Reciprocal Sum of Palindromes

Phakhinkon Phunphayap and Prapanpong Pongsriiam^{1} Department of Mathematics Faculty of Science Silpakorn University Nakhon Pathom 73000 Thailand

                     phakhinkon@gmail.com
        pongsriiam p@silpakorn.edu, prapanpong@gmail.com

Abstract

A positive integer n is a b-adic palindrome if the representation of n in base b reads the same backward as forward. Let s_{b} be the reciprocal sum of all b-adic palindromes. In this article, we obtain upper and lower bounds, and an asymptotic formula for s_{b}. We also show that the sequence (s_{b})_{b≥2} is strictly increasing and log-concave.

1 Introduction

Let n ≥ 1 and b ≥ 2 be integers. We call n a palindrome in base b (or b-adic palindrome) if the b-adic expansion of n = (a_{k}a_{k−1} ⋯ a_{0})_{b} with a_{k} ≠ 0 has the symmetric property a_{k−i} = a_{i} for 0 ≤ i ≤ ⌊\frac{k}{2}⌋. As usual, if we write a number without specifying the base, then it is always in base 10. So, for example, 9 = (1001)_{2} = (100)_{3} is a palindrome in bases 2 and 10 but not in base 3.

In recent years, there has been an increasing interest in the importance of palindromes in mathematics [1, 2, 3, 13, 17, 25], theoretical computer science [4, 9, 12], and theoretical physics [11, 14]. There are also some discussions on the reciprocal sum of palindromes on the internet but as far as we are aware, our observation has not appeared in the literature. Throughout this article, we let b ≥ 2, s_{b} the reciprocal sum of all b-adic palindromes, and s_{b,k} the reciprocal sum of all b-adic palindromes which have k digits in their b-adic expansions.

1Prapanpong Pongsriiam is the corresponding author.

The set of all b-adic palindromes is infinite but quite sparse, so it is not difficult to see that s_{b} converges. In fact, Shallit proposed the convergence of s_{b} as a problem in the Fibonacci Quarterly in 1980 [26, 27].

In this article, we obtain upper and lower bounds for s_{b}, which enable us to show that s_{b+1} > s_{b} for all b ≥ 2 and s_{b}^{2} − s_{b−1}s_{b+1} > 0 for all b ≥ 3. That is, the sequence (s_{b})_{b≥2} is strictly increasing and log-concave. Furthermore, we give an asymptotic formula for s_{b} of the form s_{b} = g(b) + O(h(b)) where the implied constant can be taken to be 1 and the order of magnitude of h(b) is \frac{log b}{b^{3}} as b → ∞. Our result s_{b+1} > s_{b} for all b ≥ 2 also implies that if b_{1} > b_{2} ≥ 2 and if we use the logarithmic measure, then we can say that the palindromes in base b_{1} occur more often than those in base b_{2}. On the other hand, if we use the usual counting measure, then we obtain from Pongsriiam and Subwattanachai’s exact formula [22] that the number of palindromes in different bases which are less than or equal to N are not generally comparable. It seems that there are races between palindromes in different bases which may be similar to races between primes in different residue classes. We will get back to this problem in the near future.

The reciprocal sum of an integer sequence is also of general interest in mathematics and theoretical physics as proposed by Bayless and Klyve [8], and by Roggero, Nardelli, and Di Noto [24]. See also the work of Nguyen and Pomerance [19] on the reciprocal sum of the amicable numbers, the preprint of Kinlaw, Kobayashi, and Pomerance [15] on the reciprocal sum of the positive integers n satisfying φ(n) = φ(n + 1), and the article by Lichtman [16] on the reciprocal sum of primitive nondeficient numbers. In addition, Banks [5], Cilleruelo, Luca, and Baxter [10], and Rajasekaran, Shallit, and Smith [23] have recently investigated some additive properties of palindromes while Banks, Hart, and Sakata [6] and Banks and Shparlinski [7] show some of their multiplicative properties. For more information concerning palindromes, we refer the reader to the entry \underline{A002113} in the On-Line Encyclopedia of Integer Sequences (OEIS) [28].

2 Results

Throughout this section, a, c, m, n, k, ℓ denote positive integers, and x, y, z denote positive real numbers. Furthermore,

• ⌊x⌋ is the greatest integer less than or equal to x;

• ⌈x⌉ is the least integer greater than or equal to x;

• log x is the natural logarithm of x;

• x_{b} = ∑_{m=1}^{b−1} \frac{1}{m};

• y_{b} = ∑_{m=b}^{b^{2}−1} \frac{1}{m}; and

• z_{b} = ∑_{m=b+1}^{b^{2}} \frac{1}{m}.

Note that x_{b} = s_{b,1} and x_{b}/(b + 1) = s_{b,2} and that

z_{b} − y_{b} = \frac{1}{b^{2}} − \frac{1}{b} = \frac{1 − b}{b^{2}}.

Theorem 1. We have

\frac{y_{b}}{b} − \frac{x_{b}}{b^{3}} ≤ s_{b,3} ≤ \frac{y_{b}}{b} and \frac{z_{b}}{b^{⌊^{\frac{k}{2}}⌋}} ≤ s_{b,k} ≤ \frac{y_{b}}{b^{⌊^{\frac{k}{2}}⌋}} for every k ≥ 4.

