Golden Ratio Base Expansions of the Logarithm and Inverse Tangent of Fibonacci and Lucas Numbers

Kunle Adegoke

Department of Physics and Engineering Physics

Obafemi Awolowo University Ile-Ife Nigeria

adegoke00@gmail.com

Jaume Oliver Lafont Conselleria d’Educacio i Cultura

Govern de les Illes Balears Palma Spain

joliverlafont@gmail.com

Abstract

Let α = (1 + \sqrt{5})/2, the golden ratio. Let F_{n} and L_{n} be the Fibonacci and Lucas numbers. We derive base-α expansions of log F_{n}, log L_{n}, arctan \frac{1}{F_{n}}, and arctan \frac{1}{L_{n}} for all positive integers n.

1 Introduction

Let α denote the golden ratio; that is, α = (1 + \sqrt{5})/2. Let β = −1/α = (1 − \sqrt{5})/2. Thus αβ = −1 and α + β = 1. Let F_{n} and L_{n} be the Fibonacci and Lucas numbers, defined by F_{n} = (α^{n} − β^{n})/\sqrt{5} and L_{n} = α^{n} + β^{n}, for all non-negative integers n.

Let b be any non-zero number whose magnitude is greater than unity. Let n and s be positive integers. A convergent series of the form

C = ∑_{k=0}^{∞} \frac{1}{b^{k}} (\frac{a_{1}}{(kn + 1)^{s}} + \frac{a_{2}}{(kn + 2)^{s}} + ⋯ + \frac{a_{n}}{(kn + n)^{s}} ),  (1)

where a_{1}, a_{2}, …, a_{n} are certain numbers, defines a base-b, length-n and degree-s expansion of the mathematical constant C.

If b is an integer and a_{k} are rational numbers, then (1) is referred to as a BBP-type formula, after the initials of the authors of the paper [4] in which such an expansion was first presented for π and some other mathematical constants. Any mathematical constant that possesses a base-b BBP-type formula has the property that one can compute its n-th digit, in base b, using only O((log n)^{c}) space for some constant c, and time O(n).

Our goal in this paper is to derive base-α expansion formulas for the logarithm and the inverse tangent of all Fibonacci and Lucas numbers. We will often give the expansion using the compact P -notation for BBP-type formulas, introduced by Bailey and Crandall [5], namely,

C = P (s, b, n, A) = ∑_{k=0}^{∞} \frac{1}{b^{k}} ∑_{j=1}^{n} \frac{a_{j}}{(kn + j)^{s}} ,

where s and n are integers and, in this present paper, b is an integer power of α and A = (a_{1}, a_{2}, …, a_{n}) is a vector of rational multiples of powers of β. For example, we will show (see (19)) that

log F_{3} = log 2 = ∑_{k=0}^{∞} \frac{1}{α^{12k}} (\frac{β^{2}}{6k + 1} + \frac{3β^{4}}{6k + 2} + \frac{4β^{6}}{6k + 3} + \frac{3β^{8}}{6k + 4} + \frac{β^{10}}{6k + 5}),

which, in the P -notation, can be written as

log F_{3} = log 2 = P (1, α^{12}, 6, (β^{2}, 3β^{4}, 4β^{6}, 3β^{8}, β^{10}, 0)).

To conclude this section we provide a brief summary of some previous studies on base-α expansions.

Bailey and Crandall [5] derived a base-2/α formula for π\sqrt{α}. Similar formulas were found by Chan [7] who also later proved [8] several formulas expressing π in terms of α. Borwein and Chamberland [6] found a base-α expansion for π^{2}.

Zhang [11] gave base-α expansions for π\sqrt{α\sqrt{5}} and some other constants.

In a previous paper [2], we established base-α formulas for π, log α, log 2 and several other mathematical constants.

Recently, Kristensen and Mathiasen [9], using an algorithm implemented in SAGE, found a base-α formula for π. They also obtained several base-α zero relations; that is, series that sum to zero.

2 Base-α expansions of logarithms

The base-α expansions of the logarithms of Fibonacci and Lucas numbers are presented in Theorems 3 and 5 but first we state a couple of lemmata upon which the results are based.

Let

Li_{1}(x) = − log(1 − x) = ∑_{k=1}^{∞} \frac{x^{k}}{k} = ∑_{k=0}^{∞} x^{k} \frac{x}{k + 1}, −1 ≤ x < 1.

Lemma 1. If |b| > 1, t > 0 and m and n are arbitrary positive integers, then

Li_{1} (\frac{1}{b^{t}} ) = ∑_{k=0}^{∞} \frac{1}{b^{tmk}} ∑_{j=1}^{m} \frac{1/b^{tj}}{(mk + j)},  (2)

Li_{1} (−\frac{1}{b^{t}} ) = ∑_{k=0}^{∞} \frac{1}{b^{2tnk}} ∑_{j=1}^{2n} \frac{(−1)^{j}/b^{tj}}{(2nk + j)}.  (3)

Proof. We have

Li_{1} (\frac{1}{b^{t}} ) = ∑_{k=0}^{∞} \frac{1}{b^{tk}} \frac{1/b^{t}}{k + 1},

from which (2) follows upon using the identity

∑_{k=0}^{∞} f_{k} = ∑_{k=0}^{∞} ∑_{j=1}^{m} f_{mk+j−1},  (4)

with

f_{k} = \frac{1}{b^{tk+t}} \frac{1}{k + 1}.

The proof of (3) is similar, with m = 2n in (4).

