# Polynomial Reduction and Super Congruences

Based on a reduction processing, we rewrite a hypergeometric term as the sum of the difference of a hypergeometric term and a reduced hypergeometric term (the reduced part, in short). We show that when the initial hypergeometric term has a certain kind of symmetry, the reduced part contains only odd or even powers. As applications, we derived two infinite families of super-congruences.

## Authors

• 4 publications
• 1 publication
• 6 publications
12/23/2014

### Symmetry in Image Registration and Deformation Modeling

We survey the role of symmetry in diffeomorphic registration of landmark...
08/10/2018

### Model Reduction with Memory and the Machine Learning of Dynamical Systems

The well-known Mori-Zwanzig theory tells us that model reduction leads t...
11/16/2020

### Polynomial-time Tests for Difference Terms in Idempotent Varieties

We consider the following practical question: given a finite algebra A i...
08/30/2021

### A New Rational Approach to the Square Root of 5

In this paper, authors construct a new type of sequence which is named a...
11/09/2017

### On First-order Cons-free Term Rewriting and PTIME

In this paper, we prove that (first-order) cons-free term rewriting with...
01/17/2022

### Effective error estimation for model reduction with inhomogeneous initial conditions

A priori error bounds have been derived for different balancing-related ...
03/27/2020

### A squarefree term not occurring in the Leech sequence

Let < a r r a y > The Leech sequence L is the squarefree sequence ...
##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 Introduction

In recent years, many super congruences involving combinatorial sequences are discovered, see for example, Sun [16]. The standard methods for proving these congruences include combinatorial identities [18], Gauss sums [5], symbolic computation [14] et al.

We are interested in the following super congruence conjectured by van Hamme [19]

 p−12∑k=0(−1)k(4k+1)((1/2)k(1)k)3≡(−1)p−12p(modp3),

where is a odd prime and is the rising factorial. This congruence was proved by Mortenson [13] Zudilin [21] and Long [12] by different methods. Sun [17] proved a stronger version for prime

 p−12∑k=0(−1)k(4k+1)((1/2)k(1)k)3≡(−1)p−12p+p3Ep−3(modp4),

where is the -th Euler number defined by

 2ex+e−x=∞∑n=0Enxnn!.

A similar congruence was given by van Hamme [19] for :

 p−12∑k=0(4k+1)((1/2)k(1)k)4≡p(modp3).

Long [12] showed that in fact the above congruence holds for arbitrary odd prime modulo . Motivated by these two congruences, Guo [8] proposed the following conjectures (corrected version).

###### Conjecture 1.1
• For any odd prime , positive integer and odd integer , there exists an integer such that

 pr−12∑k=0(−1)k(4k+1)m((1/2)k(1)k)3≡am,ppr(−1)(p−1)r2(modpr+2). (1.1)
• For any odd prime , positive integer and odd integer , there exists an integer such that

 pr−12∑k=0(4k+1)m((1/2)k(1)k)4≡bm,ppr(modpr+3). (1.2)

Liu [11] and Wang [20] confirmed the conjectures for and some initial values . Jana and Kalita [10] and Guo [9] confirmed (1.1) for and . We will prove a stronger version of (1.1) for the case of and arbitrary odd and a weaker version of (1.2) for the case of and arbitrary odd by a reduction process.

Recall that a hypergeometric term is a function of such that is a rational function of . Our basic idea is to rewrite the product of a polynomial in and a hypergeometric term as

 f(k)tk=Δ(g(k)tk)+h(k)tk=(g(k+1)tk+1−g(k)tk)+h(k)tk,

where are polynomials in such that the degree of is bounded. To this aim, we construct such that equals the product of and a polynomial and that and has the same leading term. Then we have

 f(k)tk−Δx(k)tk=(f(k)−u(k))tk

is the product of and a polynomial of degree less than . We call such a reduction process one reduction step. Continuing this reduction process, we finally obtain a polynomial with bounded degree. We will show that for , and an arbitrary polynomial of form with odd, the reduced polynomial can be taken as . This enables us to reduce the congruences (1.1) and (1.2) to the special case of , which is known for .

We notice that Pirastu-Strehl [15] and Abramov [1, 2] gave the minimal decomposition when is a rational function, Abramov-Petkovšek [3, 4] gave the minimal decomposition when is a hypergeometric term, and Chen-Huang-Kauers-Li [6] applied the reduction to give an efficient creative telescoping algorithm. These algorithms concern a general hypergeometric term. While we focus on a kind of special hypergeometric term so that the reduced part has a nice form.