Proof. We first consider the case k = 3. The b-adic palindromes which have 3 digits are of the form (aca)_{b} where 1 ≤ a ≤ b − 1 and 0 ≤ c ≤ b − 1. Since

|\frac{1}{(aca)_{b}} − \frac{1}{(ac0)_{b}} | = |\frac{−a}{(ab^{2} + cb + a)(ab^{2} + cb)}| ≤ \frac{a}{(ab^{2})^{2}} = \frac{1}{ab^{4}} ,

we obtain

\frac{1}{(ac0)_{b}} − \frac{1}{ab^{4}} ≤ \frac{1}{(aca)_{b}} ≤ \frac{1}{(ac0)_{b}} .  (1)

Observe that

∑_{\begin{matrix}1≤a≤b−1 \\ 0≤c≤b−1 \end{matrix}} \frac{1}{(ac0)_{b}} = \frac{1}{b} ∑_{\begin{matrix}1≤a≤b−1 \\ 0≤c≤b−1 \end{matrix}} \frac{1}{ab + c} = \frac{y_{b}}{b} and ∑_{\begin{matrix}1≤a≤b−1 \\ 0≤c≤b−1 \end{matrix}} \frac{1}{ab^{4}} = \frac{x_{b}}{b^{3}} .

So by summing (1) over all a = 1, 2, …, b−1 and c = 0, 1, …, b−1, we obtain the inequality \frac{y_{b}}{b} − \frac{x_{b}}{b^{3}} ≤ s_{b,3} ≤ \frac{y_{b}}{b}. For k = 4, 1 ≤ a ≤ b − 1, and 0 ≤ c ≤ b − 1, we have

\frac{1}{b^{2}(ab + c + 1)} = \frac{1}{(ac00)_{b} + b^{2}} ≤ \frac{1}{(acca)_{b}} ≤ \frac{1}{(ac00)_{b}} = \frac{1}{b^{2}(ab + c)}.

Summing over all a = 1, 2, …, b − 1 and c = 0, 1, …, b − 1 leads to

\frac{z_{b}}{b^{2}} = \frac{1}{b^{2}} ∑_{\begin{matrix}1≤a≤b−1 \\ 0≤c≤b−1 \end{matrix}} \frac{1}{ab + c + 1} ≤ s_{b,4} ≤ \frac{1}{b^{2}} ∑_{\begin{matrix}1≤a≤b−1 \\ 0≤c≤b−1 \end{matrix}} \frac{1}{ab + c} = \frac{y_{b}}{b^{2}} .

Let k ≥ 5. The b-adic palindromes which have k digits are of the form (aa_{1}a_{2} ⋯ a_{k−2}a)_{b} where 1 ≤ a ≤ b − 1, 0 ≤ a_{i} ≤ b − 1 for all i ∈ {1, 2, …, k − 2} with the usual symmetric property on a_{i}. We fix a and a_{1} and count the number of palindromes in this form. There are b choices for a_{2} ∈ {0, 1, 2, …, b−1} and so there is only 1 choice for a_{k−3} = a_{2}. Similarly, there are b choices for a_{3} and 1 choice for a_{k−4}. By continuing this counting, we see that the number of palindromes in this form (when a and a_{1} are already chosen) is equal to b^{⌈^{\frac{k−4}{2}}⌉}. Therefore the reciprocal sum of such palindromes satisfies

∑_{a_{2},...,a_{k−3}} \frac{1}{(aa_{1}a_{2} ⋯ a_{k−2}a)_{b}} ≤ \frac{b^{⌈^{\frac{k−4}{2}}⌉}}{(aa_{1}0 ⋯ 0a_{1}a)_{b}} ≤ \frac{b^{⌈^{\frac{k−4}{2}}⌉}}{b^{k−2}(ab + a_{1})} = \frac{1}{b^{⌊^{\frac{k}{2}}⌋}(ab + a_{1})},

where a_{2}, …, a_{k−3} run over all integers 0, 1, 2, …, b − 1 with the symmetric condition of palindromes. Hence

s_{b,k} = ∑_{\begin{matrix}1≤a≤b−1 \\ 0≤a_{1}≤b−1 \end{matrix}} ∑_{a_{2},...,a_{k−3}} \frac{1}{(aa_{1}a_{2} ⋯ a_{k−2}a)_{b}} ≤ ∑_{\begin{matrix}1≤a≤b−1 \\ 0≤a_{1}≤b−1 \end{matrix}} \frac{1}{b^{⌊^{\frac{k}{2}}⌋}(ab + a_{1})} = \frac{y_{b}}{b^{⌊^{\frac{k}{2}}⌋}} .

Similarly, if a and a_{1} are fixed, then

∑_{a_{2},...,a_{k−3}} \frac{1}{(aa_{1}a_{2} ⋯ a_{k−2}a)_{b}} ≥ \frac{b^{⌈^{\frac{k−4}{2}}⌉}}{ab^{k−1} + a_{1}b^{k−2} + b^{k−2}} = \frac{1}{b^{⌊^{\frac{k}{2}}⌋}(ab + a_{1} + 1)}.

Summing the above over all a = 1, 2, …, b − 1 and a_{1} = 0, 1, …, b − 1, we obtain the desired lower bound for s_{b,k}. This completes the proof.

