Theorem 1: For any prime p and positive integer r, define

A(r; p) := (∑_{n=0}^{rp−1} \binom{2n}{n}) mod p .

Then, A(1; p) = \{ \begin{matrix}1, if p ≡ 1 mod 3 \\ −1, if p ≡ 2 mod 3 . \end{matrix} Proof: Observe that

\binom{2n}{n} = CT_{x} [\frac{(1 + x)^{2n}}{x^{n}} ]

Therefore,

∑_{n=0}^{p−1} \binom{2n}{n} = ∑_{n=0}^{p−1} CT_{x} [(\frac{(1 + x)^{2n}}{x^{n}} )]
= ∑_{n=0}^{p−1} CT_{x} [(2 + x + \frac{1}{x})^{n}]
= CT_{x} [\frac{(2 + x + \frac{1}{x})^{p} − 1}{2 + x + \frac{1}{x} − 1}]
≡_{p} CT_{x} [\frac{1 + x^{p} + \frac{1}{x^{p}}}{1 + x + \frac{1}{x}} ]
≡_{p} CT_{x} [\frac{1 + x^{p} + x^{2p}}{(1 + x + x^{2})x^{p−1}} ]
≡_{p} COEF F_{x^{p−1}} [\frac{1}{1 + x + x^{2}} ]
≡_{p} COEF F_{x^{p−1}} [\frac{1 − x}{1 − x^{3}} ]

The result folllows from series expansion of the last expression.

Corollary 1: A(2; p) = \{ \begin{matrix}3, if p ≡ 1 mod 3 \\ −3, if p ≡ 2 mod 3 . \end{matrix} Proof:

∑_{n=0}^{2p−1} \binom{2n}{n} = ∑_{n=0}^{2p−1} CT_{x} [(\frac{(1 + x)^{2n}}{x^{n}} )]
= CT_{x} [\frac{(2 + x + \frac{1}{x})^{2p} − 1}{2 + x + \frac{1}{x} − 1}]
= CT_{x} [\frac{(6 + 4x + \frac{4}{x} + x^{2} + \frac{1}{x^{2}} )^{p} − 1}{2 + x + \frac{1}{x} − 1}]
≡_{p} CT_{x} [\frac{(6 + 4x^{p} + \frac{4}{x^{p}} + x^{2p} + \frac{1}{x^{2p}} ) − 1}{2 + x + \frac{1}{x} − 1}]
≡_{p} COEF F_{x^{2p−1}} [\frac{1 + 4x^{p}}{1 + x + x^{2}} ]
≡_{p} COEF F_{x^{2p−1}} [\frac{1}{1 + x + x^{2}} ] + 4COEF F_{x^{p−1}} [\frac{1}{1 + x + x^{2}} ]

The result folllows from Theorem 1 and the last congruence.

Corollary 2 (Catalan Numbers): Let a(n) be the constant term of (1 − x) (2 + x + \frac{1}{x})^{n} and let

A(r; p) := (∑_{n=0}^{rp−1} a(n)) mod p

Then, A(1; p) = \{ \begin{matrix}1, if p ≡ 1 mod 3 \\ −2, if p ≡ 2 mod 3 . \end{matrix} Proof:

∑_{n=0}^{p−1} a(n) = ∑_{n=0}^{p−1} CT_{x} [(1 − x) (2 + x + \frac{1}{x})^{n}]
= CT_{x} [\frac{(1 − x) ((2 + x + \frac{1}{x})^{p} − 1)}{2 + x + \frac{1}{x} − 1}]
≡_{p} CT_{x} [\frac{(1 − x) ((2 + x^{p} + \frac{1}{x^{p}} ) − 1)}{2 + x + \frac{1}{x} − 1}]
≡_{p} COEF F_{x^{p−1}} [\frac{1 − x}{1 + x + x^{2}} ]
≡_{p} COEF F_{x^{p−1}} [\frac{1}{1 + x + x^{2}} ] − COEF F_{x^{p−2}} [\frac{1}{1 + x + x^{2}} ]

The result follows from Theorem 1 and the last congruence.

Theorem 2 (Motzkin Numbers): Let a(n) be the constant term of (1 − x^{2}) (1 + x + \frac{1}{x})^{n} and let

A(r; p) := ∑_{n=0}^{rp−1} a(n) .

