Two Properties of

Catalan-Larcombe-French Numbers

Xiao-Juan Ji School of Mathematical Sciences Soochow University Suzhou Jiangsu 215006 P. R. China

               xiaojuanji2015@yahoo.com

Zhi-Hong Sun School of Mathematical Sciences Huaiyin Normal University Huaian Jiangsu 223001 P. R. China

                 zhihongsun@yahoo.com

Abstract

Let (P_{n}) be the Catalan-Larcombe-French numbers. The numbers P_{n} occur in the theory of elliptic integrals, and are related to the arithmetic-geometric-mean. In this paper we investigate the properties of the related sequence S_{n} = P_{n}/2^{n} instead, since S_{n} is an Apéry-like sequence. We prove a congruence and an inequality for P_{n}.

1 Introduction

Let (P_{n}) be the sequence given by

P_{0} = 1, P_{1} = 8 and (n + 1)^{2}P_{n+1} = 8(3n^{2} + 3n + 1)P_{n} − 128n^{2}P_{n−1} (n ≥ 1). (1)

The numbers P_{n} are called Catalan-Larcombe-French numbers since Catalan first defined P_{n} in [1], and Larcombe and French [6] proved that

P_{n} = 2^{n} ∑_{k=0}^{⌊n/2⌋}(−4)^{k}\binom{2n − 2k}{n − k}^{2}\binom{n − k}{k} = ∑_{k=0}^{n} \begin{matrix}\binom{2k}{k} & 2 & \binom{2n−2k}{n−k} & 2 \\ ( & {n \atop k} & ) \end{matrix} ,

where ⌊x⌋ is the greatest integer not exceeding x. The numbers P_{n} are related to the arithmetic-geometric-mean. See [6] and \underline{A053175} in Sloane’s “On-Line Encyclopedia of Integer Sequences”.

Let (S_{n}) be defined by

S_{0} = 1, S_{1} = 4 and (n + 1)^{2}S_{n+1} = 4(3n^{2} + 3n + 1)S_{n} − 32n^{2}S_{n−1} (n ≥ 1). (2)

Comparing (2) with (1), we see that

S_{n} = \frac{P_{n}}{2^{n}} .

Zagier noted that

S_{n} = ∑_{k=0}^{⌊n/2⌋} \binom{2k}{k}^{2}\binom{n}{2k}4^{n−2k}.

As observed by Jovović [7] in 2003 ,

S_{n} = ∑_{k=0}^{n} \binom{n}{k}\binom{2k}{k}\binom{2n − 2k}{n − k} (n = 0, 1, 2, …).

Recently Z. W. Sun stated that

S_{n} = ∑_{k=0}^{n} \binom{2k}{k}^{2}\binom{k}{n − k}(−4)^{n−k} = \frac{1}{(−2)^{n}} ∑_{k=0}^{n} \binom{2k}{k}\binom{2n − 2k}{n − k}\binom{k}{n − k}(−4)^{k}.

The first few values of S_{n} are shown below:

S_{0} = 1, S_{1} = 4, S_{2} = 20, S_{3} = 112, S_{4} = 676, S_{5} = 4304, S_{6} = 28496,
S_{7} = 194240, S_{8} = 1353508, S_{9} = 9593104, S_{10} = 68906320,
S_{11} = 500281280, S_{12} = 3664176400, S_{13} = 27033720640.

Let p be an odd prime. Jarvis, Larcombe, and French [3] proved that if n = a_{r}p^{r} + ⋯ + a_{1}p + a_{0} with a_{0}, a_{1}, …, a_{r} ∈ {0, 1, …, p − 1}, then

P_{n} ≡ P_{a_{r}} ⋯ P_{a_{1}}P_{a_{0}} (mod p).

Jarvis and Verrill [5] showed that

P_{n} ≡ (−1)^{\frac{p−1}{2}} 128^{n}P_{p−1−n} (mod p) for n = 0, 1, …, p − 1

and

P_{mp^{r}} ≡ P_{mp^{r−1}} (mod p^{r}) for m, r ∈ ℤ^{+},

where ℤ^{+} is the set of positive integers.

For a prime p let ℤ_{p} denote the set of those rational numbers whose denominator is not divisible by p. Let p be an odd prime, n ∈ ℤ_{p} and n ≢ 0, −16 (mod p). The second author [11] proved that

∑_{k=0}^{p−1} \binom{2k}{k}\frac{S_{k}}{(n + 16)^{k}} ≡ (\frac{n(n + 16)}{p}) ∑_{k=0}^{p−1} \frac{\binom{2k}{k}2\binom{4k}{2k}}{n^{2k}} (mod p),

where (\frac{a}{p}) is the Legendre symbol.