Lemma 2. If r is an integer, then

log L_{r} = r Li_{1} (\frac{1}{α^{2}} ) − Li_{1} (\frac{(−1)^{r+1}}{α^{2r}} ) ,  (5)
log F_{r} = (r − 2) Li_{1} (\frac{1}{α^{2}} ) + Li_{1} (\frac{1}{α^{4}} ) − Li_{1} (\frac{(−1)^{r}}{α^{2r}} ) , r ≠ 0.  (6)

Proof. We have

Li_{1} (−\frac{β^{r}}{α^{r}} ) = − log (\frac{α^{r} + β^{r}}{α^{r}} ) = − log (\frac{L_{r}}{α^{r}} ) = − log L_{r} + r log α,  (7)

in which setting r = 1 gives

log α = Li_{1} (\frac{1}{α^{2}} ) .  (8)

Using (8) in (7) gives (5).

Also,

Li_{1} (\frac{β^{r}}{α^{r}} ) = − log (\frac{α^{r} − β^{r}}{α^{r}} ) = − log (\frac{F_{r}\sqrt{5}}{α^{r}}) = − log F_{r} + r log α − log \sqrt{5}, (9)

in which setting r = 2 gives

log \sqrt{5} = 2 log α − Li_{1} (\frac{1}{α^{4}} ) = 2 Li_{1} (\frac{1}{α^{2}} ) − Li_{1} (\frac{1}{α^{4}} ) ,  (10)

where we used (8). Identity (6) follows from (9) and (10).

Theorem 3. If r is an integer, then

log F_{r} = ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{r} \frac{β^{4j−2}(r − 2 + rδ_{j,(r+1)/2})}{2rk + 2j − 1} + ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{r−1} \frac{β^{4j}r}{2rk + 2j} , r odd, (11)

log F_{r} = ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{r} \frac{(r − 2)β^{4j−2}}{2rk + 2j − 1} + ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{r−1} \frac{β^{4j}r(1 − δ_{j,r/2})}{2rk + 2j}, r even. (12)

Here and throughout this paper, δ_{mn} denotes the Kronecker delta symbol whose value is unity when m equals n and zero otherwise.

Proof. We prove (12). When r is even, (6) reads

log F_{r} = (r − 2) Li_{1} (\frac{1}{α^{2}} ) + Li_{1} (\frac{1}{α^{4}} ) − Li_{1} (\frac{1}{α^{2r}} ) , r ≠ 0.  (13)

We proceed to write the three Li_{1} terms in a common base α^{4r}, using (2) with appropriate t and m choices. Thus,

Li_{1} (\frac{1}{α^{2}} ) = ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{2r} \frac{1/α^{2j}}{2rk + j} ,  (14)

Li_{1} (\frac{1}{α^{4}} ) = ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{r} \frac{1/α^{4j}}{rk + j} ,  (15)

Li_{1} (\frac{1}{α^{2r}} ) = ∑_{k=0}^{∞} \frac{1}{α^{4rk}} (\frac{1/α^{2r}}{2k + 1} + \frac{1/α^{4r}}{2k + 2}).  (16)

Using (14), (15) and (16) in (13) gives

\begin{aligned}log F_{r} & = ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{2r} \frac{β^{2j}(r − 2)}{2rk + j} + ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{r} \frac{2β^{4j}}{2rk + 2j} \\ & − ∑_{k=0}^{∞} \frac{1}{α^{4rk}} (\frac{rβ^{2r}}{2rk + r} + \frac{rβ^{4r}}{2rk + 2r}). \end{aligned}  (17)

Using the summation identity

∑_{j=1}^{2r} f_{j} = ∑_{j=1}^{r} f_{2j} + ∑_{j=1}^{r} f_{2j−1}

to write its inner sum, the first term on the right hand side of (17) can be written as

∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{2r} \frac{β^{2j}(r − 2)}{2rk + j} = ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{r} \frac{β^{4j}(r − 2)}{2rk + 2j} + ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{r} \frac{β^{4j−2}(r − 2)}{2rk + 2j − 1}.  (18)

Using (18) in (17) yields (12).

Identities (11) and (12) written in the P -notation are

log F_{r} = P (1, α^{4r}, 2r, (a_{1}, a_{2}, …, a_{2r})),

where for 1 ≤ j ≤ r,

a_{2j−1} = β^{4j−2}(r − 2 + rδ_{j,(r+1)/2}), a_{2j} = β^{4j}r(1 − δ_{rj}), r odd;

and

a_{2j−1} = (r − 2)β^{4j−2}, a_{2j} = β^{4j}r(1 − δ_{j,r/2} − δ_{j,r}), r even.

Example 4.