Theorem 2. For every b, ℓ ≥ 2, we have

(\frac{b + 2}{b + 1}) x_{b} + ∑_{k=3}^{2ℓ−1} s_{b,k} + \frac{2z_{b}}{(b − 1)b^{ℓ−1}} ≤ s_{b} ≤ (\frac{b + 2}{b + 1}) x_{b} + ∑_{k=3}^{2ℓ−1} s_{b,k} + \frac{2y_{b}}{(b − 1)b^{ℓ−1}} .

In particular,

(\frac{b + 2}{b + 1}) x_{b} + \frac{y_{b}}{b} − \frac{x_{b}}{b^{3}} + \frac{2z_{b}}{b(b − 1)} ≤ s_{b} ≤ (\frac{b + 2}{b + 1}) x_{b} + (\frac{1}{b} + \frac{2}{b(b − 1)}) y_{b}.  (2)

Proof. For simplicity, we write x, y, z instead of x_{b}, y_{b}, z_{b}, respectively. We consider s_{b,k} for each k as follows. Obviously s_{b,1} = 1 + \frac{1}{2} + ⋯ + \frac{1}{b−1} = x. For k = 2, s_{b,k} is

∑_{a=1}^{b−1} \frac{1}{(aa)_{b}} = ∑_{a=1}^{b−1} \frac{1}{a(b + 1)} = \frac{x}{b + 1}.

By writing s_{b} = x + \frac{x}{b+1} + ∑_{k=3}^{2ℓ−1} s_{b,k} + ∑_{k=2ℓ}^{∞} s_{b,k} and applying Theorem 1, we obtain

s_{b} ≤ \frac{b + 2}{b + 1}x + ∑_{k=3}^{2ℓ−1} s_{b,k} + ∑_{k=2ℓ}^{∞} \frac{y}{b^{⌊^{\frac{k}{2}}⌋}} = \frac{b + 2}{b + 1}x + ∑_{k=3}^{2ℓ−1} s_{b,k} + \frac{2y}{(b − 1)b^{ℓ−1}} .

Similarly,

s_{b} ≥ \frac{b + 2}{b + 1}x + ∑_{k=3}^{2ℓ−1} s_{b,k} + ∑_{k=2ℓ}^{∞} \frac{z}{b^{⌊^{\frac{k}{2}}⌋}} = (\frac{b + 2}{b + 1}) x + ∑_{k=3}^{2ℓ−1} s_{b,k} + \frac{2z}{(b − 1)b^{ℓ−1}} .

This proves the first part of this theorem. The second part follows from Theorem 1 and the substitution ℓ = 2 in the first part.

Theorem 3. The sequence (s_{b})_{b≥2} is strictly increasing.

Proof. We first verify that s_{b+1} > s_{b} for 2 ≤ b ≤ 16. Myers gives the decimal expansion of s_{2} in the entry \underline{A244162} in the OEIS [28]. Myers also describe the algorithm in his calculation, which can be found in the web page [18]. So we know that s_{2} < 2.3787957. Alternatively, substituting ℓ = 3 and b = 2, 3 in Theorem 2 and running the computation in a computer, we obtain 2.32137259 ≤ s_{2} ≤ 2.44637260 and 2.60503980 ≤ s_{3} ≤ 2.62973117, which implies that s_{2} < s_{3}. Similarly, we apply (2) to obtain upper and lower bounds for s_{b} and we see that s_{b} < s_{b+1} for 3 ≤ b ≤ 16. So we assume throughout that b ≥ 16. We first observe that

y_{b} = z_{b} + \frac{1}{b} − \frac{1}{b^{2}} and \frac{y_{b}}{b} − \frac{z_{b}}{b} = \frac{b − 1}{b^{3}} ≥ \frac{x_{b}}{b^{3}} .

Therefore \frac{y_{b}}{b} − \frac{x_{b}}{b^{3}} ≥ \frac{z_{b}}{b} for all b ≥ 2. So the term \frac{y_{b}}{b} − \frac{x_{b}}{b^{3}} in (2) can be replaced by \frac{z_{b}}{b}. Therefore we obtain by (2) that s_{b+1} − s_{b} is larger than

(\frac{b + 3}{b + 2}) x_{b+1} − (\frac{b + 2}{b + 1}) x_{b} + (\frac{1}{b + 1} + \frac{2}{(b + 1)b}) z_{b+1} − (\frac{1}{b} + \frac{2}{b(b − 1)}) y_{b}. (3)

In addition, z_{b+1} − z_{b} is equal to

−\frac{1}{b + 1} + ∑_{m=b^{2}+1}^{b^{2}+2b+1} \frac{1}{m} ≥ −\frac{1}{b + 1} + \frac{2b + 1}{b^{2} + 2b + 1} = \frac{b}{(b + 1)^{2}} > 0.

Since x_{b+1} = x_{b} + \frac{1}{b} and z_{b+1} > z_{b} = y_{b} − \frac{1}{b} + \frac{1}{b^{2}} , we obtain from (3) that s_{b+1} − s_{b} is larger than or equal to

(\frac{b + 3}{b + 2}) (\frac{1}{b}) + x_{b} (\frac{b + 3}{b + 2} − \frac{b + 2}{b + 1}) + y_{b} (\frac{1}{b + 1} + \frac{2}{b(b + 1)} − \frac{1}{b} − \frac{2}{b(b − 1)})

+ (\frac{1}{b^{2}} − \frac{1}{b}) (\frac{1}{b + 1} + \frac{2}{b(b + 1)})

= \frac{1}{b} + \frac{1}{b(b + 2)} − \frac{x_{b}}{(b + 1)(b + 2)} − \frac{(b + 3)y_{b}}{b(b − 1)(b + 1)} − \frac{(b − 1)(b + 2)}{b^{3}(b + 1)}.