In 1894 Franel [2] introduced the following Franel numbers (f_{n}):

f_{n} = ∑_{k=0}^{n} \binom{n}{k}^{3} (n = 0, 1, 2, …).

The first few Franel numbers are as below:

f_{0} = 1, f_{1} = 2, f_{3} = 10, f_{4} = 56, f_{5} = 346, f_{6} = 2252, f_{7} = 15184.

Franel [2] noted that the sequence (f_{n}) satisfies the recurrence relation:

(n + 1)^{2}f_{n+1} = (7n^{2} + 7n + 2)f_{n} + 8n^{2}f_{n−1} (n ≥ 1).

Let r ∈ ℤ^{+} and p be a prime with p ≡ 5, 7 (mod 8). The second author [11] conjectured that

S_{\frac{p^{r}−1}{2}} ≡ 0 (mod p^{r}) and f_{\frac{p^{r}−1}{2}} ≡ 0 (mod p^{r}).  (3)

In this paper we prove (3) in the case r = 2. We also prove the second author’s conjecture [11]:

(1 + \frac{1}{m(m − 1)})S_{m}^{2} > S_{m+1}S_{m−1} for m = 2, 3, ….

2 Basic lemmas

Lemma 1 (Lucas’ theorem [8]). Let p be an odd prime. Suppose a = a_{r}p^{r} + ⋯ + a_{1}p + a_{0} and b = b_{r}p^{r} + ⋯ + b_{1}p + b_{0}, where a_{r}, …, a_{0}, b_{r}, …, b_{0} ∈ {0, 1, …, p − 1}. Then

\binom{a}{b} ≡ \binom{a_{r}}{b_{r}} ⋯ \binom{a_{0}}{b_{0}} (mod p).

Lucas’ theorem is often formulated as follows.

Lemma 2 ([8]). Let p be an odd prime and a, b ∈ ℤ^{+}. Suppose a_{0}, b_{0} ∈ {0, 1, …, p − 1}. Then

\binom{ap + a_{0}}{bp + b_{0}} ≡ \binom{a}{b}\binom{a_{0}}{b_{0}} (mod p).

Lemma 3 ([4, Lemma 2.7]). For any positive integer n we have

S_{n} = 2 ∑_{k=1}^{n} \binom{n − 1}{k − 1}\binom{2k}{k}\binom{2n − 2k}{n − k}.

Lemma 4 ([9]). Let p be an odd prime. Suppose n = n_{1}p + n_{0} and k = k_{1}p + k_{0} with k_{1}, n_{1} ∈ ℤ^{+} and k_{0}, n_{0} ∈ {0, 1, …, p − 1}. Then

\binom{n}{k} ≡ \binom{n_{1}}{k_{1}}((1 + n_{1})\binom{n_{0}}{k_{0}} − (n_{1} + k_{1})\binom{n_{0} − p}{k_{0}} − k_{1}\binom{n_{0} − p}{k_{0} + p}) (mod p^{2}).

Lemma 5. Let p be an odd prime. Then

∑_{t=0}^{(p−1)/2}(−1)^{t}(\binom{\frac{p−1}{2} + t}{t} − \binom{p + \frac{p−1}{2} + t}{p + t})\binom{−\frac{1}{2}}{t}^{2} ≡ 0 (mod p^{2}).

Proof. For 0 ≤ t ≤ (p − 1)/2, from Lemma 2 we have

\binom{\frac{p−1}{2} − p}{p + t} = (−1)^{t+1}\binom{p + \frac{p+1}{2} + t − 1}{p + t} ≡ (−1)^{t+1}\binom{\frac{p+1}{2} + t − 1}{t}

= −\binom{\frac{p−1}{2} − p}{t} (mod p)

and so

\binom{\frac{p−1}{2} − p}{t} + \binom{\frac{p−1}{2} − p}{p + t} = (−1)^{t}(\binom{\frac{p−1}{2} + t}{t} − \binom{p + \frac{p−1}{2} + t}{p + t}) ≡ 0 (mod p).