log F_{3} = log 2 = P (1, α^{12}, 6, (β^{2}, 3β^{4}, 4β^{6}, 3β^{8}, β^{10}, 0)),  (19)
\begin{aligned}log F_{5} = log 5 & = P (1, α^{20}, 10, (3β^{2}, 5β^{4}, 3β^{6}, 5β^{8}, 8β^{10}, \\ & 5β^{12}, 3β^{14}, 5β^{16}, 3β^{18}, 0)), \end{aligned}  (20)
log F_{4} = log 3 = P (1, α^{16}, 8, (2β^{2}, 4β^{4}, 2β^{6}, 0, 2β^{10}, 4β^{12}, 2β^{14}, 0)),  (21)

\begin{aligned}log F_{8} = log 21 & = P (1, α^{32}, 16, (6β^{2}, 8β^{4}, 6β^{6}, 8β^{8}, 6β^{10}, 8β^{12}, 6β^{14}, \\ & 0, 6β^{18}, 8β^{20}, 6β^{22}, 8β^{24}, 6β^{26}, 8β^{28}, 6β^{30}, 0)), \end{aligned}

\begin{aligned}log F_{12} = log 144 & = P (1, α^{48}, 24, (10 β^{2}, 12 β^{4}, 10 β^{6}, 12 β^{8}, 10 β^{10}, \\ & 12 β^{12}, 10 β^{14}, 12 β^{16}, 10 β^{18}, 12 β^{20}, 10 β^{22}, 0, \\ & 10 β^{26}, 12 β^{28}, 10 β^{30}, 12 β^{32}, 10 β^{34}, 12 β^{36}, \\ & 10 β^{38}, 12 β^{40}, 10 β^{42}, 12 β^{44}, 10 β^{46}, 0)). \end{aligned}  (22)

Theorem 5. If r is an integer, then

log L_{r} = ∑_{k=0}^{∞} \frac{1}{α^{2rk}} ∑_{j=1}^{r−1} \frac{β^{2j}r}{rk + j} , r odd,  (23)
log L_{r} = ∑_{k=0}^{∞} \frac{1}{α^{4rk}} ∑_{j=1}^{2r−1} \frac{β^{2j}r(1 + δ_{rj})}{2rk + j}, r even.  (24)

Proof. We prove (23). If r is an odd integer, (5) gives

log L_{r} = r Li_{1} (\frac{1}{α^{2}} ) − Li_{1} (\frac{1}{α^{2r}} ) .  (25)

With

Li_{1} (\frac{1}{α^{2}} ) = ∑_{k=0}^{∞} \frac{1}{α^{2rk}} ∑_{j=1}^{r} \frac{1/α^{2j}}{rk + j}

and

Li_{1} (\frac{1}{α^{2r}} ) = ∑_{k=0}^{∞} \frac{1}{α^{2rk}} \frac{r/α^{2r}}{rk + r}

in (25); identity (23) follows.

Identities (23) and (24) in the P -notation are

log L_{r} = P (1, α^{2r}, r, (a_{1}, a_{2}, …, a_{r})), r odd,

with

a_{j} = rβ^{2j}(1 − δ_{rj}), 1 ≤ j ≤ r,

and

log L_{r} = P (1, α^{4r}, 2r, (a_{1}, a_{2}, …, a_{2r})), r even,

with

a_{j} = rβ^{2j}(1 + δ_{jr} − δ_{j,2r}), 1 ≤ j ≤ 2r.

Example 6.

log L_{2} = log 3 = ∑_{k=0}^{∞} \frac{1}{α^{8k}} (\frac{2β^{2}}{4k + 1} + \frac{4β^{4}}{4k + 2} + \frac{2β^{6}}{4k + 3});

that is,

log 3 = P (1, α^{8}, 4, (2β^{2}, 4β^{4}, 2β^{6}, 0)).  (26)
log L_{3} = log 4 = P (1, α^{6}, 3, (3β^{2}, 3β^{4}, 0)),  (27)
log L_{4} = log 7 = 4β^{2}P (1, α^{16}, 8, (1, β^{2}, β^{4}, 2β^{6}, β^{8}, β^{10}, β^{12}, 0)),
\begin{aligned}log L_{6} = log 18 & = P (1, α^{24}, 12, (6β^{2}, 6β^{4}, 6β^{6}, 6β^{8}, 6β^{10}, \\ & 12β^{12}, 6β^{14}, 6β^{16}, 6β^{18}, 6β^{20}, 6β^{22}, 0)). \end{aligned}  (28)

3 Base-α expansions of inverse tangents

The base-α expansions of the inverse tangent of Fibonacci and Lucas numbers are stated in Theorems 10–19 but first we collect some required identities in Lemmata 7–9. Lemma 7. If r is an integer, then

α^{r} − α^{−r} = \begin{cases}F_{r}\sqrt{5}, & r even; \\ L_{r}, & r odd; \end{cases}

α^{r} + α^{−r} = \begin{cases}L_{r}, & r even; \\ F_{r}\sqrt{5}, & r odd. \end{cases}

Lemma 8. If r and m are integers, then

arctan \frac{1}{α^{r−m}} − arctan \frac{1}{α^{r+m}} = \begin{cases}arctan (L_{m}/(F_{r}\sqrt{5})) , & m odd, r odd; \\ arctan(L_{m}/L_{r}), & m odd, r even; \\ arctan(F_{m}/F_{r}), & m even, r odd; \\ arctan(F_{m}\sqrt{5}/L_{r}), & m even, r even; \end{cases}

arctan \frac{1}{α^{r−m}} + arctan \frac{1}{α^{r+m}} = \begin{cases}arctan(F_{m}\sqrt{5}/L_{r}), & m odd, r odd; \\ arctan(F_{m}/F_{r}), & m odd, r even; \\ arctan(L_{m}/L_{r}), & m even, r odd; \\ arctan (L_{m}/(F_{r}\sqrt{5})) , & m even, r even. \end{cases}

Proof. The arctangent subtraction and addition formulas give

arctan \frac{1}{α^{r−m}} − arctan \frac{1}{α^{r+m}} = arctan (\frac{α^{r}(α^{m} − α^{−m})}{α^{2r} + 1}) ,

arctan \frac{1}{α^{r−m}} + arctan \frac{1}{α^{r+m}} = arctan (\frac{α^{r}(α^{m} + α^{−m})}{α^{2r} − 1});

and hence the stated identities upon the use of Lemma 7. Lemma 9. If r is an integer, then

α^{2r} − 1 = α^{r}L_{r}, β^{2r} − 1 = β^{r}L_{r}, r odd,  (29)
α^{2r} − 1 = α^{r}F_{r}\sqrt{5}, β^{2r} − 1 = −β^{r}F_{r}\sqrt{5}, r even,  (30)
α^{2r} + 1 = α^{r}F_{r}\sqrt{5}, β^{2r} + 1 = −β^{r}F_{r}\sqrt{5}, r odd,
α^{2r} + 1 = α^{r}L_{r}, β^{2r} + 1 = β^{r}L_{r}, r even.

