# Thep-Adic Valuations of Sums of Binomial Coefficients

Authors: Yong Zhang; Peisen Yuan
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we prove three supercongruences on sums of binomial coefficients conjectured by Z.-W. Sun. Letp be an odd prime and let h∈ℤ with 2h−1≡0modp. For a∈ℤ+ and pa>3, we show that ∑k=0pa−1hpa−1k2kk−h/2k≡0modpa+1. Also, for any n∈ℤ+, we have νp∑k=0n−1hn−1k2kk−h/2k≥νpn, where νpn denotes the p-adic order of n. For any integer m≡0modp and positive integer n, we have 1/pn∑k=0pn−1pn−1k2kk/−mk−mm−4/p∑k=0n−1n−1k2kk/−mk∈ℤp, where −˙ is the Legendre symbol and ℤp is the ring of p-adic integers.


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## Body

## 1. Introduction

Letp be an odd prime. In 2006, Pan and Sun [1] proved the congruence(1)∑k=0p−12kk+d≡p−d3modp,ford=0,…,p,via a curious combinatorial identity. For any positive integer a and prime p≥5, later Sun and Tauraso [2] established the following general result:(2)∑k=0pa−12kk≡pa3modp2.Guo [3] conjectured a q-analogue of (2), which was confirmed by Liu and Petrov [4] using a q-analogue of Sun–Zhao congruence on harmonic sums and a q-series identity. Guo and Zudilin [5] also gave the q-generalizations of (2).LetA,B∈ℤ. The Lucas sequence un=unA,Bn≥0 is given by(3)u0=0,u1=1,un+1=Aun−Bun−1n≥1.Ifp is an odd prime not dividing B, then it is known that p|up−Δ/p (see, e.g., [6]). For a nonzero integer n and a prime p, let νpn denote the p-adic valuation (or p-adic order) of n, i.e., νpn is the largest integer such that pνpn|n, especially νp0=+∞ and we define νpm/n=νpm−νpn for rational number m/n. For more developments on p-adic valuation, we refer the reader to the papers [7–10].In 2011, Sun [11] proved that for any nonzero integer m and odd prime p with p∤m, there holds(4)∑k=0p−12kk1mk≡Δp+up−Δ/pm−2,1modp2,where Δ=mm−4. As a common extension of (4), Sun [12] showed that(5)1pn∑k=0pn−12kk1mk−Δp∑k=0n−12kk1mk∈ℤp,and furthermore(6)1n∑k=0pn−12kk1mk−Δp∑k=0n−12kk1mk≡2nn2mn−1up−Δ/pm−2,1modp2.Letp be an odd prime and let m be an integer with p∤m. One can easily get the following formula:(7)∑k=0p−1p−1k−1k2kk/mk≡∑k=0p−12kk1mkmodp,since for any k∈1,2,…,p−1, we get p−1k≡−1kmodp. It looks like the left-hand side of (7) has some connection with the right-hand side. Motivated by (4) and (7), Sun [13] determined the sum ∑k=0pa−1hpa−1k2kk/mk modulo p2, where h is a p-adic integer and m∈ℤ with p∤m. For example, if h≡0modp and (((2h≡1modp) or pa>3), then(8)∑k=0pa−1hpa−1k2kk−h2k≡1−2hpa1+h4−2hp−1−1modp2.It is interesting to consider whether there exists the supercongruence as (8) modulo the higher powers of p in the case 2h−1≡0modp and pa>3. Sun [13] managed to investigate the above case and made the following conjecture. The first aim of this paper is to prove the conjectured results.Theorem 1.
Letp be an odd prime and let h∈ℤ with 2h−1≡0modp. If a∈ℤ+ and pa>3, then(9)∑k=0pa−1hpa−1k2kk−h2k≡0modpa+1.Also, for anyn∈ℤ+, we have(10)νp∑k=0n−1hn−1k2kk−h2k≥νpn.On the other hand, based on (5) and (7), Sun [12] conjectured the corresponding result with p∤m. The second aim of this paper is to show the following result.Theorem 2.
Letp be an odd prime and let Δ=mm−4. For any integer m≡0modp and positive integer n, we have(11)1pn∑k=0pn−1pn−1k2kk1−mk−Δp∑k=0n−1n−1k2kk1−mk∈ℤp.The remainder of the paper is organized as follows. In the next section, we give some lemmas. The proofs of Theorems1 and 2 will be given in Section 3.