We first assume p ≡ 1 (mod 4). Applying Lemma 4 we get

(\frac{3(p−1)}{4} ; \frac{p−1}{4}) − \binom{p + \frac{3(p−1)}{4}}{p + \frac{p−1}{4}} ≡ (\frac{3(p−1)}{4} ; \frac{p−1}{2}) − (2(\frac{3(p−1)}{4} ; \frac{p−1}{2}) − \binom{\frac{3(p−1)}{4} − p}{\frac{p−1}{2}})

= −(\frac{3(p−1)}{4} ; \frac{p−1}{2}) + (−1)^{\frac{p−1}{2}} (\frac{3(p−1)}{4} ; \frac{p−1}{2}) = 0 (mod p^{2})

and

\binom{\frac{p−1}{2} + t}{t} − \binom{p + \frac{p−1}{2} + t}{p + t} + \binom{p − 1 − t}{\frac{p−1}{2} − t} − \binom{p + p − 1 − t}{p + \frac{p−1}{2} − t}

≡ −\binom{\frac{p−1}{2} + t}{\frac{p−1}{2}} + \binom{\frac{p−1}{2} − p + t}{\frac{p−1}{2}} − \binom{p − 1 − t}{\frac{p−1}{2}} + \binom{−1 − t}{\frac{p−1}{2}}

= ((−1)^{\frac{p−1}{2}} − 1)\binom{\frac{p−1}{2} + t}{\frac{p−1}{2}} + ((−1)^{\frac{p−1}{2}} − 1)\binom{p − 1 − t}{\frac{p−1}{2}}

= 0 (mod p^{2}).

Also,

(−1)^{t}\binom{−\frac{1}{2}}{t}^{2} − (−1)^{\frac{p−1}{2} −t}\binom{−\frac{1}{2}}{\frac{p−1}{2} − t}^{2} ≡ (−1)^{t}(\binom{\frac{p−1}{2}}{t}^{2} − \binom{\frac{p−1}{2}}{\frac{p−1}{2} − t}^{2}) = 0 (mod p).

By the above four congruences, we have

∑_{t=0}^{(p−1)/2}(−1)^{t}(\binom{\frac{p−1}{2} + t}{t} − \binom{p + \frac{p−1}{2} + t}{p + t})\binom{−\frac{1}{2}}{t}^{2}

= ∑_{t=0}^{(p−5)/4}(−1)^{t}\binom{−\frac{1}{2}}{t}^{2}(\binom{\frac{p−1}{2} + t}{t} − \binom{p + \frac{p−1}{2} + t}{p + t})

+ (−1)^{\frac{p−1}{4}} \binom{−\frac{1}{2}}{\frac{p−1}{4}}^{2}((\frac{3(p−1)}{4} ; \frac{p−1}{4}) − \binom{p + \frac{3(p−1)}{4}}{p + \frac{p−1}{4}})

+ ∑_{t=0}^{(p−5)/4}(−1)^{\frac{p−1}{2} −t}\binom{−\frac{1}{2}}{\frac{p−1}{2} − t}^{2}(\binom{p − 1 − t}{\frac{p−1}{2} − t} − \binom{p + p − 1 − t}{p + \frac{p−1}{2} − t})

≡ ∑_{t=0}^{(p−5)/4} ((−1)^{t}\binom{−\frac{1}{2}}{t}^{2} − (−1)^{\frac{p−1}{2} −t}\binom{−\frac{1}{2}}{\frac{p−1}{2} − t}^{2})(\binom{\frac{p−1}{2} + t}{t} − \binom{p + \frac{p−1}{2} + t}{p + t})

+ (−1)^{\frac{p−1}{4}} \binom{−\frac{1}{2}}{\frac{p−1}{4}}^{2}((\frac{3(p−1)}{4} ; \frac{p−1}{4}) − \binom{p + \frac{3(p−1)}{4}}{p + \frac{p−1}{4}}) ≡ 0 (mod p^{2}).

Thus the result is true for p ≡ 1 (mod 4).

Now we assume p ≡ 3 (mod 4). By Lemma 4,

\binom{\frac{p−1}{2} + t}{t} − \binom{p + \frac{p−1}{2} + t}{p + t} ≡ −(\binom{\frac{p−1}{2} + t}{\frac{p−1}{2}} + \binom{p − 1 − t}{\frac{p−1}{2}}) (mod p^{2}).