Theorem 10. If r is an odd integer greater than unity, then

arctan \frac{1}{F_{r}} = P (1, α^{4(r^{2}−4)}, 4(r^{2} − 4), (a_{1}, a_{2}, …, a_{4(r^{2}−4)})),  (31)

where the only non-zero constants a_{j} are given by

a_{(r−2)(4j−3)} = −β^{(r−2)(4j−3)}(r − 2), j = 1, 2, …, r + 2,
a_{(r−2)(4j−1)} = β^{(r−2)(4j−1)}(r − 2), j = 1, 2, …, r + 2,
a_{(r+2)(4j−3)} = β^{(r+2)(4j−3)}(r + 2), j = 1, 2, …, r − 2,
a_{(r+2)(4j−1)} = −β^{(r+2)(4j−1)}(r + 2), j = 1, 2, …, r − 2,
a_{(r−2)(r+2)} = (−1)^{(r+1)/2}4β^{r^{2}−4}
a_{3(r−2)(r+2)} = (−1)^{(r−1)/2}4β^{3(r^{2}−4)}.

Proof. With r an odd number, setting m = 2 in (8) gives

arctan \frac{1}{F_{r}} = arctan \frac{1}{α^{r−2}} − arctan \frac{1}{α^{r+2}} .

The following identity, proved in a previous work [1, Identity (10)]:

n\sqrt{n} arctan (\frac{1}{\sqrt{n}}) = ∑_{k=0}^{∞} \frac{1}{n^{2k}} (\frac{n}{4k + 1} − \frac{1}{4k + 3})  (32)

gives

arctan \frac{1}{α^{r−2}} = ∑_{k=0}^{∞} \frac{1}{α^{(r−2)(4k+3)}} (\frac{α^{2r−4}}{4k + 1} − \frac{1}{4k + 3})

and

arctan \frac{1}{α^{r+2}} = ∑_{k=0}^{∞} \frac{1}{α^{(r+2)(4k+3)}} (\frac{α^{2r+4}}{4k + 1} − \frac{1}{4k + 3}),

or, by (4),

arctan \frac{1}{α^{r−2}} = ∑_{k=0}^{∞} \frac{1}{α^{(4r^{2}−16)k}} ∑_{j=1}^{r+2} (\frac{α^{−(r−2)(4j−3)}}{4(r + 2)k + 4j − 3} − \frac{α^{−(r−2)(4j−1)}}{4(r + 2)k + 4j − 1})

and

arctan \frac{1}{α^{r+2}} = ∑_{k=0}^{∞} \frac{1}{α^{(4r^{2}−16)k}} ∑_{j=1}^{r−2} (\frac{α^{−(r+2)(4j−3)}}{4(r − 2)k + 4j − 3} − \frac{α^{−(r+2)(4j−1)}}{4(r − 2)k + 4j − 1}).

Thus,

\begin{aligned}arctan \frac{1}{F_{r}} & = ∑_{k=0}^{∞} \frac{1}{α^{(4r^{2}−16)k}} × \\ ( & ∑_{j=1}^{r+2} (\frac{−β^{(r−2)(4j−3)}(r − 2)}{4(r^{2} − 4)k + (r − 2)(4j − 3)} + \frac{β^{(r−2)(4j−1)}(r − 2)}{4(r^{2} − 4)k + (r − 2)(4j − 1)}) \\ & + ∑_{j=1}^{r−2} (\frac{β^{(r+2)(4j−3)}(r + 2)}{4(r^{2} − 4)k + (r + 2)(4j − 3)} − \frac{β^{(r+2)(4j−1)}(r + 2)}{4(r^{2} − 4)k + (r + 2)(4j − 1)})) . \end{aligned} ; (33)

Identity (31) is (33) expressed in the P -notation. Example 11.

arctan \frac{1}{F_{3}} = arctan \frac{1}{2} = P (1, α^{20}, 20, (−β, 0, β^{3}, 0, 4 β^{5}, 0, β^{7}, 0, −β^{9}, 0, β^{11},
0, −β^{13}, 0, −4 β^{15}, 0, −β^{17}, 0, β^{19}, 0)),

(34)

arctan \frac{1}{F_{5}} = arctan \frac{1}{5} = P (1, α^{84}, 84, (0, 0, −3 β^{3}, 0, 0, 0, 7 β^{7}, 0, 3 β^{9}, 0, 0, 0, 0, 0, −3 β^{15},
0, 0, 0, 0, 0, −4 β^{21}, 0, 0, 0, 0, 0, −3 β^{27}, 0, 0, 0, 0, 0, 3 β^{33},
0, 7 β^{35}, 0, 0, 0, −3 β^{39}, 0, 0, 0, 0, 0, 3 β^{45}, 0, 0, 0, −7 β^{49},
0, −3 β^{51}, 0, 0, 0, 0, 0, 3 β^{57}, 0, 0, 0, 0, 0, 4 β^{63}, 0, 0, 0, 0, 0,