Then, A(1; p) = \{ \begin{matrix}2, if p ≡ 1 mod 4 \\ −2, if p ≡ 3 mod 4 . \end{matrix} Proof:

∑_{n=0}^{p−1} a(n) = ∑_{n=0}^{p−1} CT [(1 − x^{2}) (1 + x + \frac{1}{x})^{n}]
= CT [\frac{(1 − x^{2}) ((1 + x + \frac{1}{x})^{p} − 1)}{1 + x + \frac{1}{x} − 1}]
≡_{p} COEF F_{x^{p−1}} [\frac{1 − x^{2}}{1 + x^{2}} ]
≡_{p} COEF F_{x^{p−1}} [\frac{1}{1 + x^{2}} ] − COEF F_{x^{p−3}} [\frac{1}{1 + x^{2}} ]

The result follows from series expansion of \frac{1}{1 + x^{2}} and the last congruence.

Corollary 3: Let a(n) be the constant term of (1 − x^{2}) (1 + x + \frac{1}{x})^{n} and let

A(r; p) := ∑_{n=0}^{rp−1} a(n) .

Then, A(1; p) = \{ \begin{matrix}4, if p ≡ 1 mod 4 \\ −4, if p ≡ 3 mod 4 . \end{matrix} Proof:

∑_{n=0}^{2p−1} a(n) = ∑_{n=0}^{2p−1} CT_{x} [(1 − x^{2}) (1 + x + \frac{1}{x})^{n}]
= CT_{x} [\frac{(1 − x^{2}) ((1 + x + \frac{1}{x})^{2p} − 1)}{1 + x + \frac{1}{x} − 1}]
= CT_{x} [\frac{(1 − x^{2}) ((3 + 2x + x^{2} + \frac{2}{x} + \frac{1}{x^{2}} )^{p} − 1)}{1 + x + \frac{1}{x} − 1}]
= CT_{x} [\frac{(1 − x^{2}) ((3 + 2x^{p} + x^{2p} + \frac{2}{x^{p}} + \frac{1}{x^{2p}} ) − 1)}{1 + x + \frac{1}{x} − 1}]
≡_{p} COEF F_{x^{2p−1}} [\frac{(1 − x^{2})(1 + 2x^{p})}{1 + x^{2}}]
≡_{p} COEF F_{x^{2p−1}} [\frac{1}{1 + x^{2}} ] + 2COEF F_{x^{p−1}} [\frac{1}{1 + x^{2}} ]
− COEF F_{x^{2p−3}} [\frac{1}{1 + x^{2}} ] − 2COEF F_{x^{p−3}} [\frac{1}{1 + x^{2}} ] .

The result follows from Theorem 2 and the last congruence.

Theorem 3: Let P (x) be a Laurent polynomial in x and let p be a prime. Let R(x) the denominator after simplifying the expression

\frac{P (x^{p}) − 1}{P (x) − 1}.

Then, for any positive integer r and Laurent polynomial Q(x),

(∑_{n=0}^{rp−1} CT_{x} [P (x)^{n}Q(x)]) mod p

is congruent to a p-finite sequence and can be expressed as linear combinations of finite terms of the coefficients of \frac{1}{R(x)}.

Theorem 4: Let p be a prime number and let r be a positive integer. Let

A(r, s; p) := ∑_{n=0}^{rp−1} ∑_{m=0}^{sp−1} \binom{m + n}{m} mod p .

Then,

A(r, s; p^{α}) = \binom{r + s}{s} − 1 , where α = 1, 2, 3 .

Proof: Let P (x) = 1 + \frac{1}{x} and Q(x) = 1 + x. Then,

∑_{n=0}^{rp−1} ∑_{m=0}^{sp−1} \binom{m + n}{m} = ∑_{n=0}^{rp−1} ∑_{m=0}^{sp−1} CT_{x} [P (x)^{m}Q(x)^{n}]
= ∑_{n=0}^{sp−1} CT_{x} [\frac{P (x)^{rp} − 1)Q(x)^{n}}{P (x) − 1} ]
= CT_{x} [(\frac{P (x)^{rp} − 1}{P (x) − 1}) (\frac{Q(x)^{sp} − 1}{Q(x) − 1})]
= CT_{x} [((1 + \frac{1}{x})^{rp} − 1) ((1 + x)^{sp} − 1)]
= CT_{x} [((1 + \frac{1}{x})^{rp} − 1) ((1 + x)^{sp} − 1)]
= CT_{X} [(1 + \frac{1}{x})^{rp} (1 + x)^{sp} − (1 + \frac{1}{x})^{rp} − (1 + x)^{sp} + 1]
= CT_{X} [\frac{(1 + x)^{rp+sp}}{x^{sp}} − 1 − 1 + 1]
= \binom{p(r + s)}{ps} − 1
≡_{p^{α}} \binom{r + s}{s} − 1, where α = 1, 2, 3 .