As

\binom{\frac{p−1}{2} + t}{\frac{p−1}{2}} + \binom{p − 1 − t}{\frac{p−1}{2}} ≡ \binom{\frac{p−1}{2} + t}{\frac{p−1}{2}} + \binom{−1 − t}{\frac{p−1}{2}}

= \binom{\frac{p−1}{2} + t}{\frac{p−1}{2}} + (−1)^{\frac{p−1}{2}} \binom{t + \frac{p−1}{2}}{\frac{p−1}{2}} = 0 (mod p)

and

(−1)^{t}\binom{−\frac{1}{2}}{t}^{2} + (−1)^{\frac{p−1}{2} −t}\binom{−\frac{1}{2}}{\frac{p−1}{2} − t}^{2} ≡ (−1)^{t}(\binom{\frac{p−1}{2}}{t}^{2} − \binom{\frac{p−1}{2}}{\frac{p−1}{2} − t}^{2}) = 0 (mod p),

we obtain

∑_{t=0}^{(p−1)/2}(−1)^{t}(\binom{\frac{p−1}{2} + t}{t} − \binom{p + \frac{p−1}{2} + t}{p + t})\binom{−\frac{1}{2}}{t}^{2}

≡ − ∑_{t=0}^{(p−1)/2}(−1)^{t}(\binom{\frac{p−1}{2} + t}{\frac{p−1}{2}} + \binom{p − 1 − t}{\frac{p−1}{2}})\binom{−\frac{1}{2}}{t}^{2}

= − ∑_{t=0}^{(p−3)/4} (\binom{\frac{p−1}{2} + t}{\frac{p−1}{2}} + \binom{p − 1 − t}{\frac{p−1}{2}})((−1)^{t}\binom{−\frac{1}{2}}{t}^{2} + (−1)^{\frac{p−1}{2} −t}\binom{−\frac{1}{2}}{\frac{p−1}{2} − t}^{2})

≡ 0 (mod p^{2}). Hence the result is also true in this case. The proof is now complete. Lemma 6 ([10, Theorem 3.3]). Let p be a prime with p ≡ 5, 7 (mod 8). Then

∑_{k=0}^{\frac{p−1}{2}} \frac{\binom{2k}{k}3}{(−64)^{k}} ≡ 0 (mod p^{2}).

Lemma 7 ([4, Lemma 2.8]). Let m ∈ ℤ and k, p ∈ ℤ^{+}. Then

\binom{mp^{r} − 1}{k} = (−1)^{k−⌊ \frac{k}{p} ⌋}\binom{mp^{r−1} − 1}{⌊k/p⌋} ∏_{i=1, p∤i}^{k} (1 − \frac{mp^{r}}{i} ).

3 Congruences for S_{\frac{p^{2}−1}{2}} and f_{\frac{p^{2}−1}{2}} (mod p^{2})

Theorem 8. Let p be a prime with p ≡ 5, 7 (mod 8). Then

S_{\frac{p^{2}−1}{2}} ≡ f_{\frac{p^{2}−1}{2}} ≡ 0 (mod p^{2}).

Moreover,

S_{\frac{p^{2}−1}{2}} ≡ f_{\frac{p^{2}−1}{2}} (mod p^{3}).

Proof. For \frac{p−1}{2} < t < p and 0 ≤ s ≤ \frac{p−1}{2}, from Lemma 2 we see that

\binom{p^{2} − 1}{\frac{p^{2}−1}{2}} ≡ \binom{p − 1}{\frac{p−1}{2}}^{2} ≡ 1 (mod p),

\binom{\frac{p−1}{2}p + \frac{p−1}{2}}{sp + t} ≡ \binom{\frac{p−1}{2}}{s}\binom{\frac{p−1}{2}}{t} = 0 (mod p),

\binom{2sp + 2t}{sp + t} = \binom{(2s + 1)p + 2t − p}{sp + t} ≡ \binom{2s + 1}{s}\binom{2t − p}{t} = 0 (mod p)

and

\binom{p^{2} − 1 − 2sp − 2t}{\frac{p^{2}−1−2sp−2t}{2}} = \binom{(p − 2s − 2)p + 2p − 2t − 1}{(\frac{p−1}{2} − s − 1)p + p + \frac{p−1}{2} − t}

≡ \binom{p − 2s − 2}{\frac{p−1}{2} − s − 1}\binom{2p − 2t − 1}{p + \frac{p−1}{2} − t} = 0 (mod p).