The paper is organized as follows. In Section 2, we consider the reduction process for a general hypergeometric term . Then in Section 3 we consider those with the property is a shift of , where . As an application, we prove a stronger version of (1.1) for the case . Finally, we consider the case of is a shift of , which corresponds to (1.2). In this case, we show that there is a rational number instead of an integer such that (1.2) holds when .

## 2 The Difference Space and Polynomial Reduction

Let be a field and be the ring of polynomials in with coefficients in . Let be a hypergeometric term. Suppose that

 tk+1tk=a(k)b(k),

where . It is straightforward to verify that

 Δk(b(k−1)x(k)tk)=(a(k)x(k+1)−b(k−1)x(k))tk. (2.1)

We thus define the difference space corresponding to and to be

 Sa,b={a(k)x(k+1)−b(k−1)x(k):x(k)∈K[k]}.

We see that for , we have for a certain polynomial .

Let denote the set of nonnegative integers and the set of integers, respectively. Given , we denote

 u(k)=a(k)−b(k−1), (2.2)
 d=max{degu(k),dega(k)−1}, (2.3)

and

 m0=−lcu(k)/lca(k), (2.4)

where denotes the leading coefficient of .

We first introduce the concept of degeneration.

###### Definition 2.1

Let and be given by (2.2) and (2.4). If

 degu(k)=dega(k)−1andm0∈N,

we say that the pair is degenerated.

We will see that the degeneration is closely related to the degrees of the elements in .

###### Lemma 2.2

Let and be given by (2.3) and (2.4). For any polynomial , let

 p(k)=a(k)x(k+1)−b(k−1)x(k).

If is degenerated and , then ; Otherwise, .

Proof. Notice that

 p(k)=u(k)x(k)+a(k)(x(k+1)−x(k)).

If the leading terms of and do not cancel, the degree of is . Otherwise, we have and

 lcu(k)+lca(k)⋅degx(k)=0,

i.e., .

It is clear that is a subspace of , but is not a sub-ring of in general. Let denote the coset of a polynomial . We see that the quotient space is finite dimensional.

###### Theorem 2.3

Let and be given by (2.3) and (2.4). We have

 K[k]/Sa,b={⟨[k0],[k1],…,[kd−1],[kd+m0]⟩,if (a(k),b(k)) is degenerated,⟨[k0],[k1],…,[kd−1]⟩,otherwise.

Proof. For any nonnegative integer , let

 ps(k)=a(k)(k+1)s−b(k−1)ks.

We first consider the case when the pair is not degenerated. By Lemma 2.2, we have

 degps(k)=d+s,∀s≥0.

Suppose that is a polynomial of degree . Then

 p′(k)=p(k)−lcp(k)lcpm−d(k)pm−d(k) (2.5)

is a polynomial of degree less than and . By induction on , we derive that for any polynomial of degree , there exists a polynomial of degree such that . Therefore,

 K[k]/Sa,b=⟨[k0],[k1],…,[kd−1]⟩.

Now assume that is degenerated. By Lemma 2.2,

 degps(k)=d+s,∀s≠m0anddegpm0(k)

The above reduction process (2.5) works well except for the polynomials of degree . But in this case,

 p(k)−lcp(k)⋅kd+m0

is a polynomial of degree less than . Then the reduction process continues until the degree is less than . We thus derive that

 K[k]/Sa,b=⟨[k0],[k1],…,[kd−1],[kd+m0]⟩,

completing the proof.

###### Example 2.1

Let be a positive integer and

 tk=(−n)k/k!,

where is the raising factorial. Then

 a(k)=k−n,b(k)=k+1,

and

 Sa,b={(k−n)⋅x(k+1)−k⋅x(k):x(k)∈K[k]}.

We have

 K[k]/Sa,b=⟨[kn]⟩

is of dimension one.

## 3 The case when a(k)=−b(k+α)

In this section, we consider the case when and has a symmetric property. We will show that in this case, the reduction process maintains the symmetric property. Notice that in this case

 u(k)=a(k)−b(k−1)=−b(k+α)−b(k−1)

has the same degree as , the pair is not degenerated.

We first consider the relation between the symmetric property and the expansion of a polynomial.

###### Lemma 3.1

Let and . Then the following two statements are equivalent.

• (, respectively).

• is the linear combination of (, respectively).

Proof. Suppose that

 p(β+k)=∑iciki.

Then

 p(β−k)=∑ici(−k)i.

Therefore,

 p(β+k)=p(β−k)⟺c2i+1=0, i=0,1,….

The case of can be proved in a similar way.

Now we are ready to state the main theorem.