## 2. Some Lemmas

In the following section, for an assertionA, we adopt the notation:(12)A=1,ifAholds,0,otherwise.We know thatm=n coincides with the Kronecker symbol δm,n.Lemma 1.
Letn,k,α be positive integers and p be a prime. Then,(13)pαn−1k≡pα−1n−1kp−1k−k/p1−npα∑j=1p∤jk1jmodp2α.Proof.
Note that(14)pαn−1k=∏j=1kpαn−jj=pα−1n−1kp∏j=1p∤jkpαn−jj≡pα−1n−1kp−1k−k/p1−∑j=1p∤jkpαnjmodp2α.
This proves (13). The congruence (13) is a result of Beukers [[14], Lemma 2].Lemma 2.
Letp be an odd prime. Then, for any integers a,b and positive integers r,s, we have(15)prapsbpr−1aps−1b≡1modpr+s+minr,s−δp,3.This lemma is a well-known congruence due to Osburn et al., see, e.g., [[15], (19)].The following curious result is due to Sun [16].Lemma 3 (see [16], Theorem 1]).
Letm∈ℤ and n∈ℤ+. Suppose that p is an odd prime dividing m−4. Then,(16)νp∑k=0n−12kk1mk≥νpn,νp∑k=0n−1n−1k−1k2kk1mk≥νpn.Furthermore,(17)1n∑k=0n−12kk1mk≡2n−1n−14n−1+δp,33|nm−432n3ν3n−1n3ν3n−1modpνpm−4,and also(18)1n∑k=0n−1n−1k−1k2kkmk≡Cn−14n−1modpνpm−4−δp,3,where Ck denotes the Catalan number 1/k+12kk=2kk−2kk+1. Thus, for a∈ℤ+, we have(19)1pa∑k=0pa−12kk1mk≡1+δp,3m−43≡m−13modp.Lemma 4.
Letp be an odd prime and let h∈ℤ with 2h−1≡0modp. Let l,α be nonnegative integers. If p≥5, then we have(20)∑k/pα=l2kkh2k≡pα4l2llmodpα+1.Ifp=3, then(21)∑k/pα=l2kkh2k≡0modpα.Ifp=3 and l≡1mod3, then(22)∑k/pα=l2kkh2k≡pα4l2ll+2α≥1pα−11−2hh2llmodpα+1.Proof.
Observe that(23)∑k/pα=l2kkh2k=∑k=0pαl+pα−12kkh2k−∑k=0pαl−12kkh2k.
Since2/h−4=21−2h/h≡0modp and l≡1mod3, by (17), we obtain that(24)∑h/pα=l2kkh2k≡pαl+pα2l+2pαl+1pα4pαl+pα−12+δp,33|pαl+pα21−2h3h2pαl+pα3ν3pαl+pα−1pαl+pα3ν3pαl+pα−1−pαl2pαlpαl14pαl−12+δp,33|pαl21−2h3h2pαl3ν3pαl−1pαl3ν3pαl−1modpα+1.
Ifp≥5, by (15) and (24), we get(25)∑k/pα=l2kkh2k≡pαl+12l+2l+14l2−pαl2ll4l−12=pα2l+12ll4l−2pαl2ll4l=pα2ll4lmodpα+1.
Thus, (20) is proved. The congruence (21) is easily deduced from (24).
Finally, we will prove (22). The congruence (22) is trivial when α=0. Now we may assume α≥1. With the help of (15), (17), and (24), for any nonnegative integer l with l≡1mod3, we have(26)∑k/3α=l2kkh2k≡3αl+12l+2l+14l2+21−2h3h2l+1l+1−3αl2ll4l−12+1−2h3h2ll≡3α4l2ll+2⋅3α−11−2h2llhmod3α+1.
This concludes the proof.Lemma 5.
Letp be an odd prime and l,s be nonnegative integers. Let Δ=mm−4. For any integer m≡0modp, we have(27)∑k=pslpsl+ps−12kk1mk−Δp∑k=ps−1lps−1l+ps−1−12kk1mk≡ps−12llml2−m2l+1up−Δpm−2,1modps+1.Proof.
The proof is very similar to (20). Clearly,(28)∑k=pslpsl+ps−12kk1mk=∑k=0psl+ps−12kk1mk−∑k=0psl−12kk1mk.
Substitutingn=ps−1l+1 and n=ps−1l in (6), we obtain (27).Lemma 6.
Letp be an odd prime and h∈ℤ with 2h−1≡0modp. Let a be a positive integer. Then,(29)∑k=0pa−12kkh2k∑j=1,p∤jk1j≡0modp,ifa≥2,2modp,ifa=1.Proof.
Note that(30)∑k=0pa−12kkh2k∑j=1,p∤jk1j=∑s=0pa−1−1∑t=0p−12ps+tps+th2ps+t∑j=1,p∤jps+t1j,∑j=1,p∤jps+t1j=∑j=1,p∤jps−11j+∑j=ps+1ps+t1j=∑j=1,p∤jps−11ps−j+∑j=1t1ps+j≡∑j=1t1jmodp.
With the help of Lucas’ theorem (cf. [17], p. 44]), it follows that(31)∑k=0pa−12kkh2k∑j=1,p∤jk1j≡∑s=0pa−1−1∑t=1p−1/22ss2tth2s+t∑j=1t1j+∑s=0pa−1−1∑t=p+1/2p−12s+1s2t−pth2s+t∑j=1t1j≡∑s=0pa−1−12ssh2s∑t=1p−1/22tth2t∑j=1t1jmodp.
Since2/h−4≡0modp, in view of (16), then(32)∑s=0pa−1−12ssh2s≡0modpa−1.
The congruence (29) with a≥2 holds by (31) and (32).
Now suppose thata=1. In fact, for k=0,…,p−1/2, we clearly have(33)2kk=−12k−4k≡p−12k−4kmodp.
Observing that2h≡1modp, we obtain(34)∑t=1p−1/22tth2t∑j=1t1j≡∑t=1p−1/2p−12t−2ht∑j=1t1j≡∑j=1p−1/2−1jj∑t=jp−1/2p−12t−1t−jmodp.
Recall the Chu–Vandermonde identity (see, e.g., [18], p. 169])(35)∑k=0nxkyn−k=x+yn.
Thus,(36)∑j=1p−1/2−1jj∑t=jp−1/2p−12t−1t−j=∑j=1p−1/2−1jjp−32j−1=2p−1∑j=1p−1/2p−12j−1j=−2p−1≡2modp.
Therefore, (29) with a=1 is proved by (31), (34), and (36).
By the above, we have completed the proof of Lemma6.