Now we assert that

∑_{t=0}^{(p−1)/2} \binom{\frac{p−1}{2}p + \frac{p−1}{2}}{sp + t}^{3} ≡ 0 (mod p^{2}) for s = 0, 1, 2, ….  (4)

We prove the result by induction on s. For 0 ≤ t ≤ (p − 1)/2 we see that

\binom{\frac{p^{2}−1}{2}}{t} ≡ \binom{−\frac{1}{2}}{t} = \frac{\binom{2t}{t}}{(−4)^{t}} (mod p^{2}).

From Lemma 6 we know that the result is true for s = 0. Suppose that (4) holds for s = k. For s = k + 1, applying Lemma 4 we have

∑_{t=0}^{(p−1)/2} \binom{\frac{p−1}{2}p + \frac{p−1}{2}}{(k + 1)p + t}^{3}

≡ \binom{\frac{p−1}{2}}{k + 1}^{3} ∑_{t=0}^{\frac{p−1}{2}} (\frac{p + 1}{2}\binom{\frac{p−1}{2}}{t} − (\frac{p − 1}{2} + k)\binom{\frac{p−1}{2} − p}{t}

− k\binom{\frac{p−1}{2} − p}{t + p} − (\binom{\frac{p−1}{2} − p}{t} + \binom{\frac{p−1}{2} − p}{t + p}))^{3} (mod p^{2}).

Hence ∑_{t=0}^{(p−1)/2} \binom{\frac{p−1}{2} p+ \frac{p−1}{2}}{(k+1)p+t}^{3} ≡ 0 (mod p^{2}) for k ≥ \frac{p−1}{2}. For k < \frac{p−1}{2}, by the inductive hypothesis and Lemma 4 we have

∑_{t=0}^{(p−1)/2} (\frac{p + 1}{2}\binom{\frac{p−1}{2}}{t} − (\frac{p − 1}{2} + k)\binom{\frac{p−1}{2} − p}{t} − k\binom{\frac{p−1}{2} − p}{t + p})^{3} ≡ 0 (mod p^{2}).

Also, \binom{\frac{p−1}{2}}{t} ≡ \binom{\frac{p−1}{2} −p}{t} ≡ \binom{− \frac{1}{2}}{t} (mod p) and \binom{\frac{p−1}{2} −p}{t}+\binom{\frac{p−1}{2} −p}{t+p} = (−1)^{t}(\binom{\frac{p−1}{2} +t}{t}−\binom{p+ \frac{p−1}{2} +t}{t+p}) ≡ 0 (mod p) for t ∈ \{0, 1, …, \frac{p−1}{2}\}. By Lemma 5,

∑_{t=0}^{(p−1)/2} \binom{\frac{p−1}{2}p + \frac{p−1}{2}}{(k + 1)p + t}^{3}

≡ \binom{\frac{p−1}{2}}{k + 1}^{3}( ∑_{t=0}^{(p−1)/2} (\frac{p + 1}{2}\binom{\frac{p−1}{2}}{t} − (\frac{p − 1}{2} + k)\binom{\frac{p−1}{2} − p}{t} − k\binom{\frac{p−1}{2} − p}{t + p})^{3}

+ 3 ∑_{t=0}^{(p−1)/2} (\frac{p + 1}{2}\binom{\frac{p−1}{2}}{t} − (\frac{p − 1}{2} + k)\binom{\frac{p−1}{2} − p}{t} − k\binom{\frac{p−1}{2} − p}{t + p})

× (\binom{\frac{p−1}{2} − p}{t} + \binom{\frac{p−1}{2} − p}{t + p})^{2}

− 3 ∑_{t=0}^{(p−1)/2} (\frac{p + 1}{2}\binom{\frac{p−1}{2}}{t} − (\frac{p − 1}{2} + k)\binom{\frac{p−1}{2} − p}{t} − k\binom{\frac{p−1}{2} − p}{t + p})^{2}

× (\binom{\frac{p−1}{2} − p}{t} + \binom{\frac{p−1}{2} − p}{t + p})

− ∑_{t=0}^{(p−1)/2} (\binom{\frac{p−1}{2} − p}{t} + \binom{\frac{p−1}{2} − p}{t + p})^{3})

≡ −3\binom{\frac{p−1}{2}}{k + 1}^{3} ∑_{t=0}^{(p−1)/2} \binom{−\frac{1}{2}}{t}^{2}(\binom{\frac{p−1}{2} − p}{t} + \binom{\frac{p−1}{2} − p}{t + p})

= −3\binom{\frac{p−1}{2}}{k + 1}^{3} ∑_{t=0}^{(p−1)/2}(−1)^{t}\binom{−\frac{1}{2}}{t}^{2}(\binom{\frac{p−1}{2} + t}{t} − \binom{p + \frac{p−1}{2} + t}{t + p})

≡ 0 (mod p^{2}).