###### Theorem 3.2

Let such that

 a(k)=−b(k+α)andb(β+k)=±b(β−k),

for some . Then for any non-negative integer , we have

 [(k+γ)2m]∈⟨[(k+γ)2i]:0≤2i

and

 [(k+γ)2m+1]∈⟨[(k+γ)2i+1]:0≤2i+1

where

 γ=−β+α−12. (3.1)

Proof. We only prove the case of . The case of can be proved in a similar way. By Lemma 3.1, we may assume that

 b(k)=br(k−β)r+br−2(k−β)r−2+⋯+b0,

where is even and are the coefficients.

Since is not degenerated, taking

 x(k)=xs(k)=−12(k+γ−12)s (3.2)

in Lemma 2.2, we derive that

 ps(k)=a(k)xs(k+1)−b(k−1)xs(k) (3.3)

is a polynomial of degree . More explicitly, we have

 ps(k)=12(b(k+α)(k+γ+12)s+b(k−1)(k+γ−12)s)

is a polynomial with leading term .

Notice that

 ps(−γ+k)=12(b(k+α−γ)(k+12)s+b(k−γ−1)(k−12)s)

and

 ps(−γ−k) =12(b(−k+α−γ)(−k+12)s+b(−k−γ−1)(−k−12)s) =(−1)s2(b(−k+α−γ)(k−12)s+b(−k−γ−1)(k+12)s).

Since , i.e., , we deduce that

 ps(−γ−k) =(−1)s2(b(2β+k−α+γ)(k−12)s+b(2β+k+γ+1)(k+12)s).

By the relation (3.1), we derive that

 ps(−γ−k)=(−1)sps(−γ+k).

Suppose that is a linear combination of the even powers of and . By Lemma 3.1, we have and thus

 p′(k)=p(k)−lcp(k)br⋅pdegp(k)−r(k)

also satisfies since and are both even. It is clear that and the degree of is less than the degree of . Continuing this reduction process, we finally derive that for some polynomial with degree and satisfying . Therefore,

 [p(k)]∈⟨[(k+γ)2i]:0≤2i

Suppose that is a linear combination of the odd powers of and . Then we have and thus

 p′(k)=p(k)−lcp(k)br⋅pdegp(k)−r(k)

also satisfies . Continuing this reduction process, we finally derive that

 [p(k)]∈⟨[(k+γ)2i+1]:0≤2i+1

This completes the proof.

We may further require to express as an integral linear combination of when .

###### Theorem 3.3

Let

 tk=(−1)k((α)kk!)r,

where is a positive integer and is a rational number with denominator . Then for any positive integer , there exist integers and a polynomial such that

 (2Dk+Dα)mtk=r−1∑i=0ai(2Dk+Dα)itk+Δk(2r−1(Dk)rx(2Dk)tk).

Moreover, if .

Proof. We have

 tk+1tk=−(k+α)r(k+1)r.

Let

 a(k)=−(k+α)randb(k)=(k+1)r.

We see that it is the case of and of Theorem 3.2. From (2.1), we derive that

 Δk(krxs(k)tk)=ps(k)tk, (3.4)

where and are given by (3.2) and (3.3) respectively. Multiplying on both sides, we obtain

 Δk(2r−1(Dk)r~xs(2Dk)tk)=~ps(k′)tk, (3.5)

where ,

 ~xs(k)=−(k+Dα−D)s, (3.6)

and

 ~ps(k)=12((k+Dα)r(k+D)s+(k−Dα)r(k−D)s). (3.7)

Notice that and is a monic polynomial of degree . Moreover, contains only even powers of or only odd powers of . Using to do the reduction (2.5), we derive that there exists integers such that

 p(k)=km−cm~pm−r(k)−cm−2~pm−r−2(k)−⋯

becomes a polynomial of degree less than . Clearly, . Replacing by and multiplying , we derive that

 (k′)mtk=p(k′)tk+Δk(2r−1(Dk)r(cm~xm−r(2Dk)+cm−2~xm−r−2(2Dk)+⋯)tk),

completing the proof.

As an application, we confirm Conjecture 6 of [11].

###### Theorem 3.4

Let

 Sm=p−12∑k=0(−1)k(4k+1)m((1/2)k(1)k)3.

For any positive odd integer , there exist integers and such that

 Sm≡am(p(−1)p−12+p3Ep−3)+p3cm(modp4)

holds for any prime .