The last equality follows from the beautiful result that \binom{pa}{pb} ≡_{p^{α}} \binom{a}{b} for α = 1, 2, 3.

Theorem 5: Let p be a prime number and let r be a positive integer. Let

A(r, s; p) := (∑_{n=0}^{rp−1} ∑_{m=0}^{sp−1} \binom{n + m}{m}^{2}) mod p .

Then, A(1, 1; p) = \{ \begin{matrix}0, if p ≡ 0 mod 3 \\ 1, if p ≡ 1 mod 3 \\ −1, if p ≡ 2 mod 3 . \end{matrix}

Furthermore, A(2, 2; p) ≡_{p} COEF F_{[x^{2p−1}y^{2p−1}]} \frac{1 + 4x^{p} + 4y^{p} + 16x^{p}y^{p}}{(1 + x + xy)(1 + y + xy)}.

Proof: Let P (x, y) = (1 + y) (1 + \frac{1}{x}) and Q(x, y) = (1 + x) (1 + \frac{1}{y}). First observe that

\binom{n + m}{m}^{2} = \binom{n + m}{m}\binom{n + m}{n} = CT_{xy} (P (x, y)^{n}Q(x, y)^{m}) .

Then,

∑_{m=0}^{p−1} ∑_{n=0}^{p−1} \binom{m + n}{m}^{2} = ∑_{m=0}^{p−1} ∑_{n=0}^{p−1} CT_{xy} [P (x, y)^{n}Q(x, y)^{m}]

= ∑_{m=0}^{p−1} CT_{xy} [\frac{P (x, y)^{p} − 1)Q(x, y)^{m}}{P (x, y) − 1} ]

= CT_{xy} [(\frac{P (x, y)^{p} − 1}{P (x, y) − 1}) (\frac{Q(x, y)^{p} − 1}{Q(x, y) − 1})]

Using the fact that (a + b)^{p} ≡_{p} a^{p} + b^{p}, we get

∑_{m=0}^{p−1} ∑_{n=0}^{p−1} \binom{m + n}{m}^{2} ≡_{p} CT_{xy} [(\frac{P (x^{p}, y^{p}) − 1}{P (x, y) − 1}) (\frac{Q(x^{p}, y^{p}) − 1}{Q(x, y) − 1})]

≡_{p} CT_{xy} [\frac{(1 + y^{p} + x^{p}y^{p})(1 + x^{p} + x^{p}y^{p})}{(1 + y + xy)(1 + x + xy)x^{p−1}y^{p−1}} ]
≡_{p} COEF F [x^{p−1}y^{p−1}] [\frac{(1 + y^{p} + x^{p}y^{p})(1 + x^{p} + x^{p}y^{p})}{(1 + y + xy)(1 + x + xy)}]
≡_{p} COEF F [x^{p−1}y^{p−1}] [\frac{1}{(1 + y + xy)(1 + x + xy)}]

IUsing the Almkvist-Apagodu-Zeilberger algorithm, the diagonal coefficients satisfy the recurrence N^{2} + N + 1 = 0 with the initial conditions a(0) = 1, a(1) = 1, a(2) = 0. The result now follows from the fact that the recurrence is equivalent to N^{3} − 1 = 0.

Finally, if we apply the above method to A(2, 2; p) we get,

A(2, 2; p) ≡_{p} COEF F_{[x^{2p−1}y^{2p−1}]} \frac{1 + 4x^{p} + 4y^{p} + 16x^{p}y^{p}}{(1 + x + xy)(1 + y + xy)} .