Recall that if a and b are integers, a < b, and f is monotone on [a, b], then

min{f(a), f(b)} ≤ ∑_{n=a}^{b} f(n) − ∫_{a}^{b} f(t)dt ≤ max{f(a), f(b)}.  (4)

From (4), we obtain

x_{b} = ∑_{m=1}^{b−1} \frac{1}{m} ≤ 1 + log(b − 1) ≤ \frac{3}{2} log b,

y_{b} = −\frac{1}{b − 1} + ∑_{m=b−1}^{b^{2}−1} \frac{1}{m} ≤ log(b + 1) ≤ \frac{5}{4} log b.

In addition, it is straightforward to verify that

\frac{1}{b(b + 2)} − \frac{(b − 1)(b + 2)}{b^{3}(b + 1)} > −\frac{1}{b^{2}} .

Therefore s_{b+1} − s_{b} is larger than

\frac{1}{b} − \frac{3 log b}{2(b + 1)(b + 2)} − \frac{5(b + 3) log b}{4b(b − 1)(b + 1)} − \frac{1}{b^{2}} > \frac{1}{b} − \frac{3 log b}{2b^{2}} − \frac{3 log b}{2b^{2}} − \frac{1}{b^{2}} = \frac{1}{b} − \frac{1}{b^{2}} − \frac{3 log b}{b^{2}}.

(5)

Observe that the function x ↦ \frac{log x}{x} is decreasing on [3, ∞). Since b ≥ 16, we obtain

\frac{3 log b}{b} ≤ \frac{3 log 16}{16} < \frac{7}{10} and \frac{1}{b} < \frac{1}{10}.

Hence we obtain from (5) that

s_{b+1} − s_{b} > \frac{1}{b} − \frac{1}{b^{2}} − \frac{3 log b}{b^{2}} > \frac{1}{b} − \frac{1}{10b} − \frac{7}{10b} = \frac{1}{5b} > 0.

This completes the proof.

Recall that if we write f(b) = g(b) + O^{∗}(h(b)), then it means that f(b) = g(b) + O(h(b)) and the implied constant can be taken to be 1. In addition, f(b) = g(b) + Ω_{+}(h(b)) means lim sup_{b→∞} \frac{f(b)−g(b)}{h(b)} > 0. From this point on, we use (4) without further reference.

Theorem 4. Uniformly for b ≥ 2,

s_{b} = (\frac{b + 2}{b + 1}) x_{b} + (\frac{1}{b} + \frac{2}{b^{2}} ) y_{b} + O^{∗} (\frac{5 log b}{b^{3}}) .

This estimate is sharp in the sense that O^{∗} (\frac{5 log b}{b^{3}}) can be replaced by Ω_{+} (\frac{log b}{b^{3}}).

Proof. Let g(b) = (\frac{b+2}{b+1}) x_{b} + (\frac{1}{b} + \frac{2}{b^{2}} ) y_{b} be the main term above. Since y_{b} ≤ log(b + 1), we obtain by (2) that

s_{b} − g(b) ≤ \frac{2y_{b}}{b^{2}(b − 1)} ≤ \frac{2 log(b + 1)}{b^{2}(b − 1)}.

If b = 2, then it is easy to check that \frac{2y_{b}}{b^{2}(b−1)} = \frac{5}{12} < \frac{5 log b}{b^{3}}. If b ≥ 3, then we assert that

\frac{2 log(b+1)}{b^{2}(b−1)} ≤ \frac{5 log b}{b^{3}}. To verify this assertion, we observe that it is equivalent to (\frac{5}{2} log b) (\frac{b−1}{b}) ≥ log(b + 1). Since b ≥ 3, we obtain

(\frac{5}{2} log b) (\frac{b − 1}{b}) ≥ \frac{5}{3} log b ≥ log 2 + log b = log(2b) ≥ log(b + 1), as desired.

So in any case,

s_{b} − g(b) ≤ \frac{5 log b}{b^{3}}.  (6)

We also obtain by (2) that s_{b} − g(b) is larger than or equal to

\frac{2z_{b}}{b(b − 1)} − \frac{2y_{b}}{b^{2}} − \frac{x_{b}}{b^{3}} = \frac{2z_{b}}{b(b − 1)} − \frac{2 (z_{b} + \frac{1}{b} − \frac{1}{b^{2}} )}{b^{2}} − \frac{x_{b}}{b^{3}} = \frac{2z_{b}}{b^{2}(b − 1)} − \frac{x_{b}}{b^{3}} − \frac{2}{b^{3}} + \frac{2}{b^{4}} . (7)

We have

x_{b} = ∑_{m=1}^{b−1} \frac{1}{m} ≤ 1 + log(b − 1) ≤ 2 log b,  (8)

z_{b} = ∑_{m=b}^{b^{2}} \frac{1}{m} − \frac{1}{b} ≥ ∫_{b}^{b^{2}} \frac{1}{t}dt + \frac{1}{b^{2}} − \frac{1}{b} ≥ log b − \frac{1}{b} ≥ \frac{log b}{4}.  (9)

