More Congruences for Central Binomial Sums with Fibonacci and Lucas Numbers

Roberto Tauraso Dipartimento di Matematica Università di Roma Tor Vergata 00133 Roma Italy

                    tauraso@mat.uniroma2.it

Abstract We mainly determine ∑_{k=1}^{p−1} \binom{2k}{k}h_{k}x^{k} modulo a prime p with h_{k} = ∑_{j=1}^{k} \frac{1}{2j−1}. We also provide some applications of this polynomial congruence for some special values of x which involve the Fibonacci and Lucas numbers.

1 Introduction

In [4], Lehmer presented a long list of interesting series involving the central binomial coefficient \binom{2k}{k} such as

∑_{k=1}^{∞} \binom{2k}{k}\frac{x^{k}}{k} = −2 ln (\frac{1 + \sqrt{1 − 4x}}{2}) .  (1)

Since then, using many different approaches and methods, several other authors have actively investigated this kind of sums. Among them we would like to mention [3, 1, 2]. In particular, Chen in [2, Theorem 8] obtained

∑_{k=1}^{∞} \binom{2k}{k}h_{k}x^{k} = −\frac{ln(1 − 4x)}{2\sqrt{1 − 4x}}  (2)

with h_{k} = ∑_{j=1}^{k} \frac{1}{2j−1}. In this work, as we did in [5, 7, 6], we are going to consider the sums of the first p − 1 terms of the above series, with a prime p, and evaluate them modulo p.

The paper is organized as follows. The next section is devoted to the proofs of two polynomial congruences which are the finite analogues of the infinite series (1) and (2). In the third section, we specialize those polynomial congruences to some particular values (u_{n})_{n≥1} and (v_{n})_{n≥1} related to Fibonacci numbers F_{n} (\underline{A000045}) and Lucas numbers L_{n} (\underline{A000032}):

u_{n} = \frac{(−1)^{n}}{L_{n}^{2}} = \frac{(−1)^{n}}{L_{2n} + 2(−1)^{n}} → −1, \frac{1}{9}, −\frac{1}{16}, \frac{1}{49}, −\frac{1}{121}, \frac{1}{324}, −\frac{1}{841}, …

v_{n} = \frac{(−1)^{n−1}}{5F_{n}^{2}} = \frac{(−1)^{n−1}}{L_{2n} − 2(−1)^{n}} → \frac{1}{5}, −\frac{1}{5}, \frac{1}{20}, −\frac{1}{45}, \frac{1}{125}, −\frac{1}{320}, \frac{1}{845}, …

Notice that, up to sign, u_{n} and v_{n} are the reciprocals of \underline{A001254} and \underline{A099921} in the On-Line Encyclopedia of Integer Sequences (OEIS).

For instance, after extending a result by Williams [8], we show that for any integer n ≥ 1, and for any prime p > 2 not dividing 5F_{n}L_{n},

∑_{k=1}^{p−1} \binom{2k}{k}\frac{u_{n}^{k}}{k} ≡ 2q_{p}(L_{n}) − \frac{5nF_{n}}{L_{n}} f_{p} (mod p),

∑_{k=1}^{p−1} \binom{2k}{k}h_{k}v_{n}^{k} ≡ − (\frac{5}{p}) (\frac{1}{2}q_{p} (\frac{L_{n}^{2}}{5F_{n}^{2}} ) + (−1)^{n} \frac{2n}{F_{2n}} f_{p}) (mod p).

Here (\frac{5}{p}) stands for a Legendre symbol, and let

q_{p}(x) = \frac{x^{p−1} − 1}{p}, f_{p} = \frac{F_{p−(\frac{5}{p})}}{p}

denote the Fermat quotient and the Fibonacci quotient, respectively.

In order to demonstrate the similarities of the above formulas with the corresponding series evaluations, notice that, by (1) and (2), we obtain

∑_{k=1}^{∞} \binom{2k}{k}\frac{u_{n}^{k}}{k} = 2 ln(L_{n}) − 2n ln (\frac{1 + \sqrt{5}}{2}) ,

∑_{k=1}^{∞} \binom{2k}{k}h_{k}v_{n}^{k} = −\frac{\sqrt{5}F_{n}}{2L_{n}} ln (\frac{L_{n}^{2}}{5F_{n}^{2}} ) .