## 3. Proofs of Theorems1 and 2

Proof of Theorem 1.
We first prove (10). Let νpn=a. (10) is evidently trivial when a=0. Next, we suppose that a≥1. With the help of (13), we have(37)∑k=0n−1hn−1k2kk−h2k≡∑k=0n−1nph−1kp−1k−k/p2kk−h2k=∑l=0n/p−1nph−1l−1l∑k/p=l2kkh2kmodpa.
By (20) and (21), for any odd prime p, we get(38)∑k/p=l2kkh2k≡0modp.
Repeating the above processa−1 times, we obtain that(39)∑l=0n/p−1nph−1l−1l∑k/p=l2kkh2k≡∑m=0n/p2−1np2h−1m−1m∑l/p=m∑k/p=l2kkh2kby2.1≡∑l=0n/pa−1npah−1l−1l∑k/pa=l2kkh2k≡0modpa.
Let us turn to (9). We assume that p≥5. In view of (13), we obtain(40)∑k=0pa−1hpa−1k2kk−h2k≡∑k=0pa−1hpa−1−1kp−1k−k/p1−hpa∑j=1,p∤jk1j2kk−h2k=∑l=0pa−1−1hpa−1−1l−1l−1∑k/p=l2kkh2k1−hpa∑j=1,p∤jk1j+∑l=0pa−1−1∑k/p=l2kkh2k1−hpa∑j=1,p∤jk1jmodpa+1.
For any positive integera, we have(41)hpa−1−1l−1l−1≡0modp.
Therefore,(42)∑k=0pa−1hpa−1k2kk−h2k≡∑l=0pa−1−1hpa−1−1l−1l∑k/p=l2kkh2k−hpa∑l=0pa−1−1∑k/p=l2kkh2k∑j=1,p∤jk1jmodpa+1.
With the help of Lemma6, for a≥2, we get(43)∑l=0pa−1−1∑k/p=l2kkh2k∑j=1,p∤jk1j=∑k=0pa−12kkh2k∑j=1,p∤jk1j≡0modp.
Ifa=1, by (20), (29), and (42), then(44)∑k=0pa−1hpa−1k2kk−h2k≡∑k=0p−12kkh2k−2hp≡p1−2h≡0modp2.
The congruence (9) holds with a=1 and p≥5. If a≥2, combining (42) and (43), we have modulo pa+1,(45)∑k=0pa−1hpa−1k2kk−h2k≡∑l=0pa−1−1hpa−1−1l−1l∑k/p=l2kkh2k≡∑l=0pa−s−1hpa−s−1l−1l∑k/ps=l2kkh2k.
Repeating this processs−1 times, we have(46)∑l=0pa−s−1hpa−s−1l−1l∑k/ps=l2kkh2k≡∑m=0pa−s−1−1hpa−s−1−1m−1m∑l/p=m∑k/ps=l2kkh2k−hpa−s∑m=0pa−s−1−1∑l/p=m∑j=1,p∤jl1j∑k/ps=l2kkh2k.
By (20) and (29), we get(47)hpa−s∑l=1pa−s−1∑j=1,p∤jl1j∑k/ps=l2kkh2k≡hpa∑l=1pa−s−12ll4l∑j=1,p∤jl1j≡hpa∑l=1pa−s−12llh2l∑j=1,p∤jl1j≡2hpamodpa+1,ifs=a−1,0modpa+1,ifs<a−1.
At last, we only need to think about the cases=a−1. From (45)–(47), we have(48)∑k=0pa−1hpa−1k2kk−h2k≡∑k=0pa−12kkh2k−2hpa≡1−2hpa≡0modpa+1.