Therefore (7) implies that

s_{b} − g(b) ≥ \frac{log b}{2b^{2}(b − 1)} − \frac{2 log b}{b^{3}} − \frac{2}{b^{3}} ≥ \frac{log b}{2b^{3}} − \frac{2 log b}{b^{3}} − \frac{3 log b}{b^{3}} > −\frac{5 log b}{b^{3}}. (10)

By (6) and (10), we obtain |s_{b} − g(b)| ≤ \frac{5 log b}{b^{3}}. This proves the first part of this theorem. For the Ω_{+} result, we only need to observe that as b → ∞, (7) and the inequalities x_{b} ≤ 1 + log(b − 1) and z_{b} ≥ log b − \frac{1}{b} given in (8) and (9) imply that

s_{b} − g(b) ≥ \frac{2 (log b − \frac{1}{b})}{b^{2}(b − 1)} − \frac{1 + log(b − 1)}{b^{3}} − \frac{2}{b^{3}} + \frac{2}{b^{4}} > \frac{3 log b}{2b^{2}(b − 1)} − \frac{11 log b}{10b^{3}} − \frac{2}{b^{3}} + \frac{2}{b^{4}} ,

so

lim sup_{b→∞} \frac{s_{b} − g(b)}{(\frac{log b}{b^{3}})} ≥ \frac{3}{2} − \frac{11}{10} > 0.

This completes the proof.

Recall that by applying Euler-Maclaurin summation formula, we get

∑_{m≤n} \frac{1}{m} = log n + γ + \frac{1}{2n} − \frac{1}{12n^{2}} + \frac{θ_{n}}{60n^{4}} ,  (11)

where γ is Euler’s constant and θ_{n} ∈ [0, 1]. The calculation of (11) can be found in Tenenbaum [30, p. 6]. From this, we obtain another form of Theorem 4 as follows.

Theorem 5. Uniformly for b ≥ 2,

s_{b} = log b + γ + (\frac{1}{b} + \frac{1}{b + 1}) log b + \frac{γ}{b + 1} − \frac{1}{2b} + \frac{2 log b}{b^{2}} − \frac{1}{12b(b + 1)} + O^{∗} (\frac{6 log b}{b^{3}}) . (12)

This estimate is sharp in the sense that O^{∗} (\frac{6 log b}{b^{3}}) is also Ω_{+} (\frac{log b}{b^{3}}).

Proof. By (11), we have

x_{b} = ∑_{m≤b} \frac{1}{m} − \frac{1}{b} = log b + γ − \frac{1}{2b} − \frac{1}{12b^{2}} + \frac{θ_{b}}{60b^{4}} ,

z_{b} = ∑_{m≤b^{2}} \frac{1}{m} − ∑_{m≤b} \frac{1}{m} = (log b^{2} + γ + \frac{1}{2b^{2}} − \frac{1}{12b^{4}} + \frac{θ_{b^{2}}}{60b^{8}} ) − (log b + γ + \frac{1}{2b} − \frac{1}{12b^{2}} + \frac{θ_{b}}{60b^{4}} )

= log b − \frac{1}{2b} + \frac{7}{12b^{2}} − \frac{5 + θ_{b}}{60b^{4}} + \frac{θ_{b^{2}}}{60b^{8}} ,

y_{b} = z_{b} + \frac{1}{b} − \frac{1}{b^{2}} = log b + \frac{1}{2b} − \frac{5}{12b^{2}} − \frac{5 + θ_{b}}{60b^{4}} + \frac{θ_{b^{2}}}{60b^{8}} .

Writing \frac{b+2}{b+1} = 1 + \frac{1}{b+1} and substituting x_{b} and y_{b} in Theorem 4, we obtain

s_{b} = h(b) + h_{1}(b) + O^{∗} (\frac{5 log b}{b^{3}})  (13)

where h(b) is the main term given in (12) and

h_{1}(b) = \frac{11b^{2} − 3b − 10}{12b^{4}(b + 1)} + \frac{θ_{b}}{60b^{4}} + \frac{θ_{b}}{60b^{4}(b + 1)} + \frac{θ_{b^{2}}}{60b^{9}} + \frac{θ_{b^{2}}}{30b^{10}} − \frac{5 + θ_{b}}{60b^{5}} − \frac{5 + θ_{b}}{30b^{6}} .

It is not difficult to see that h_{1}(b) ≥ 0 and

h_{1}(b) ≤ \frac{11b^{2}}{12b^{4}(b + 1)} + \frac{1}{60b^{4}} + \frac{1}{60b^{4}(b + 1)} ≤ \frac{11 log b}{12b^{3}} + \frac{log b}{60b^{3}} + \frac{log b}{60b^{3}} ≤ \frac{log b}{b^{3}}.  (14)

Therefore (13) implies that s_{b} = h(b) + O^{∗} (\frac{6 log b}{b^{3}}), which is the same as (12). In addition, by the first inequality given in (14), we see that h_{1}(b) ≪ \frac{1}{b^{3}} . Since O^{∗} (\frac{5 log b}{b^{3}}) in (13) is Ω_{+} (\frac{log b}{b^{3}}) and h_{1}(b) ≪ \frac{1}{b^{3}} , h_{1}(b) + O^{∗} (\frac{5 log b}{b^{3}}) in (13) is Ω_{+} (\frac{log b}{b^{3}}). This completes the proof.

Corollary 6. The sequence (s_{b})_{b≥2} diverges to +∞ and the sequence (s_{b} −s_{b−1})_{b≥3} converges to zero as b → ∞.