3 β^{69}, 0, 0, 0, 0, 0, −3 β^{75}, 0, −7 β^{77}, 0, 0, 0, 3 β^{81}, 0, 0, 0)),

arctan \frac{1}{F_{7}} = arctan \frac{1}{13} = P (1, α^{180}, 180, (0, 0, 0, 0, −5 β^{5}, 0, 0, 0, 9 β^{9}, 0, 0, 0, 0, 0, 5 β^{15},
0, 0, 0, 0, 0, 0, 0, 0, 0, −5 β^{25}, 0, −9 β^{27}, 0, 0, 0, 0, 0, 0, 0,
5 β^{35}, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4 β^{45}, 0, 0, 0, 0, 0, 0, 0, 0, 0,
5 β^{55}, 0, 0, 0, 0, 0, 0, 0, −9 β^{63}, 0, −5 β^{65}, 0, 0, 0, 0, 0,
0, 0, 0, 0, 5 β^{75}, 0, 0, 0, 0, 0, 9 β^{81}, 0, 0, 0, −5 β^{85}, 0, 0,
0, 0, 0, 0, 0, 0, 0, 5 β^{95}, 0, 0, 0, −9 β^{99}, 0, 0, 0, 0, 0,
− 5 β^{105}, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5 β^{115}, 0, 9 β^{117}, 0, 0, 0,
0, 0, 0, 0, −5 β^{125}, 0, 0, 0, 0, 0, 0, 0, 0, 0, −4 β^{135},
0, 0, 0, 0, 0, 0, 0, 0, 0, −5 β^{145}, 0, 0, 0, 0, 0, 0, 0, 9 β^{153}, 0,

5 β^{155}, 0, 0, 0, 0, 0, 0, 0, 0, 0, −5 β^{165}, 0, 0, 0, 0, 0, −9 β^{171},

0, 0, 0, 5 β^{175}, 0, 0, 0, 0, 0)).

Theorem 12. If r is a positive even integer, then

arctan \frac{1}{F_{r}} = P (1, α^{4(r^{2}−1)}, 4(r^{2} − 1), (a_{1}, a_{2}, …, a_{4(r^{2}−1)})),

where the only non-zero constants a_{j} are given by

a_{(r−1)(4j−3)} = −β^{(r−1)(4j−3)}(r − 1), j = 1, 2, …, r + 1,
a_{(r−1)(4j−1)} = β^{(r−1)(4j−1)}(r − 1), j = 1, 2, …, r + 1,
a_{(r+1)(4j−3)} = −β^{(r+1)(4j−3)}(r + 1), j = 1, 2, …, r − 1,
a_{(r+1)(4j−1)} = β^{(r+1)(4j−1)}(r + 1), j = 1, 2, …, r − 1,
a_{(r−1)(r+1)} = (−1)^{r/2}2β^{r^{2}−1}
a_{3(r−1)(r+1)} = (−1)^{(r+2)/2}2β^{3(r^{2}−1)}.

Proof. Setting m = 1 in (8) gives

arctan \frac{1}{F_{r}} = arctan \frac{1}{α^{r−1}} + arctan \frac{1}{α^{r+1}} , r even.

The proof now proceeds as in that of Theorem 10. Example 13.

arctan \frac{1}{F_{2}} = \frac{π}{4} = P (1, α^{12}, 12, (−β, 0, −2 β^{3}, 0, −β^{5}, 0, β^{7}, 0, 2 β^{9}, 0, β^{11}, 0)),

arctan \frac{1}{F_{4}} = arctan \frac{1}{3} = P (1, α^{60}, 60, (0, 0, −3 β^{3}, 0, −5 β^{5}, 0, 0, 0, 3 β^{9}, 0, 0, 0, 0, 0, 2 β^{15},

0, 0, 0, 0, 0, 3 β^{21}, 0, 0, 0, −5 β^{25}, 0, −3 β^{27}, 0, 0, 0, 0, 0, 3 β^{33}, 0, 5 β^{35}, 0, 0, 0, −3 β^{39}, 0, 0, 0, 0, 0, −2 β^{45}, 0, 0, 0, 0, 0, −3 β^{51},

0, 0, 0, 5 β^{55}, 0, 3 β^{57}, 0, 0, 0)).

(35)

Theorem 14. If r is a positive even integer, then

arctan \frac{1}{L_{r}} = P (1, α^{4(r^{2}−1)}, 4(r^{2} − 1), (a_{1}, a_{2}, …, a_{4(r^{2}−1)})),

where the only non-zero constants a_{j} are given by

a_{(r−1)(4j−3)} = −β^{(r−1)(4j−3)}(r − 1), j = 1, 2, …, r + 1,
a_{(r−1)(4j−1)} = β^{(r−1)(4j−1)}(r − 1), j = 1, 2, …, r + 1,
a_{(r+1)(4j−3)} = β^{(r+1)(4j−3)}(r + 1), j = 1, 2, …, r − 1,
a_{(r+1)(4j−1)} = −β^{(r+1)(4j−1)}(r + 1), j = 1, 2, …, r − 1,
a_{(r−1)(r+1)} = (−1)^{(r+2)/2} 2r β^{r^{2}−1}
a_{3(r−1)(r+1)} = (−1)^{r/2} 2r β^{3(r^{2}−1)}.