Hence

f_{\frac{p^{2}−1}{2}} ≡ ∑_{s=0}^{(p−1)/2} ∑_{t=0}^{(p−1)/2} \binom{\frac{p−1}{2}p + \frac{p−1}{2}}{sp + t}^{3} ≡ 0 (mod p^{2}).

Set H_{0} = H_{0}(1, 1) = 0, H_{k} = ∑_{i=1}^{k} \frac{1}{k} and H_{k}(1, 1) = ∑_{1≤i<j≤k} \frac{1}{ij} for k ∈ ℤ^{+}. For 0 ≤ s ≤ (p − 1)/2, it is easily seen that H_{p−1} ≡ 0 (mod p), \binom{p−1}{2s} ≡ 1 − pH_{2s} + p^{2}H_{2s}(1, 1) (mod p^{3}) and so \frac{1}{\binom{p−1}{2s}} ≡ 1 + pH_{2s} + p^{2}(H_{2s}^{2} − H_{2s}(1, 1)) (mod p^{3}). By Lemma 7, for 0 ≤ t ≤ (p − 1)/2 we see that

\binom{p^{2} − 1}{2sp + 2t} = \binom{p − 1}{2s} ∏_{i=1,p∤i}^{2sp+2t} (1 − \frac{p^{2}}{i} ) ≡ \binom{p − 1}{2s}(1 − p^{2} ∑_{i=1,p∤i}^{2sp+2t} \frac{1}{i}) (mod p^{3}).

Applying (4), Lemma 6 and the identity

\binom{a − b}{c − d}\binom{b}{d} = \binom{a}{c}\binom{c}{d}\binom{a − c}{b − d}/\binom{a}{b}

we derive that

S_{\frac{p^{2}−1}{2}} ≡ ∑_{s=0}^{(p−1)/2} ∑_{t=0}^{(p−1)/2} \binom{\frac{p^{2}−1}{2}}{sp + t}\binom{2sp + 2t}{sp + t}\binom{p^{2} − 1 − 2sp − 2t}{\frac{p^{2}−1}{2} − sp − t}

= \binom{p^{2} − 1}{\frac{p^{2}−1}{2}} ∑_{s=0}^{(p−1)/2} ∑_{t=0}^{(p−1)/2} \begin{matrix}\binom{(p^{2}−1)/2}{sp+t} & 3 \\ ( & {p^{2}−1 \atop 2sp+2t} & ) \end{matrix}

≡ \binom{p^{2} − 1}{\frac{p^{2}−1}{2}} ∑_{s=0}^{(p−1)/2} \frac{1}{\binom{p−1}{2s}} ∑_{t=0}^{(p−1)/2} \binom{\frac{p−1}{2}p + \frac{p−1}{2}}{sp + t}^{3}(1 + p^{2} ∑_{i=1,p∤i}^{2sp+2t} \frac{1}{i})

≡ \binom{p^{2} − 1}{\frac{p^{2}−1}{2}} ∑_{s=0}^{(p−1)/2} \frac{1}{\binom{p−1}{2s}} ∑_{t=0}^{(p−1)/2} \binom{\frac{p−1}{2}p + \frac{p−1}{2}}{sp + t}^{3}

+ p^{2}\binom{p^{2} − 1}{\frac{p^{2}−1}{2}} ∑_{s=0}^{(p−1)/2} \frac{\binom{2s}{s}3}{(−64)^{s}} ∑_{t=0}^{(p−1)/2} \frac{\binom{2t}{t}3}{(−64)^{t}} H_{2t}

≡ \binom{p^{2} − 1}{\frac{p^{2}−1}{2}}( ∑_{s=0}^{(p−1)/2} ∑_{t=0}^{(p−1)/2} \binom{\frac{p−1}{2}p + \frac{p−1}{2}}{sp + t}^{3}

+ p ∑_{s=0}^{(p−1)/2} (H_{2s} + p(H_{2s}^{2} − H_{2s}(1, 1))) ∑_{t=0}^{(p−1)/2} \binom{\frac{p−1}{2}p + \frac{p−1}{2}}{sp + t}^{3})

≡ ∑_{s=0}^{(p−1)/2} ∑_{t=0}^{(p−1)/2} \binom{\frac{p−1}{2}p + \frac{p−1}{2}}{sp + t}^{3}

≡ f_{\frac{p^{2}−1}{2}} (mod p^{3}).