Proof. The first assertion follows immediately from Theorem 5. Recall that log(b − 1) = log b + O (\frac{1}{b}). So we obtain by Theorem 5 that as b → ∞,

0 < s_{b} − s_{b−1} = log b − log(b − 1) + O (\frac{log b}{b}) ≪ \frac{log b}{b},

which implies our assertion.

Recall that a sequence (a_{n})_{n≥0} is said to be log-concave if a_{n}^{2} − a_{n−1}a_{n+1} > 0 for every n ≥ 1 and is said to be log-convex if a_{n}^{2} − a_{n−1}a_{n+1} < 0 for every n ≥ 1. For a survey article concerning the log-concavity and log-convexity of sequences, we refer the reader to Stanley [29]. See also Pongsriiam [21] for some combinatorial sequences which are log-concave or log-convex, and some open problems concerning the log-properties of a certain sequence.

Theorem 7. The sequence (s_{b})_{b≥2} is log-concave.

Proof. For each b = 2, 3, …, 15, we use Theorem 2 with ℓ = 4 to get an upper bound C_{b} and a lower bound D_{b} for s_{b}. In addition, for each b ≥ 13, let U_{b} and L_{b} be the upper and lower bounds of s_{b} given in (2), respectively. Then

s_{b}^{2} − s_{b−1}s_{b+1} > D_{b}^{2} − C_{b−1}C_{b+1} for 3 ≤ b ≤ 14 and s_{b}^{2} − s_{b−1}s_{b+1} > L_{b}^{2} − U_{b−1}U_{b+1} for b ≥ 14.

We use MATLAB to check that D_{b}^{2} − C_{b−1}C_{b+1} > 0 for 3 ≤ b ≤ 14 and L_{b}^{2} − U_{b−1}U_{b+1} > 0 for 14 ≤ b ≤ 1500. So s_{b}^{2} − s_{b−1}s_{b+1} > 0 for 3 ≤ b ≤ 1500. So we assume throughout that b > 1500. Then

U_{b−1}U_{b+1} = \frac{(b + 1)(b + 3)}{b(b + 2)}x_{b−1}x_{b+1} + \frac{b + 2}{(b − 2)(b − 1)(b + 1)}y_{b−1}y_{b+1}

+ \frac{b(b + 3)}{(b − 2)(b − 1)(b + 2)}x_{b+1}y_{b−1} + \frac{b + 2}{b^{2}}x_{b−1}y_{b+1}

= A_{1} + A_{2} + A_{3} + A_{4}, say.  (15)

Since

z_{b} = y_{b} + \frac{1}{b^{2}} − \frac{1}{b}, b^{4} + 2b^{3} − b − 1 ≥ b^{4} + 2b^{3} − 2b − 1 = (b − 1)(b + 1)^{3}, and
L_{b} = \frac{b^{4} + 2b^{3} − b − 1}{b^{3}(b + 1)}x_{b} + \frac{y_{b}}{b} + \frac{2z_{b}}{b(b − 1)},

we obtain

L_{b} = \frac{b^{4} + 2b^{3} − b − 1}{b^{3}(b + 1)}x_{b} + \frac{b + 1}{b(b − 1)}y_{b} − \frac{2}{b^{3}} ≥ \frac{(b − 1)(b + 1)^{2}}{b^{3}} x_{b} + \frac{b + 1}{b(b − 1)}y_{b} − \frac{2}{b^{3}} .

Therefore L_{b}^{2} is larger than or equal to

\frac{(b − 1)^{2}(b + 1)^{4}}{b^{6}} x_{b}^{2} + \frac{(b + 1)^{2}}{b^{2}(b − 1)^{2}} y_{b}^{2} + \frac{4}{b^{6}} + \frac{2(b + 1)^{3}}{b^{4}} x_{b}y_{b} − \frac{4(b − 1)(b + 1)^{2}}{b^{6}} x_{b} − \frac{4(b + 1)}{b^{4}(b − 1)}y_{b}

≥ \frac{(b − 1)^{2}(b + 1)^{4}}{b^{6}} x_{b}^{2} + \frac{(b + 1)^{2}}{b^{2}(b − 1)^{2}} y_{b}^{2} + \frac{2(b + 1)^{3}}{b^{4}} x_{b}y_{b} − \frac{4(b − 1)(b + 1)^{2}}{b^{6}} x_{b} − \frac{4(b + 1)}{b^{4}(b − 1)}y_{b}

= B_{1} + B_{2} + B_{3} − B_{4} − B_{5}, say.  (16)

In addition, we see that

y_{b−1} = y_{b} + \frac{1}{b − 1} − ∑_{m=(b−1)^{2}}^{b^{2}−1} \frac{1}{m} ≤ y_{b} + \frac{1}{b − 1} − \frac{2b − 1}{b^{2} − 1} ≤ y_{b} − \frac{b − 2}{b^{2}},

y_{b+1} = y_{b} − \frac{1}{b} + ∑_{m=b^{2}}^{b^{2}+2b} \frac{1}{m} ≤ y_{b} − \frac{1}{b} + \frac{2b + 1}{b^{2}} = y_{b} + \frac{b + 1}{b^{2}},
x_{b−1} = x_{b} − \frac{1}{b − 1}, and x_{b+1} = x_{b} + \frac{1}{b}.