2 Polynomial congruences

The next congruence is equivalent to [6, (32)]. Here we give a new and self-contained proof.

Theorem 1. Let p > 2 be a prime. Then

∑_{k=1}^{p−1} \binom{2k}{k}\frac{x^{k}}{k} ≡ 2\frac{1 − α^{p} − (1 − α)^{p}}{p} (mod p)  (3)

where x = α(1 − α). Proof. Let β = 1 − α. Then, for 1 ≤ k ≤ p − 1, it follows that

\binom{p}{k} = \frac{p(p − 1) ⋯ (p − (k − 1))}{k!} ≡ \frac{p(−1)^{k−1}}{k} (mod p^{2}),

and we find

(1 − αz)^{p}(1 − βz)^{p} = ∑_{k=0}^{p} \binom{p}{k}(−αz)^{k} ∑_{k=0}^{p} \binom{p}{k}(−βz)^{k}

≡ (1 − α^{p}z^{p} − p ∑_{k=1}^{p−1} \frac{α^{k}z^{k}}{k} ) (1 − β^{p}z^{p} − p ∑_{k=1}^{p−1} \frac{β^{k}z^{k}}{k} )

≡ 1 − (α^{p} + β^{p})z^{p} + (αβ)^{p}z^{2p}

− p(1 − α^{p}z^{p}) ∑_{k=1}^{p−1} \frac{β^{k}z^{k}}{k} − p(1 − β^{p}z^{p}) ∑_{k=1}^{p−1} \frac{α^{k}z^{k}}{k} (mod p^{2}).

Hence, we have

[z^{p}](1 − αz)^{p}(1 − βz)^{p} ≡ −α^{p} − β^{p} (mod p^{2}).  (4)

On the other hand,

[z^{p}](1 − αz)^{p}(1 − βz)^{p} = [z^{p}](1 − z + xz^{2})^{p}

= [z^{p}] ∑_{k=0}^{p} \binom{p}{k}x^{k}z^{2k}(1 − z)^{p−k}

= [z^{p}] ∑_{k=0}^{p} \binom{p}{k}x^{k}z^{2k} ∑_{j=0}^{p−k} \binom{p − k}{j}(−z)^{j}

= ∑_{k=0}^{\frac{p−1}{2}} \binom{p}{k}\binom{p − k}{p − 2k}(−1)^{p−2k}x^{k}

= − ∑_{k=0}^{\frac{p−1}{2}} \binom{p}{k}\binom{p − k}{k}x^{k}.

Therefore, we get

[z^{p}](1 − αz)^{p}(1 − βz)^{p} ≡ −1 + p ∑_{k=1}^{\frac{p−1}{2}}(−1)^{k}\binom{p − k}{k}\frac{x^{k}}{k}
≡ −1 + \frac{p}{2} ∑_{k=1}^{p−1} \binom{2k}{k}\frac{x^{k}}{k} (mod p^{2})  (5)

where we used the fact that, for 1 ≤ k ≤ \frac{p−1}{2},

\binom{p − k}{k} = \frac{(p − k) … (p − (2k − 1))}{k!} ≡ \frac{(−1)^{k}}{2} \binom{2k}{k} (mod p),

and p divides \binom{2k}{k} for \frac{p−1}{2} < k < p. Combining (4) with (5) yields the desired result.