Expanding, we get

COEF F_{[x^{2p−1}y^{2p−1}]} \frac{1}{(1 + x + xy)(1 + y + xy)} + 4COEF F_{[x^{p−1}y^{2p−1}]} \frac{1}{(1 + x + xy)(1 + y + xy)}

+4COEF F_{[x^{2p−1}y^{p−1}]} \frac{1}{(1 + x + xy)(1 + y + xy)}+16COEF F_{[x^{p−1}y^{p−1}]} \frac{1}{(1 + x + xy)(1 + y + xy)}

By symmetry in the second and third term, we have

A(2, 2; p) ≡_{p} COEF F_{[x^{2p−1}y^{2p−1}]} \frac{1}{(1 + x + xy)(1 + y + xy)}+

8COEF F_{[x^{p−1}y^{2p−1}]} \frac{1}{(1 + x + xy)(1 + y + xy)} + 16COEF F_{[x^{p−1}y^{p−1}]} \frac{1}{(1 + x + xy)(1 + y + xy)}

Let p be a prime number. Then,

C(r, s; p) := (∑_{n=0}^{rp−1} ∑_{m=0}^{sp−1} \binom{n + m}{m}^{3}) mod p .

Since

\binom{n + m}{m}^{3} = CT_{xyz} [\frac{(1 + x)^{m+n}}{x^{m}} \frac{(1 + y)^{m+n}}{y^{m}} \frac{(1 + z)^{m+n}}{z^{n}} ] .

If we let P (x, y, z) = (1 + \frac{1}{x}) (1 + \frac{1}{y}) (1 + z) and Q(x, y, z) = (1 + x) (1 + y) (1 + \frac{1}{z}), then we have

\binom{n + m}{m}^{3} = CT_{xyz} ((P (x, y)^{m}Q(x, y)^{n}) .

Repeating the above arguement verbatim, we arive at

C(1, 1,; p) ≡_{p} COEF F [x^{p−1}y^{p−1}z^{p−1}] [\frac{1}{p(x, y, z)q(x, y, z)}] ,

where p(x, y, z) = 1+x+y +xy +xz +yz +xyz and q(x, y, z) = 1+y +z +xy +xz +yz +xyz.

Theorem 6: Let p be a prime number and let

A(r, s, t; p) := ∑_{m_{1}=0}^{rp−1} ∑_{m_{2}=0}^{sp−1} ∑_{m_{3}=0}^{tp−1} \binom{m_{1} + m_{2} + m_{3}}{m_{1}, m_{2}, m_{3}} .

Then, A(1, 1, 1; p) ≡_{p} 1 for p > 2.

Proof (α = 1 case): Observe that

\binom{m_{1} + m_{2} + m_{3}}{m_{1}, m_{2}, m_{3}} = CT_{xyz} [\frac{(1 + x + y + z)^{m_{1}+m_{2}+m_{3}}}{x^{m_{1}}y^{m_{2}}z^{m_{3}}} ]

As in the proof of Theorem 2 using the fact that (x + y)^{p} ≡_{p} x^{p} + y^{p}, we get

∑_{m1=0}^{p−1} ∑_{m2=0}^{p−1} ∑_{m_{3}=0}^{p−1} \binom{m_{1} + m_{2} + m_{3}}{m_{1}, m_{2}, m_{3}} = ∑_{m_{1}=0}^{p−1} ∑_{m_{2}=0}^{p−1} ∑_{m_{3}=0}^{p−1} CT_{xyz} [\frac{(1 + x + y + z)^{m_{1}+m_{1}+m_{3}}}{x^{m_{1}}y^{m_{2}}z^{m_{3}}} ]

≡_{p} CT_{xyz} \frac{(1 + x^{p}y^{p} + x^{p}z^{p})(1 + y^{p}x^{p} + y^{p}z^{p})(1 + z^{p}x^{p} + z^{p}y^{p})}{(xyz)^{p−1}(1 + x + y)(1 + x + z)(1 + y + z)}
≡_{p} COEF F_{[x^{p−1}y^{p−1}z^{p−1}]} \frac{1}{(1 + x + y)(1 + x + z)(1 + y + z)}

Using Almkvist-Apagodu-Zeilberger algorithm, the diagonal coefficients, a(n) := a(n, n, n), satisfy the recurrence

24(5n + 11)(3n + 5)(3n + 4)a(n) + (295n^{3} + 1614n^{2} + 2855n + 1620)a(n + 1)
+2(2n + 5)(n + 2)(5n + 6)a(n + 2) = 0 .

Passing to mod p, the terms with powers of p are zero and the recurrence reduces to the much simpler second order recurrence,

288a(p) + 84a(p + 1) + 6a(p + 2) = 0

with initial conditions a(0) = 1 and a(1) = −14, whose solution is given by

a(p) ≡_{p} 4(−8)^{p} − 3(−6)^{p} .