Summarizing the above proves the theorem.

4 An inequality involving (S_{m})

Theorem 9. For m = 2, 3, 4, … we have

(1 + \frac{1}{m(m − 1)})S_{m}^{2} > S_{m+1}S_{m−1}.

Proof. It is easily seen that

(1 + \frac{1}{(m − 1)(m − 2)})S_{m−1}^{2} > S_{m}S_{m−2} for m = 3, 4, …, 13.

Now suppose m ≥ 14 and (1 + \frac{1}{(m−1)(m−2)})S_{m−1}^{2} > S_{m}S_{m−2}. By (2), Lemma 3 and the inductive hypothesis we have

(1 + \frac{1}{m(m − 1)})S_{m}^{2} − S_{m+1}S_{m−1}

= (1 + \frac{1}{m(m − 1)})S_{m}^{2} − \frac{4(3m^{2} + 3m + 1)}{(m + 1)^{2}}S_{m}S_{m−1} + \frac{32m^{2}}{(m + 1)^{2}} S_{m−1}^{2}

> (\frac{m^{2} − m + 1}{m(m − 1)}S_{m} − \frac{4(3m^{2} + 3m + 1)}{(m + 1)^{2}}S_{m−1} + \frac{32m^{2}(m − 1)(m − 2)}{(m + 1)^{2}(m^{2} − 3m + 3)}S_{m−2})S_{m}

= ((20m^{5} − 60m^{4} + 52m^{3} + 28m^{2} − 36m + 12)S_{m−1}

+ (−128m^{5} + 320m^{4} − 256m^{3} − 32m^{2} + 192m − 96)S_{m−2})

× \frac{S_{m}}{(m + 1)^{2}(m^{2} − 3m + 3)m^{3}(m − 1)}

= \frac{16S_{m}}{(m + 1)^{2}(m^{2} − 3m + 3)m^{3}(m − 1)} ∑_{k=0}^{m−2} \binom{m − 2}{k}\binom{2k}{k}\binom{2m − 4 − 2k}{m − 2 − k}F (m, k),

where

F (m, k) = (5m^{5} − 15m^{4} + 13m^{3} + 7m^{2} − 9m + 3)\frac{2k + 1}{k + 1}
− 8m^{5} + 20m^{4} − 16m^{3} − 2m^{2} + 12m − 6.

For m ≥ 14 it is easily seen that 3 < \frac{(2m−7)(2m−5)}{(m−3)(m−2)} < 4, 5m^{5}−15m^{4}+13m^{3}+7m^{2}−9m+3 > 0, −8m^{5}+20m^{4}−16m^{3}−2m^{2}+12m−6 < 0, 6m^{7}−75m^{6}+223m^{5}−283m^{4}−61m^{3}+427m^{2}−87m− 42 > 0, and F (m, k + 1) > F (m, k) for k = 0, 1, …, m − 3. Thus, F (m, m − 3) + F (m, 1) > F (m, 5) + F (m, 1) > 0 and

F (m, k) ≥ F (m, 2) = \frac{5}{3}(5m^{5} − 15m^{4} + 13m^{3} + 7m^{2} − 9m + 3)
− 8m^{5} + 20m^{4} − 16m^{3} − 2m^{2} + 12m − 6 > 0 for k ≥ 2.

From the above we derive that

(1 + \frac{1}{m(m − 1)})S_{m}^{2} − S_{m+1}S_{m−1}

> \frac{16S_{m}}{(m + 1)^{2}(m^{2} − 3m + 3)m^{3}(m − 1)}( ∑_{k=0}^{2} \binom{m − 2}{k}\binom{2k}{k}\binom{2m − 4 − 2k}{m − 2 − k}F (m, k)

+ ∑_{k=m−4}^{m−2} \binom{m − 2}{k}\binom{2k}{k}\binom{2m − 4 − 2k}{m − 2 − k}F (m, k))

= \frac{16S_{m}}{(m + 1)^{2}(m^{2} − 3m + 3)m^{3}(m − 1)}(\binom{2m − 4}{m − 2}(F (m, m − 2) + F (m, 0))