Once the preliminary congruence (3) is obtained, we are in the position to state the main theorem of this section. Theorem 2. If p > 2 is a prime then

∑_{k=1}^{p−1} \binom{2k}{k}h_{k}x^{k} ≡ \frac{1}{2}(1 − 4x)^{\frac{p−1}{2}} (∑_{k=1}^{p−1} \binom{2k}{k}\frac{x^{k}}{k} − ∑_{k=1}^{p−1} \binom{2k}{k}\frac{y^{k}}{k} ) (mod p)  (6)

where y = −\frac{x}{1−4x}. Proof. For 1 ≤ k ≤ \frac{p−1}{2},

\binom{\frac{p−1}{2}}{k} = \frac{(p − 1)(p − 3) … (p − (2k − 1))}{2^{k}k!} ≡ \frac{(−1)^{k}(2k − 1)!!}{2^{k}k!} = \binom{2k}{k}\frac{1}{(−4)^{k}} (mod p)

which implies that

∑_{k=0}^{p−1} \binom{2k}{k}x^{k} ≡ ∑_{k=0}^{\frac{p−1}{2}} \binom{\frac{p−1}{2}}{k}(−4x)^{k} ≡ (1 − 4x)^{\frac{p−1}{2}} (mod p).  (7)

Furthermore, by [7, (14)], we have that

(∑_{k=0}^{p−1} \binom{2k}{k}x^{k}) · (∑_{k=1}^{p−1} \binom{2k}{k}\frac{x^{k}}{k} ) ≡ 2 ∑_{k=1}^{p−1} \binom{2k}{k}(H_{2k} − H_{k})x^{k} (mod p)  (8)

where H_{k} = ∑_{j=1}^{k} \frac{1}{j} is the k-th harmonic number.

In view of the known identity ∑_{k=1}^{n} \binom{n}{k} \frac{(−1)^{k}}{k} = −H_{n}, it follows that

∑_{k=1}^{p−1} \binom{2k}{k}H_{k}x^{k} ≡ ∑_{k=1}^{\frac{p−1}{2}} \binom{\frac{p−1}{2}}{k}H_{k}(−4x)^{k}

= − ∑_{k=1}^{\frac{p−1}{2}} \binom{\frac{p−1}{2}}{k}(−4x)^{k} ∑_{j=1}^{k} \binom{k}{j}\frac{(−1)^{j}}{j}

= − ∑_{j=1}^{\frac{p−1}{2}} \frac{(−1)^{j}}{j} ∑_{k=j}^{\frac{p−1}{2}} \binom{\frac{p−1}{2}}{k}\binom{k}{j}(−4x)^{k}

= − ∑_{j=0}^{\frac{p−1}{2}} \frac{(−1)^{j}}{j} \binom{\frac{p−1}{2}}{j}(−4x)^{j}(1 − 4x)^{\frac{p−1}{2} −j}

≡ −(1 − 4x)^{\frac{p−1}{2}} ∑_{j=1}^{p−1} \binom{2j}{j}\frac{y^{j}}{j} (mod p).  (9)

Finally, given that h_{k} = H_{2k} − \frac{1}{2}H_{k} and combining (7), (8), and (9), we obtain

∑_{k=1}^{p−1} \binom{2k}{k}h_{k}x^{k} = ∑_{k=1}^{p−1} \binom{2k}{k}(H_{2k} − H_{k})x^{k} + \frac{1}{2} ∑_{k=1}^{p−1} \binom{2k}{k}H_{k}x^{k}

≡ \frac{1}{2}(1 − 4x)^{\frac{p−1}{2}} (∑_{k=1}^{p−1} \binom{2k}{k}\frac{x^{k}}{k} − ∑_{k=1}^{p−1} \binom{2k}{k}\frac{y^{k}}{k} ) (mod p).

If x = \frac{1}{2} then y = −\frac{x}{1−4x} = \frac{1}{2} and congruence (6) immediately implies the following corollary.

Corollary 3. For any prime p > 2,

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

It is worth pointing out that, based on a numerical experiment, the above congruence seems to have the following conjectural q-analogue: for any odd integer n,

∑_{k=1}^{n−1} \left[{2k \atop k}\right]_{q} \frac{q^{k}}{∏_{j=1}^{k}(1 + q^{j})} ∑_{j=1}^{k} \frac{q^{2j−1}}{[2j − 1]_{q}} ≡ (−1)^{\frac{n−1}{2}} \frac{(n − 1)(q − 1)}{2}q^{\frac{n^{2}−1}{4}} (mod Φ_{n}(q)) (10)

where [n]_{q} = (1 − q^{n})/(1 − q), for 0 ≤ k ≤ n,

\left[{n \atop k}\right]_{q} = \frac{(q; q)_{n}}{(q; q)_{k}(q; q)_{n−k}}

is the q-binomial coefficient with (z; q)_{n} = (1 − z)(1 − zq) ⋯ (1 − zq^{n−1}), and Φ_{n}(q) denotes the n-th cyclotomic polynomial.