From the above equation, we know that A(1, 1, 1; p) ≡ a(p − 1) (mod p). Thus,

A(1, 1, 1; p) = a(p − 1) ≡_{p} 4(−8)^{p−1} − 3(−6)^{p−1} ≡_{p} 2^{p−1}, .

. Now it suffices to show that 2^{p−1} − 1 ≡_{p} 0. But this follows fom

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

Observation 1: A(1, 1, 1; p) ≡_{p^{α}} 1 for p > 2, for α = 2, 3. Observation 2: The above computer data suggests that A(2, 2, 2; p) ≡_{p^{α}} 16 for p ≥ 17, for α = 1, 2, 3

In general, if we denote,

A(r_{1}, r_{2}, …, r_{a}; p) := ∑_{m_{1}=0}^{r_{1}p−1} ∑_{m_{2}=0}^{r_{2}p−1} … ∑_{m_{a}=0}^{r_{a}p−1} \binom{m_{1} + m_{2} … + m_{a}}{m_{1}, m_{2}, …, m_{a}} .

Then,

(1) A(1, 1, …, 1; p) ≡_{p} Coeff_{[x_{1}^{p−1}x_{2}^{p−1}...x_{a}^{p−1}]} \frac{1}{∏_{j=1}^{a}(1 + ∑_{i=1,i≠j}^{a} x_{i})}.

(2) Conjecture: A(1, 1, …, 1; p) ≡_{p} 1 for p > 2 and A(1, 1, …, 1; 2) ≡_{p} 0. Proof (of i) : Observe that

\binom{m_{1} + m_{2} … + m_{a}}{m_{1}, m_{2}, …, m_{a}} = CT_{x_{1}x_{2}...x_{a}} [\frac{(1 + x_{1} + x_{2} + … + x_{a})^{m_{1}+m_{2}+...+m_{a}}}{∏_{i=1}^{a} x_{i}^{m_{i}}} ]

As in the proof of Theorem 4,

∑_{m_{1}=0}^{p−1} … ∑_{m_{a}=0}^{p−1} \binom{m_{1} + m_{2} … + m_{a}}{m_{1}, m_{2}, …, m_{a}} = ∑_{m_{1}=0}^{p−1} … ∑_{m_{a}=0}^{p−1} CT_{x_{1}x_{2}...x_{a}} [\frac{(1 + x_{1} + x_{2} + … + x_{a})^{∑_{j=1}^{a} m_{j}}}{∏_{i=1}^{a} x_{i}^{m_{i}}} ]
≡_{p} COEF F_{[x_{1}^{p−1}x_{2}^{p−1}...x_{a}^{p−1}]} \frac{1}{∏_{j=1}^{a} (1 + ∑_{i=1,i≠j}^{a} x_{i})}.

Apery’s Number: Apery’s number is defined by

A(n) := ∑_{k=0}^{n} \binom{n}{k}^{2}\binom{n + k}{k}^{2}

and it is known that

A(n) = CT_{xyz} [(\frac{(1 + x)(1 + y)(1 + z)(1 + y + z + yz + xyz)}{xyz})^{n}]

Let p be a prime number and define

a(p) := ∑_{n=0}^{p−1} A(n) .

We can express a(p) as

∑_{n=0}^{p−1} ∑_{k=0}^{n} \binom{n}{k}^{2}\binom{n + k}{k}^{2} = ∑_{n=0}^{p−1} CT_{xyz} [(\frac{(1 + x)(1 + y)(1 + z)(1 + y + z + yz + xyz)}{xyz})^{n}]
= CT_{xyz} [R1(x, y, z) + R2(x, y, z)]
= CT_{xyz} [\frac{R1(x, y, z)}{x^{p−1}y^{p−1}z^{p−1}} ]
= CT_{xyz} [\frac{P (x, y, z)}{x^{p−1}y^{p−1}z^{p−1}Q(x, y, z)}]
≡_{p} COEF F_{x^{p−1}y^{p−1}z^{p−1}} \frac{P (x^{p}, y^{p}, z^{p})}{Q(x, y, z)}
≡_{p} COEF F_{x^{p−1}y^{p−1}z^{p−1}} [\frac{1}{Q(x, y, z)}]

where, R2(x, y, z) = \frac{xyz}{Q(x,y,z)} and has zero constant term and R1(x, y, z) = \frac{P (x,y,z)}{Q(x,y,z)}, where