+ 3(m − 2)(m − 3)\binom{2m − 8}{m − 4}(F (m, m − 4) + F (m, 2))

+ 2(m − 2)\binom{2m − 6}{m − 3}(F (m, 1) + F (m, m − 3)))

> (3(m^{2} − 5m + 6)F (m, m − 4) + \frac{4(2m − 7)(2m − 5)}{(m − 3)(m − 2)}F (m, 0))

× \frac{16S_{m}\binom{2m−8}{m−4}}{(m + 1)^{2}(m^{2} − 3m + 3)m^{3}(m − 1)}

= \frac{S_{m}\binom{2m−8}{m−4}}{(m + 1)^{2}(m^{2} − 3m + 3)m^{3}(m − 1)}((6(m − 2)(2m − 7) + \frac{8(2m − 7)(2m − 5)}{(m − 3)(m − 2)})

× (40m^{5} − 120m^{4} + 104m^{3} + 56m^{2} − 72m + 24) + (−128m^{5} + 320m^{4} − 256m^{3} − 32m^{2} + 192m − 96)

× (3(m − 2)(m − 3) + \frac{4(2m − 7)(2m − 5)}{(m − 3)(m − 2)}))

> ((6(m − 2)(2m − 7) + 24)(40m^{5} − 120m^{4} + 104m^{3} + 56m^{2} − 72m + 24) + (3(m − 2)(m − 3) + 16)(−128m^{5} + 320m^{4} − 256m^{3} − 32m^{2} + 192m − 96))

× \frac{S_{m}\binom{2m−8}{m−4}}{(m + 1)^{2}(m^{2} − 3m + 3)m^{3}(m − 1)}

= (6m^{7} − 75m^{6} + 223m^{5} − 283m^{4} − 61m^{3} + 427m^{2} − 87m − 42)

× \frac{16S_{m}\binom{2m−8}{m−4}}{(m + 1)^{2}(m^{2} − 3m + 3)m^{3}(m − 1)}

> 0. Hence the inequality is proved by induction.

Corollary 10. For m = 2, 3, 4, … we have

(1 + \frac{1}{m(m − 1)})P_{m}^{2} > P_{m+1}P_{m−1}.

Proof. Since P_{m} = 2^{m}S_{m}, the result follows from Theorem 9.

References

[1] E. Catalan, Sur les nombres de Segner, Rend. Circ. Mat. Palermo 1 (1887), 190–201.

[2] J. Franel, On a question of Laisant, L’Intermédiaire des Mathématiciens 1 (1894), 45– 47.

[3] A. F. Jarvis, P. J. Larcombe, and D. R. French, On small prime divisibility of the Catalan-Larcombe-French sequence, Indian J. Math. 47 (2005), 159–181.

[4] X. J. Ji and Z. H. Sun, Congruences for Catalan-Larcombe-French numbers, preprint,

   2015, http://arxiv.org/abs/1505.00668v2.

[5] F. Jarvis and H. A. Verrill, Supercongruences for the Catalan-Larcombe-French numbers, Ramanujan J. 22 (2010), 171–186.

[6] P. Larcombe and D. French, On the “other” Catalan numbers: a historical formulation re-examined, Congr. Numer. 143 (2000), 33–64.

[7] P. J. Larcombe and D. R. French, A new generating function for the Catalan-Larcombe-French sequence: proof of a result by Jovovic, Congr. Numer. 166 (2004), 161–172.

[8] R. Meštrović, Lucas’ theorem: its generalizations, extensions and applications (1878–

   2014), preprint, 2014, http://arxiv.org/abs/1409.3820.

[9] B. Sagan, Congruences via Abelian groups, J. Number Theory. 20 (1985), 210–237.

[10] Z. H. Sun, Congruences concerning Legendre polynomials II, J. Number Theory. 133 (2013), 1950–1976.

[11] Z. H. Sun, Congruences involving Franel and Catalan-Larcombe-French numbers,

   preprint, 2015, http://arxiv.org/abs/1502.02499.

2010 Mathematics Subject Classification: Primary 11A07; Secondary 05A10, 05A19, 05A20. Keywords: congruence, inequality, Catalan-Larcombe-French number.

(Concerned with sequence \underline{A053175}.)

Received November 21 2015; revised versions received November 24 2015; February 23 2016. Published in Journal of Integer Sequences, April 6 2016.

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