Proof. Taking and in Theorem 3.3, there exist an integer and a polynomial such that

 (4k+1)mtk−am(4k+1)tk=Δk(32k3qm(4k)tk),

where . Summing over from to , we derive that

 Sm−amS1=32ω3qm(4ω)(−1)ω((1/2)ω(1)ω)3,

where . Noting that

 (1/2)ω(1)ω=p1p+1p−12∏i=12i−12i

and

 1p+1p−12∏i=12i−12i=1p+1p−12∏i=1p−2i2i≡(−1)p−12(modp),

we have

 ((1/2)ω(1)ω)3≡p3(−1)p−12(modp4).

Hence

 Sm−amS1≡−32p3ω3qm(4ω)(modp4)

Let . We then have

 Sm≡amS1+p3cm(modp4).

Sun [17] proved that for any prime ,

 S1≡(−1)p−12p+p3Ep−3(modp4).

Therefore,

 Sm≡am(p(−1)p−12+p3Ep−3)+p3cm(modp4).

Remark 1. The coefficient and the polynomial can be computed by the extended Zeilberger’s algorithm [7].

## 4 The case when a(k)=b(k+α)

We first give a criterion on the degeneration of .

###### Lemma 4.1

Let such that . Suppose that . Then is not degenerated.

Proof. Let and

 u(k)=a(k)−b(k−1)=b(k+α)−b(k−1).

It is clear that the coefficient of in is and the coefficient of in is . Since , we derive that . Thus,

 −lcu(k)/lca(k)=−lcu(k)/lcb(k)=−(α+1)r.

Since , the pair is not degenerated.

When is a shift of , we have a result similar to Theorem 3.2.

###### Theorem 4.2

Let such that

 a(k)=b(k+α)andb(β+k)=±b(β−k),

for some . Assume further that . Then for any non-negative integer , we have

 (k+γ)2m∈⟨[(k+γ)2i]:0≤2i

and

 (k+γ)2m+1∈⟨[(k+γ)2i+1]:0≤2i+1

where

 γ=−β+α−12.

Proof. The proof is parallel to the proof of Theorem 3.2. Instead of (3.2), we take

 x(k)=xs(k)=(k+γ−12)s

in Lemma 2.2. By Lemma 4.1, is not degenerated and

 deg(a(k)−b(k−1))=dega(k)−1.

Hence the polynomial

 ps(k)=a(k)xs(k+1)−b(k−1)xs(k)

satisfies

 degps(k)=s+dega(k)−1.

Moreover, we have

 ps(−γ−k)=(−1)s+1ps(−γ+k),

so that the reduction process maintains the symmetric property. Therefore, the reduction process continues until the degree is less than .

Similar to Theorem 3.3, we have the following result.

###### Theorem 4.3

Let

 tk=((α)kk!)r,

where is a positive integer and is a rational number with denominator . Suppose that . Then for any positive integer , there exist integers and a polynomial such that

 (2Dk+Dα)mtk=1Cmr−2∑i=0ai(2Dk+Dα)itk+1CmΔk(2r−1(Dk)rx(2Dk)tk),

where

 Cm=∏0≤2i≤m−r+1((αr+m−r+1−2i)⋅D).

Moreover, if .

Proof. The proof is parallel to the proof of Theorem 3.3. Instead of (3.6) and (3.7), we take

 ~xs(k)=(k+Dα−D)s (4.1)

and

 ~ps(k)=12((k+Dα)r(k+D)s−(k−Dα)r(k−D)s), (4.2)

so that (3.5) still holds. It is clear that . But in this case, is not monic. The leading term of is

 (αr+s)D⋅ks+r−1.

Now let us consider the reduction process. Let be a polynomial of degree . Assume further that contains only even powers of or only odd powers of . Setting

 p′(k) =lc~pℓ−r+1(k)⋅p(k)−lcp(k)⋅~pℓ−r+1(k) =(αr+ℓ−r+1)D⋅p(k)−lcp(k)⋅~pℓ−r+1(k),

we see that and . Since contains only even powers of or only odd powers of , so does . Therefore, .

Continuing this reduction process until , we finally obtain that there exist integers such that

 Cmkm−cm~pm−r+1(k)−cm−2~pm−r−1(k)−⋯,

is a polynomial of degree less than and with integral coefficients, where is the product of the leading coefficient of

 Cm=∏0≤2i≤m−r+1((αr+m−r+1−2i)D),

as desired.

For the special case of , we may further reduce the factor .

###### Lemma 4.4

Let be a positive integer and

 tk=(1/2)4k(1)4k.
• If is odd, then there exist an integer and a polynomial such that

 (4k+1)mtk=cC′m(4k+1)tk+1C′mΔk(32k4x(4k)tk),

where .

• If is even, then there exist integers and a polynomial