 k + 1] ( q ; q ) k ( − q ; q ) k ( q ; q ) k ≡ d ( q )) . Thus, we may apply Lemma 3.2 to deduce that n − X k =0 q − ( k +12 )[3 k + 1] ( q ; q ) k ( − q ; q ) k ( q ; q ) k ≡ A n ( q ) C n ( q )) . (3.2)Multiplying the left-hand side of (3.2) by ( − q ; q ) n − , we conclude that (1.10) is truemodulo A n ( q ) C n ( q ).It remains to show that (1.10) is also true modulo Y d | nd > Φ d ( q ) · Y d ∈D n − ,n − d is even Φ d ( q ) . This is exactly same as the proof of Theorem 1.1 and is omitted here. ✷ Acknowledgments.
The first author was partially supported by the National NaturalScience Foundation of China (grant 11771175). The second author was partially supportedby the Natural Science Foundation of Inner Mongolia, China (grant 2020BS01012).
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