3 Congruences involving Fibonacci and Lucas numbers

We first establish the next two congruences modulo p^{2}, which extend [8, (2.9) and (2.10)].

Lemma 4. Let ε = (\frac{5}{p}) and let p be a prime different from 2 and 5. Then for any integer n ≥ 0,

L_{np} ≡ L_{n} + \frac{5n}{2}F_{n}F_{p−ε} (mod p^{2}),  (11)
F_{np} ≡ ε (F_{n} + \frac{n}{2}L_{n}F_{p−ε}) (mod p^{2}).  (12)

Proof. We show both congruences by induction with respect to n. They are are clearly true for n = 0. For n = 1, by [8, (2.9)], we have L_{p−ε} ≡ 2ε (mod p^{2}). Therefore

L_{p} = \frac{1}{2} (L_{p−ε}L_{ε} + 5F_{p−ε}F_{ε}) ≡ εL_{ε} + \frac{5}{2}F_{p−ε}F_{ε} = L_{1} + \frac{5}{2}F_{1}F_{p−ε} (mod p^{2})

and

F_{p} = \frac{1}{2} (L_{p−ε}F_{ε} + F_{p−ε}L_{ε}) ≡ εF_{ε} + \frac{ε}{2}L_{1}F_{p−ε} ≡ ε (F_{1} + \frac{1}{2}L_{1}F_{p−ε}) (mod p^{2})

where we applied the identities

2L_{n+m} = L_{n}L_{m} + 5F_{n}F_{m}, 2F_{n+m} = L_{n}F_{m} + F_{n}L_{m},

and εL_{ε} = 1 = L_{1}, F_{ε} = 1 = F_{1}.

Inductive step for (11): for n ≥ 3, by the multiple-angle recurrence for L_{np}, and because p divides F_{p−ε}, we get

L_{np} = L_{p}L_{(n−1)p} − (−1)^{p}L_{(n−2)p}

≡ (1 + \frac{5}{2}F_{p−ε}) (L_{n−1} + \frac{5(n − 1)}{2}F_{n−1}F_{p−ε}) + (L_{n−2} + \frac{5(n − 2)}{2}F_{n−2}F_{p−ε})

≡ (L_{n−1} + L_{n−2}) + \frac{5}{2}F_{p−ε} (L_{n−1} + (n − 1)F_{n−1} + (n − 2)F_{n−2})

≡ L_{n} + \frac{5}{2}F_{p−ε} (F_{n} + (n − 1)F_{n−1} + (n − 1)F_{n−2})

≡ L_{n} + \frac{5n}{2}F_{n}F_{p−ε} (mod p^{2}).

A similar argument shows the inductive step for (12):

F_{np} = L_{p}F (n − 1)p − (−1)^{p}F_{(n−2)p}

≡ (1 + \frac{5}{2}F_{p−ε}) (F_{n−1} + \frac{(n − 1)}{2}L_{n−1}F_{p−ε}) + (F_{n−2} + \frac{(n − 2)}{2}L_{n−2}F_{p−ε})

≡ (F_{n−1} + F_{n−2}) + \frac{5}{2}F_{p−ε} (F_{n−1} + (n − 1)L_{n−1} + (n − 2)L_{n−2})

≡ F_{n} + \frac{5}{2}F_{p−ε} (L_{n} + (n − 1)L_{n−1} + (n − 1)L_{n−2})

≡ F_{n} + \frac{5n}{2}L_{n}F_{p−ε} (mod p^{2}).

In the next two theorems we specialize the polynomial congruences (3) and (6) obtained in the previous section to the numbers u_{n} and v_{n}.