P (x, y, z) = (x + 1)^{p}(y + 1)^{p}(1 + z)^{p}((1 + y + z + yz + xyz)^{p}

and

Q(x, y, z) = x^{2}y^{2}z^{2} + x^{2}y^{2}z + x^{2}yz^{2} + 2xy^{2}z^{2} + x^{2}yz + 3xy^{2}z + 3xyz^{2} + y^{2}z^{2} + xy^{2}

+4xyz + xz^{2} + 2y^{2}z + 2yz^{2} + 2xy + 2xz + y^{2} + 4yz + z^{2} + x + 2y + 2z + 1 .

Therefore,

a(p) ≡ α(p − 1, p − 1, p − 1)(mod p)

where

Q(x, y, z) = ∑_{m,n,k} α(m, n, k)x^{n}y^{m}z^{k} .

Toward Z-W Sun’s Conjecture: Z-W Sun conjectures that

∑_{n=0}^{p−1} ∑_{k=0}^{n}(−1)^{n+k}8^{k}\binom{n}{k}^{3} ≡ (\frac{p}{3}) (mod p^{2}) .

Using the fact that

\binom{n}{k}^{3} = CT_{xyz} [\frac{(1 + x)^{n}(1 + y)^{n}(1 + z)^{n}}{x^{k}y^{k}z^{k}} ] .

we get

∑_{n=0}^{p−1} ∑_{k=0}^{n}(−1)^{n+k}8^{k}\binom{n}{k}^{3} = ∑_{n=0}^{p−1} ∑_{k=0}^{n}(−1)^{n+k}8^{k}CT_{xyz} [\frac{(1 + x)^{n}(1 + y)^{n}(1 + z)^{n}}{x^{k}y^{k}z^{k}} ]
= CT_{xyz} [R1(x, y, z) + R2(x, y, z) + R3(x, y, z) + R4(x, y, z)]

where

R1(x, y, z) = −\frac{(−1)^{p}((x + 1)(y + 1)(1 + z))^{p}xyz}{(xyz + xy + xz + yz + x + y + z + 2)(xyz + 8)}

R2(x, y, z) = \frac{(−8)^{p+1}((x + 1)(y + 1)(1 + z))^{p}}{x^{p−1}y^{p−1}z^{p−1}(7xyz + 8xy + 8xz + 8yz + 8x + 8y + 8z + 8)(xyz + 8)}

R3(x, y, z) = \frac{xyz}{(xyz + xy + xz + yz + x + y + z + 2)(xyz + 8)}

and

R4(x, y, z) = −\frac{8xyz}{(7xyz + 8xy + 8xz + 8yz + 8x + 8y + 8z + 8)(xyz + 8)} .

Since the constant terms of R1(x, y, z), R3(x, y, z) and R4(x, y, z) are zero, we get,

∑_{n=0}^{p−1} ∑_{k=0}^{n}(−1)^{n+k}8^{k}\binom{n}{k}^{3} = COEF F_{x^{p−1}y^{p−1}z^{p−1}} [\frac{(−8)^{p+1}((1 + x)(1 + y)(1 + z))^{p}}{(7xyz + 8(xy + xz + yz + x + y + z + 1))(xyz + 8)}]
≡_{p} COEF F_{x^{p−1}y^{p−1}z^{p−1}} [\frac{(−8)^{p+1}}{(7xyz + 8(xy + xz + yz + x + y + z + 1))(xyz + 8)}]

Theorem 7: Let P (x, y, z) and Q(x, y, z) be Laurent polynomials in x, y and z. Let T (x, y, z) be the denominator (after clearing!) of

\frac{(P (x^{p}, y^{p}, z^{p}) − 1)(Q(x^{p}, y^{p}, z^{p}) − 1)}{(P (x, y, z) − 1)(Q(x, y, z) − 1)} .

Then, for a prime p, Laurent polynomial R(x, y, z), and positive integers r, s, t,

(∑_{k=0}^{rp−1} ∑_{n=0}^{sp−1} ∑_{n=0}^{tp−1} CT_{xyz} (P (x, y, z)^{n}Q(x, y, z)^{k}R(x, y, z)) ) mod p

is congruent to a P -finite sequence and can be expressed as a finite linear combination of the coefficients of \frac{1}{T (x, y, z)}.