From these, we obtain the following inequalities:

A_{1} = \frac{(b + 1)(b + 3)}{b(b + 2)}x_{b}^{2} − \frac{(b + 1)(b + 3)}{b^{2}(b − 1)(b + 2)}x_{b} − \frac{(b + 1)(b + 3)}{b^{2}(b − 1)(b + 2)}

≤ \frac{(b + 1)(b + 3)}{b(b + 2)}x_{b}^{2} − \frac{(b + 1)(b + 3)}{b^{2}(b − 1)(b + 2)}x_{b} − \frac{1}{b^{2}} ,

A_{2} ≤ \frac{b + 2}{(b − 2)(b − 1)(b + 1)}y_{b}^{2} + \frac{3(b + 2)}{b^{2}(b − 2)(b − 1)(b + 1)}y_{b} − \frac{b + 2}{b^{4}(b − 1)},
A_{3} ≤ \frac{b(b + 3)}{(b − 2)(b − 1)(b + 2)}x_{b}y_{b} + \frac{b + 3}{(b − 2)(b − 1)(b + 2)}y_{b} − \frac{b + 3}{b(b − 1)(b + 2)}x_{b} − \frac{b + 3}{b^{2}(b − 1)(b + 2)}

≤ \frac{b(b + 3)}{(b − 2)(b − 1)(b + 2)}x_{b}y_{b} + \frac{b + 3}{(b − 2)(b − 1)(b + 2)}y_{b} − \frac{b + 3}{b(b − 1)(b + 2)}x_{b},

A_{4} ≤ \frac{b + 2}{b^{2}}x_{b}y_{b} − \frac{b + 2}{b^{2}(b − 1)}y_{b} + \frac{(b + 1)(b + 2)}{b^{4}}x_{b} − \frac{(b + 1)(b + 2)}{b^{4}(b − 1)}

≤ \frac{b + 2}{b^{2}}x_{b}y_{b} − \frac{b + 2}{b^{2}(b − 1)}y_{b} + \frac{(b + 1)(b + 2)}{b^{4}}x_{b}.

Since b > 1500, it is not difficult to verify that

B_{1} − B_{4} − A_{1} ≥ −\frac{(b + 1)(6b^{3} + 3b^{2} − 3b − 2)}{b^{6}(b + 2)}x_{b}^{2} + \frac{b^{6} − 5b^{4} + 8b^{3} + 16b^{2} − 4b − 8}{b^{6}(b − 1)(b + 2)}x_{b} + \frac{1}{b^{2}}

≥ −\frac{7}{b^{3}} x_{b}^{2} + \frac{1}{b^{2}} ,

B_{2} − B_{5} − A_{2} ≥ −\frac{b^{2} + 5b + 2}{b^{2}(b − 2)(b − 1)^{2}(b + 1)}y_{b}^{2} − \frac{7b^{3} + 6b^{2} − 12b − 8}{b^{4}(b − 2)(b − 1)(b + 1)}y_{b} + \frac{b + 2}{b^{4}(b − 1)}

≥ −\frac{1}{b^{3}} y_{b}^{2} − \frac{1}{b^{3}} y_{b},

B_{3} − A_{3} − A_{4} ≥ −\frac{2(b^{4} + 8b^{3} + 5b^{2} − 8b − 4)}{b^{4}(b − 2)(b − 1)(b + 2)}x_{b}y_{b} − \frac{b^{2} + 4b + 8}{b^{2}(b − 2)(b − 1)(b + 2)}y_{b}

− \frac{b^{3} + 3b^{2} − 4b − 4}{b^{4}(b − 1)(b + 2)}x_{b}

≥ −\frac{4}{b^{3}} x_{b}y_{b} − \frac{3}{b^{3}} y_{b} − \frac{3}{b^{3}} x_{b}.

From (15), (16), and the above inequalities, we obtain

L_{b}^{2} − U_{b−1}U_{b+1} ≥ \frac{1}{b^{2}} − \frac{7}{b^{3}} x_{b}^{2} − \frac{1}{b^{3}} y_{b}^{2} − \frac{4}{b^{3}} x_{b}y_{b} − \frac{4}{b^{3}} y_{b} − \frac{3}{b^{3}} x_{b}.

Since x_{b} ≤ \frac{4}{3} log b and y_{b} ≤ \frac{5}{4} log b,

L_{b}^{2} − U_{b−1}U_{b+1} ≥ \frac{1}{b^{2}} − \frac{2977}{144} \frac{(log b)^{2}}{b^{3}} − \frac{9 log b}{b^{3}} ≥ \frac{b − 23(log b)^{2}}{b^{3}} .

Observe that the function x ↦ x − 23(log x)^{2} is strictly increasing on [300, ∞). Since b ≥ 1500,

b − 23(log b)^{2} ≥ 1500 − 23(log 1500)^{2} > 0.

Therefore s_{b}^{2} − s_{b−1}s_{b+1} ≥ L_{b}^{2} − U_{b−1}U_{b+1} > 0. Hence (s_{b})_{b≥2} is log-concave, as desired.

Remark 8. We have uploaded the numerical data on the computation of s_{b} in the second author’s ResearchGate account [20] which are freely downloadable by everyone.

3 Acknowledgments

We are grateful to the referee for pointing out References [15, 16, 19] and for his/her suggestions which improve the presentation of this article. We also thank the Editor-in-Chief for letting us know about the problem in the Fibonacci Quarterly [26, 27] and the algorithm of Myers [18]. Furthermore, we would like to thank Wannarut Rungrottheera for her support. Phakhinkon Phunphayap receives a scholarship from Science Achievement Scholarship of Thailand(SAST). Prapanpong Pongsriiam received financial support jointly from the Thailand Research Fund and Faculty of Science Silpakorn University, grant number RSA5980040. Prapanpong Pongsriiam is the corresponding author.