Theorem 5. Let n ≥ 1 and let p > 2 be a prime not dividing 5L_{n}F_{n}. Then

∑_{k=1}^{p−1} \binom{2k}{k}\frac{u_{n}^{k}}{k} ≡ −q_{p}(u_{n}) − \frac{5nF_{n}}{L_{n}} f_{p} (mod p)  (13)

∑_{k=1}^{p−1} \binom{2k}{k}\frac{v_{n}^{k}}{k} ≡ −q_{p}(v_{n}) − \frac{nL_{n}}{F_{n}} f_{p} (mod p).  (14)

Proof. Let φ_{±} = (1 ± \sqrt{5})/2. For α = φ_{+}^{n}/L_{n}, we have that

α(1 − α) = \frac{φ_{+}^{n}(L_{n} − φ_{+}^{n})}{L_{n}^{2}} = \frac{φ_{+}^{n}(φ_{−}^{n})}{L_{n}^{2}} = \frac{(−1)^{n}}{L_{n}^{2}} = u_{n}.

Let x = u_{n} in (3). Then we find

∑_{k=1}^{p−1} \binom{2k}{k}\frac{u_{n}^{k}}{k} ≡ 2\frac{1 − α^{p} − (1 − α)^{p}}{p}

= \frac{2}{pL_{n}^{p}} (L_{n}^{p} − φ_{+}^{np} − (L_{n} − φ_{+}^{n})^{p})

= \frac{2}{L_{n}^{p}} · \frac{L_{n}^{p} − L_{np}}{p}

= \frac{2}{L_{n}^{p−1}} q_{p}(L_{n}) − \frac{2}{L_{n}^{p}} · \frac{L_{np} − L_{n}}{p}

≡ 2q_{p}(L_{n}) − \frac{2}{L_{n}} · \frac{5nF_{n}F_{p−ε}}{2p}

≡ −q_{p}(u_{n}) − \frac{5nF_{n}}{L_{n}} f_{p} (mod p)

where we applied Fermat’s little theorem, (11), and F_{n}L_{n} = F_{2n}. Also recall that q_{p}(x · y) ≡ q_{p}(x) + q_{p}(y) (mod p).

For α = φ_{+}^{n}/(\sqrt{5}F_{n}), we have

α(1 − α) = \frac{φ_{+}^{n}(\sqrt{5}F_{n} − φ_{+}^{n})}{5F_{n}^{2}} = \frac{φ_{+}^{n}(−φ_{−}^{n})}{5F_{n}^{2}} = \frac{(−1)^{n−1}}{5F_{n}^{2}} = v_{n}.

In the same manner, setting x = v_{n} in (3), we obtain

∑_{k=1}^{p−1} \binom{2k}{k}\frac{v_{n}^{k}}{k} ≡ 2\frac{1 − α^{p} − (1 − α)^{p}}{p}

= \frac{2}{p5^{p/2}F_{n}^{p}} (5^{p/2}F_{n}^{p} − φ_{+}^{np} − (\sqrt{5}F_{n} − φ_{+}^{n})^{p})

= \frac{2}{5^{p/2}F_{n}^{p}} · \frac{5^{p/2}F_{n}^{p} − \sqrt{5}F_{np}}{p}

= \frac{2}{F_{n}^{p−1}} q_{p}(F_{n}) − \frac{2}{5^{\frac{p−1}{2}} F_{n}^{p}} · \frac{F_{np} − 5^{\frac{p−1}{2}} F_{n}}{p}

≡ 2q_{p}(F_{n}) + q_{p}(5) − \frac{2ε}{F_{n}} · \frac{F_{np} − εF_{n}}{p}

≡ −q_{p}(v_{n}) − \frac{nL_{n}}{F_{n}} f_{p} (mod p),

where we used (12), and

5^{\frac{p−1}{2}} ≡ (\frac{5}{p}) (1 + \frac{p}{2}q_{p}(5)) (mod p^{2}),

which is a consequence of [5, Lemma 4.1].