Proof. Setting m = 1 in (8) gives

arctan \frac{1}{L_{r}} = arctan \frac{1}{α^{r−1}} − arctan \frac{1}{α^{r+1}} , r even.

The proof now proceeds as in that of Theorem 10. Example 15.

arctan \frac{1}{L_{2}} = arctan \frac{1}{3} = (1, α^{12}, 12, (−β, 0, 4 β^{3}, 0, −β^{5}, 0, β^{7}, 0, −4 β^{9}, 0, β^{11}, 0)), (36)
arctan \frac{1}{L_{4}} = arctan \frac{1}{7} = (1, α^{60}, 60, (0, 0, −3 β^{3}, 0, 5 β^{5}, 0, 0, 0, 3 β^{9}, 0, 0, 0, 0, 0, −8 β^{15},
0, 0, 0, 0, 0, 3 β^{21}, 0, 0, 0, 5 β^{25}, 0, −3 β^{27}, 0, 0, 0, 0, 0, 3 β^{33},
0, −5 β^{35}, 0, 0, 0, −3 β^{39}, 0, 0, 0, 0, 0, 8 β^{45}, 0, 0, 0, 0, 0,
− 3 β^{51}, 0, 0, 0, −5 β^{55}, 0, 3 β^{57}, 0, 0, 0)).

Theorem 16. If r is an integer, then

∑_{k=0}^{∞} \frac{1}{α^{12rk}} (\frac{β^{r}}{12k + 1} + \frac{2β^{3r}}{12k + 3} + \frac{β^{5r}}{12k + 5} − \frac{β^{7r}}{12k + 7} − \frac{2β^{9r}}{12k + 9} − \frac{β^{11r}}{12k + 11})
= \begin{cases}− arctan (\frac{1}{L_{r}} ) , & r odd; \\ arctan (\frac{1}{F_{r}\sqrt{5}}) , & r even; \end{cases}

(37)

that is,

P (1, α^{12r}, 12, (β^{r}, 0, 2β^{3r}, 0, β^{5r}, 0, −β^{7r}, 0, −2β^{9r}, 0, −β^{11r}, 0))
= \begin{cases}− arctan (\frac{1}{L_{r}} ) , & r odd; \\ arctan (\frac{1}{F_{r}\sqrt{5}}) , & r even. \end{cases}

Proof. In an earlier work [1, Identity (27)], we showed that

n^{2}\sqrt{n} arctan (\frac{\sqrt{n}}{n − 1}) = ∑_{k=0}^{∞} \frac{1}{(−n^{3})^{k}} (\frac{n^{2}}{6k + 1} + \frac{2n}{6k + 3} + \frac{1}{6k + 5}).

In base n^{6}, length 12, this is

\begin{matrix}n^{2}\sqrt{n} arctan (\frac{\sqrt{n}}{n − 1}) \\ = ∑_{k=0}^{∞} \frac{1}{n^{6k}} (\frac{n^{2}}{12k + 1} + \frac{2n}{12k + 3} + \frac{1}{12k + 5} − \frac{1/n}{12k + 7} − \frac{2/n^{2}}{12k + 9} − \frac{1/n^{3}}{12k + 11}). \end{matrix} (38)

Identity (37) follows upon setting n = α^{2r} in (38) and making use of (29) and (30).

Example 17.

\frac{π}{4} = P (1, α^{12}, 12, (−β, 0, −2β^{3}, 0, −β^{5}, 0, β^{7}, 0, 2β^{9}, 0, β^{11}, 0)),  (39)

arctan (\frac{1}{4}) = P (1, α^{12}, 12, (−β^{3}, 0, −2β^{9}, 0, −β^{15}, 0, β^{21}, 0, 2β^{27}, 0, β^{33}, 0)),

arctan (\frac{1}{\sqrt{5}}) = P (1, α^{12}, 12, (β^{2}, 0, 2β^{6}, 0, β^{10}, 0, −β^{14}, 0, −2β^{18}, 0, −β^{22}, 0)),

arctan (\frac{1}{3\sqrt{5}}) = P (1, α^{12}, 12, (β^{4}, 0, 2β^{12}, 0, β^{20}, 0, −β^{28}, 0, −2β^{36}, 0, −β^{44}, 0)).

Remark 18. Identity (39) is the same golden ratio base expansion of π that was obtained in Theorem 12. Theorem 19. If r is an integer, then

∑_{k=0}^{∞} \frac{1}{α^{4rk}} (\frac{2β^{r}}{4k + 1} − \frac{2β^{3r}}{4k + 3}) = \begin{cases}− arctan (\frac{2}{L_{r}} ) , & r odd; \\ arctan (\frac{2}{F_{r}\sqrt{5}}) , & r even; \end{cases}  (40)

that is

P (1, α^{4r}, 4, (2β^{r}, 0, −2β^{3r}, 0)) = \begin{cases}− arctan (\frac{2}{L_{r}} ) , & r odd; \\ arctan (\frac{2}{F_{r}\sqrt{5}}) , & r even. \end{cases}

Proof. Setting x = β^{r} in the identity

2 arctan x = arctan (\frac{2x}{1 − x^{2}} )

and using (29) and (30), we have

2 arctan \frac{1}{α^{r}} = \begin{cases}arctan (\frac{2}{L_{r}} ) , & r odd; \\ arctan (\frac{2}{F_{r}\sqrt{5}}) , & r even. \end{cases}  (41)

Setting n = α^{2r} in (32) and comparing with (41), we obtain (40). Example 20.

arctan \frac{2}{L_{3}} = arctan \frac{1}{2} = P (1, α^{12}, 4, (−2β^{3}, 0, 2β^{9}, 0)),  (42)

arctan (\frac{2}{F_{2}\sqrt{5}}) = arctan \frac{2}{\sqrt{5}} = P (1, α^{8}, 4, (2β^{2}, 0, −2β^{6}, 0)).