So, we obtain (9) with a≥2 and p≥5. When a≥2 and p=3, the proof of (9) is very similar to the case p≥5, only requiring a few additional discussions. Without loss of generality, it suffices to study the following sum in (46). Note that(49)h3a−s∑m=03a−s−1−1∑k/3s=m∑j=1,3∤jl1j∑k/3s=l2kkh2k=h3a−s∑m=03a−s−1−1∑l=3m3m+2∑j=1,3∤jl1j∑k/3s=l2kkh2k≡h3a−s∑m=03a−s−1−1∑j=1,3∤j3m+11j∑k/3s=3m+12kkh2k+∑j=1,3∤j3m+21j∑k/3s=3m+22kkh2kmod3a+1,∑j=1,3∤j3m+11j=∑j=1,3∤j3m−11j+13m+1≡1mod3,∑j=1,3∤j3m+21j≡1+12=32≡0mod3.
With the help of (22), we obtain(50)h3a−s∑m=03a−s−1−1∑l/3=m∑j=1,3∤jl1j∑k/3s=l2kkh2k≡h3a−s∑m=03a−s−1−1∑k/3s=3m+12kkh2k≡h3a−s∑m=03a−s−1−13s43m+16m+23m+1+2⋅3s−11−2hh6m+23m+1mod3a+1.
In light of Lucas’ theorem,(51)143m+16m+23m+1=26m+143m+13m+16m3m≡22mmmod3.
Recall that Strauss et al. [19] proved that for any positive integer n,(52)ν3∑k=0n−12kk=2ν3n+ν32nn.
Guo and Zeng [20] deduced a partial q-analogue of (52). Combining (50)–(52), we have modulo mod3a+1,(53)h3a−s∑m=03a−s−1−1∑l/3=m∑j=1,3∤jl1j∑k/3s=l2kkh2k≡22−h3a3∑m=03a−s−1−12mm≡0mod3a+1,ifs≤a−2,22−h3a3mod3a+1,ifs=a−1.
Since2h≡1≡−2mod3, then h≡2mod3. From (45), (46), and (53),(54)∑k=03a−1h3a−1k2kk−h2k≡∑k=03a−12kkh2k−22−h3a3≡2−h3a3h−22−h3a3by2.5=1−2h2−h3a3h≡0mod3a+1.
Proofof Theorem 1 is now complete.Proof of Theorem 2.
Letn=pa−1d with p∤d and a≥1. The proof of (11) is very similar to (10). With the help of (13), for any integer s∈0,a−1 and nonnegative integer k, now we have(55)−1kpa−sd−1k−−1k/ppa−s−1d−1kp∈pa−sdℤp.
Therefore,(56)1d∑k=0pad−1pad−1k2kk1−mk−Δp∑k=0pa−1d−1pa−1d−1k2kk1−mk≡1d∑l=0pa−1d−1pa−1d−1l−1l∑k/p=l2kk1mk−Δp2llmlmodpa.
In light of Lemma5, we have(57)1d∑l=0pa−1d−1pa−1d−1l−1l∑k/p=l2kk1mk−Δp2llml=1d∑l=0pa−1d−1pa−1d−1l−1l−pa−2d−1lp−1l/p∑k/p=l2kk1mk−Δp2llml+1d∑l=0pa−2d−1pa−2d−1l−1l∑k=p2lp2l+p2−12kk1mk−Δp∑k=plpl+p−12llml.
By Lemma5 and (55), we get(58)1d∑l=0pa−1d−1pa−1d−1l−1l∑k/p=l2kk1mk−Δp2llml≡1d∑l=0pa−2d−1pa−2d−1l−1l∑k=p2lp2l+p2−12kk1mk−Δp∑k=plpl+p−12llml≡1d∑l=0d−1d−1l−1l∑k=palpal+pa−12kk1mk−Δp∑k=pa−1lpa−1l+pa−1−12llml≡0modpa,where the last result comes from Lemma 5.
In view of the above, we have completed the Proofof Theorem 2.



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