Theorem 6. Let n ≥ 1 and let p > 2 be a prime not dividing 5L_{n}F_{n}. Then

∑_{k=1}^{p−1} \binom{2k}{k}h_{k}u_{n}^{k} ≡ (\frac{5}{p}) (\frac{1}{2}q_{p} (\frac{v_{n}}{u_{n}} ) + (−1)^{n} \frac{2n}{F_{2n}} f_{p}) (mod p)  (15)

and

∑_{k=1}^{p−1} \binom{2k}{k}h_{k}v_{n}^{k} ≡ − ∑_{k=1}^{p−1} \binom{2k}{k}h_{k}u_{n}^{k} (mod p).  (16)

Proof. Setting x = u_{n}, the identity L_{n}^{2} − 5F_{n}^{2} = 4(−1)^{n} yields

1 − 4x = \frac{L_{n}^{2} − 4(−1)^{n}}{L_{n}^{2}} = \frac{5F_{n}^{2}}{L_{n}^{2}} ,

and therefore

y = \frac{−x}{1 − 4x} = −\frac{(−1)^{n}}{L_{n}^{2}} · \frac{L_{n}^{2}}{5F_{n}^{2}} = \frac{(−1)^{n−1}}{5F_{n}^{2}} = v_{n}.

Moreover

(1 − 4x)^{\frac{p−1}{2}} = 5^{\frac{p−1}{2}} \frac{F_{n}^{p−1}}{L_{n}^{p−1}} ≡ (\frac{5}{p}) (mod p).

Thus, by (6), (13), and (14), we obtain (15). Indeed,

∑_{k=1}^{p−1} \binom{2k}{k}h_{k}u_{n}^{k} ≡ \frac{1}{2}(1 − 4u_{n})^{\frac{p−1}{2}} (∑_{k=1}^{p−1} \binom{2k}{k}\frac{u_{n}^{k}}{k} − ∑_{k=1}^{p−1} \binom{2k}{k}\frac{v_{n}^{k}}{k} )

≡ \frac{1}{2} (\frac{5}{p}) (q_{p}(v_{n}) − q_{p}(u_{n}) + (\frac{L_{n}}{F_{n}} − \frac{5F_{n}}{L_{n}} ) nf_{p})

≡ \frac{1}{2} (\frac{5}{p}) (q_{p} (\frac{v_{n}}{u_{n}} ) + \frac{4(−1)^{n}}{F_{2n}} nf_{p}) (mod p).

It remains to show (16). Since

1 − 4v_{n} = 1 − 4y = 1 + \frac{4x}{1 − 4x} = (1 − 4x)^{−1} = (1 − 4u_{n})^{−1},

it follows that

(1 − 4v_{n})^{\frac{p−1}{2}} = (1 − 4u_{n})^{− \frac{p−1}{2}} ≡ (\frac{5}{p})^{−1} ≡ (\frac{5}{p}) (mod p)

and we are done because of (6).

4 Acknowledgment

The author wishes to thank the anonymous referee for the valuable comments and corrections. We are indebted to him/her for pointing out the conjectural congruence (10). We are also grateful to Prof. Hongwei Chen for his suggestions that have made this paper more readable.

References

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[4] D. H. Lehmer, Interesting series involving the central binomial coefficient, Amer. Math. Monthly 92 (1985), 449–457.

[5] S. Mattarei and R. Tauraso, Congruences for central binomial sums and finite polylogarithms, J. Number Theory 133 (2013), 131–157.

[6] S. Mattarei and R. Tauraso, From generating series to polynomial congruences, J. Number Theory 182 (2018), 179–205.

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[8] H. C. Williams, A note on the Fibonacci quotient F_{p−ε}/p, Can. Math. Bull. 25 (1982), 366–370.

2020 Mathematics Subject Classification: Primary 11A07; Secondary 05A10, 11B39. Keywords: congruence, central binomial coefficient, harmonic number, Fibonacci number, Lucas number.

(Concerned with sequences \underline{A000032}, \underline{A000045}, \underline{A001254}, and \underline{A099921}.)

Received June 17 2023; revised versions received June 18 2023; October 1 2023; October 2 2023. Published in Journal of Integer Sequences, October 3 2023.

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