4 Zero relations

Zero relations are expansion formulas that evaluate to zero. They are useful in the determination and classification of new expansion formulas. A base-α expansion is not considered new if it can be written as a linear combination of existing formulas and known zero relations.

4.1 Zero relations arising from the logarithm formulas

4.1.1 Zero relation from log(F_{3}^{2}/L_{3}) = 0

Theorem 21. We have

∑_{k=0}^{∞} \frac{1}{α^{12k}} (\frac{1}{6k + 1} − \frac{3β^{2}}{6k + 2} − \frac{8β^{4}}{6k + 3} − \frac{3β^{6}}{6k + 4} + \frac{β^{8}}{6k + 5}) = 0;

that is,

0 = P (1, α^{12}, 6, (1, −3β^{2}, −8β^{4}, −3β^{6}, β^{8}, 0)).

Proof. We have

2 log F_{3} − log L_{3} = 0.  (43)

The expansion of log L_{3} given in (27) has the following base-α^{12}, length-6 version:

log L_{3} = P (1, α^{12}, 6, (3β^{2}, 3β^{4}, 0, 3β^{8}, 3β^{10}, 0)).  (44)

Use of (19) and (44) in (43) yields the zero relation stated in Theorem 21.

4.1.2 Zero relation from log(L_{6}/(F_{4}^{2}F_{3})) = 0

Theorem 22. We have

0 = P (1, α^{48}, 24, (1, −5 β^{2}, −2 β^{4}, 3 β^{6}, β^{8}, 4 β^{10}, β^{12}, 3 β^{14},
− 2 β^{16}, −5 β^{18}, β^{20}, 0, β^{24}, −5 β^{26}, −2 β^{28}, 3 β^{30}, β^{32},
4 β^{34}, β^{36}, 3 β^{38}, −2 β^{40}, −5 β^{42}, β^{44}, 0)).

Proof. Write log F_{3}, log F_{4} and log L_{6}, that is, identities (19), (21) and (28), respectively, in the common base α^{48} and common length 24 and use

log L_{6} − 2 log F_{4} − log F_{3} = 0.

4.1.3 Zero relation from log(F_{12}/(F_{3}^{4}L_{2}^{2})) = 0 Theorem 23. We have

\begin{aligned}0 & = P (1, α^{48}, 24, (1, −4 β^{2}, −5 β^{4}, 0, β^{8}, 2 β^{10}, β^{12}, 0, −5 β^{16}, −4 β^{18}, β^{20}, 0, \\ & β^{24}, −4 β^{26}, −5 β^{28}, 0, β^{32}, 2 β^{34}, β^{36}, 0, −5 β^{40}, −4 β^{42}, β^{44}, 0)). \end{aligned}

Proof. Write log F_{3}, log F_{12} and log L_{2} from (19), (22) and (26), in common base α^{48} and consider

log F_{12} − 4 log F_{3} − 2 log L_{2} = 0.

4.2 Zero relations arising from the inverse tangent formulas

4.2.1 Zero relation from 2 arctan(2/L_{3}) + arctan(2/L_{5}) − arctan(2/L_{1}) = 0 Theorem 24. We have

0 = P (1, α^{60}, 60, (1, 0, −7 β^{2}, 0, −4 β^{4}, 0, −β^{6}, 0, 7 β^{8}, 0, −β^{10}, 0, β^{12}, 0, −2 β^{14},
0, β^{16}, 0, −β^{18}, 0, 7 β^{20}, 0, −β^{22}, 0, −4 β^{24}, 0, −7 β^{26}, 0, β^{28}, 0, −β^{30}, 0,
7 β^{32}, 0, 4 β^{34}, 0, β^{36}, 0, −7 β^{38}, 0, β^{40}, 0, −β^{42}, 0, 2 β^{44}, 0, −β^{46}, 0,
β^{48}, 0, −7 β^{50}, 0, β^{52}, 0, 4 β^{54}, 0, 7 β^{56}, 0, −β^{58}, 0)).

Proof. Using the addition and subtraction formulas for inverse tangents, it is easy to verify that

arctan (\frac{2}{L_{1}} ) − arctan (\frac{2}{L_{3}} ) = arctan (\frac{3}{4})

and

arctan (\frac{2}{L_{5}} ) + arctan (\frac{2}{L_{3}} ) = arctan (\frac{3}{4});

so that

2 arctan (\frac{2}{L_{3}} ) + arctan (\frac{2}{L_{5}} ) − arctan (\frac{2}{L_{1}} ) = 0,

from which the zero relation follows upon use of (40). Remark 25. The zero relation stated in Theorem 24 can also be obtained directly from

arctan \frac{1}{F_{3}} = arctan \frac{2}{L_{3}} ,

by writing (34) and (42) in the common base α^{60} and common length 60; or from

arctan \frac{1}{F_{4}} = arctan \frac{1}{L_{2}} ,

using (35) and (36).

4.2.2 Zero relation from 2 arctan(1/L_{1}) − 2 arctan(2/(F_{2}\sqrt{5})) − arctan(2/(F_{6}\sqrt{5})) = 0 Theorem 26. We have

\begin{aligned}0 & = P (1, α^{24}, 24, (1, 4 β, 2 β^{2}, 0, β^{4}, 2 β^{5}, −β^{6}, 0, −2 β^{8}, 4 β^{9}, −β^{10}, 0, β^{12}, −4 β^{13}, \\ & 2 β^{14}, 0, β^{16}, −2 β^{17}, −β^{18}, 0, −2 β^{20}, −4 β^{21}, −β^{22}, 0)). \end{aligned}

Proof. The identity

\frac{π}{2} − arctan (\frac{2}{\sqrt{5}}) = arctan (\frac{1}{4\sqrt{5}}) + arctan (\frac{2}{\sqrt{5}})

= arctan (\frac{\sqrt{5}}{2})

can be arranged as

2 arctan (\frac{1}{L_{1}} ) − 2 arctan (\frac{2}{F_{2}\sqrt{5}}) − arctan (\frac{2}{F_{6}\sqrt{5}}) = 0

which, on account of (40), gives the stated zero relation.

5 Other degree 1 base-α expansions and zero relations

5.1 Base-α expansions of log α

Theorem 27.

log α = P (1, α, 2, (0, −β)).  (45)

Proof. We have

log α = \frac{1}{2} Li_{1} (\frac{1}{α}) = \frac{1}{2} ∑_{k=0}^{∞} \frac{1}{α^{k}} \frac{1/α}{k + 1} = ∑_{k=0}^{∞} \frac{1}{α^{k}} \frac{−β}{2k + 2}.

Theorem 28.

log α = P (1, α^{2}, 2, (0, 2β^{2})).  (46)

Proof. We have

log α = Li_{1} (\frac{1}{α^{2}} ) = ∑_{k=0}^{∞} \frac{1}{α^{2k}} \frac{1/α^{2}}{k + 1} = ∑_{k=0}^{∞} \frac{1}{α^{2k}} \frac{2β^{2}}{2k + 2}.

5.2 Another base-α expansion of log 2

Theorem 29.

log 2 = P (1, α^{3}, 3, (−β, β^{2}, 2β^{3})).  (47)

Proof. A straightforward consequence of the identity

log 2 = Li_{1} (\frac{1}{α}) − Li_{1} (\frac{1}{α^{3}} ) .

5.3 Another base-α expansion of log 5

Theorem 30.

log 5 = P (1, α^{4}, 2, (4 β^{2}, 0)).  (48)

Proof. A consequence of the identity

log 5 = 2 Li_{1} (\frac{1}{α^{2}} ) − 2 Li_{1} (−\frac{1}{α^{2}} ) .

5.4 A length 2, base-α zero relation

Theorem 31.

P (1, α^{2}, 2, (1, 3 β)) = 0.  (49)

Proof. Follows from

Li_{1} (\frac{1}{α^{2}} ) + Li_{1} (−\frac{1}{α}) = 0.

Remark 32. Relation (49) also follows from (45) and (46).

5.5 A length 12, base-α zero relation

Theorem 33.

P (1, α^{12}, 12, (1, β, −2 β^{2}, 5 β^{3}, β^{4}, 10 β^{5}, β^{6}, 5 β^{7}, −2 β^{8}, β^{9}, β^{10}, 2 β^{11})) = 0. Proof. Follows fron (19) and (47).

5.6 A length 10, base-α zero relation

Theorem 34.

P (1, α^{20}, 10, (1, −5 β^{2}, β^{4}, −5 β^{6}, −4 β^{8}, −5 β^{10}, β^{12}, −5 β^{14}, β^{16}, 0)) = 0.

Proof. Ensues from (20) and (48).

5.7 A length 5, base-α zero relation

Theorem 35.

P (1, α^{5}, 5, (β, 1, −β, −β^{4}, −2 β^{4})) = 0.  (50)

Proof. Setting p = 2 cos x in the identity

∑_{k=1}^{∞} \frac{p^{k} cos(kx)}{k} = −\frac{1}{2} log(1 − 2p cos x + p^{2})

produces

∑_{k=1}^{∞} \frac{(2 cos x)^{k} cos(kx)}{k} = 0.  (51)

Now 2 cos(2π/5) = −β.

Thus, setting x = 2π/5 in (51) gives

∑_{k=0}^{∞} \frac{1}{α^{5k}} (\frac{β}{5k + 1} + \frac{1}{5k + 2} − \frac{β}{5k + 3} − \frac{β^{4}}{5k + 4} − \frac{2β^{4}}{5k + 5}) = 0;

since

cos (\frac{2π}{5} (5j − 4)) = \frac{−β}{2} = cos (\frac{2π}{5} (5j − 1)) , j = 1, 2, …

and

cos (\frac{2π}{5} (5j − 2)) = \frac{1}{2β} = cos (\frac{2π}{5} (5j − 3)) , j = 1, 2, …

6 Acknowledgments

We thank the anonymous referee for a speedy, thorough and careful review in which several useful suggestions that helped improve the paper were offered.

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2020 Mathematics Subject Classification: Primary 11Y60; Secondary 30B99. Keywords: Fibonacci number, Lucas number, logarithm, arctangent, inverse tangent, golden ratio, non-integer base expansion, BBP-type formula.

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

Received March 19 2023; revised version received March 30 2023. Published in Journal of Integer Sequences, June 6 2023.

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