References

[1] B. Adamczewski and J.-P. Allouche, Reversal and palindromes in continued fractions, Theoret. Comput. Sci. 380 (2007), 220–237.

[2] B. Adamczewski and Y. Bugeaud, Palindromic continued fractions, Ann. Inst. Fourier 57 (2007), 1557–1574.

[3] B. Adamczewski and Y. Bugeaud, Transcendence measure for continued fractions involving repetitive or symmetric patterns, J. Eur. Math. Soc. 12 (2010), 883–914.

[4] P. Ambrož, C. Frougny, Z. Masáková, and E. Pelantová, Palindromic complexity of infinite words associated with simple Parry numbers, Ann. Inst. Fourier 56 (2006), 2131–2160.

[5] W. D. Banks, Every natural number is the sum of forty-nine palindromes, Integers 16 (2016), Paper No. A3, 1–9.

[6] W. D. Banks, D. N. Hart, and M. Sakata, Almost all palindromes are composite, Math. Res. Lett. 11 (2004), 853–868.

[7] W. D. Banks and I. Shparlinski, Prime divisors of palindromes, Period. Math. Hungar. 51 (2005), 1–10.

[8] J. Bayless and D. Klyve, Reciprocal sums as a knowledge metric: theory, computation, and perfect numbers, Amer. Math. Monthly 120 (2013), 822–831.

[9] S. Brlek, S. Hamel, M. Nivat, and C. Reutenauer, On the palindromic complexity of infinite words, Internat. J. Found. Comput. Sci. 15 (2004), 293–306.

[10] J. Cilleruelo, F. Luca, and L. Baxter, Every positive integer is a sum of three palindromes, Math. Comp. 87 (2018), 3023–3055.

[11] D. Damanik and L. Q. Zamboni, Combinatorial properties of Arnoux-Rauzy subshifts and applications to Schrödinger operators, Rev. Math. Phys. 15 (2003), 745–763.

[12] G. Ficia and L. Q. Zamboni, On the least number of palindromes contained in an infinite word, Theoret. Comput. Sci. 481 (2013), 1–8.

[13] S. Fischler, Palindromic prefixes and diophantine approximation, Monatsh. Math. 151 (2007), 11–37.

[14] A. Hof, O. Knill, and B. Simon, Singular continuous spectrum for palindromic schrödinger operators, Comm. Math. Phys. 174 (1995), 149–159.

[15] P. Kinlaw, M. Kobayashi, and C. Pomerance, On the equation φ(n) = φ(n+1), preprint,

2019. Available at https://math.dartmouth.edu/~carlp/Updated_Euler_Function_
Project.pdf.

[16] J. Lichtman, The reciprocal sum of primitive nondeficient numbers, J. Number Theory, 191 (2018), 104–118.

[17] F. Luca, Palindromes in various sequences, Gac. R. Soc. Mat. Esp. 20 (2017), 49–56.

[18] J. S. Myers, Palindrome reciprocal sums, http://list.seqfan.eu/pipermail/

seqfan/2014-June/013186.html.

[19] H. M. Nguyen and C. Pomerance, The reciprocal sum of the amicable numbers, Math. Comp. 88 (2019), 1503–1526.

[20] P. Phunphayap and P. Pongsriiam, Estimates for the reciprocal sum of b-adic palin-

dromes, preprint, 2019. Available at https://www.researchgate.net/publication/
336304247_Estimates_for_the_Reciprocal_Sum_of_b-adic_Palindromes.

[21] P. Pongsriiam, Local behaviors of the number of relatively prime sets, Int. J. Number Theory 12 (2016), 1575–1593.

[22] P. Pongsriiam and K. Subwattanachai, Exact formulas for the number of palindromes up to a given positive integer, Int. J. Math. Comput. Sci. 14 (2019), 27–46.

[23] A. Rajasekaran, J. Shallit, and T. Smith, Sums of palindromes: an approach via automata, in 35th Symposium on Theoretical Aspects of Computer Science (STACS 2018), Article no. 54, pp. 54:1–54:12.

[24] P. Roggero, M. Nardelli, and F. Di Noto, Sum of the reciprocals of famous series: mathematical connections with some sectors of theoretical physics and string theory,

    preprint, 2016. Available at http://empslocal.ex.ac.uk/people/staff/mrwatkin/
    zeta/nardelli2017a.pdf.

[25] D. Roy, Approximation to real numbers by cubic algebraic integers, II, Ann. of Math. 158 (2003), 1081–1087.

[26] J. Shallit, Problem B-441, Fibonacci Quart. 18 (1980), 370.

[27] J. Shallit and H. Klauser, Sum of base-b palindrome reciprocals, Fibonacci Quart. 19 (1981), 469.

[28] N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences, https://oeis. org, 2019.

[29] R. P. Stanley, Log-concave and unimodal sequences in algebra, combinatorics, and geometry, Ann. New York Acad. Sci. 576 (1989), 500–535.

[30] G. Tenenbaum, Introduction to Analytic and Probabilistic Number Theory, Cambridge University Press, 1995.

2010 Mathematics Subject Classification: Primary 11A63; Secondary 11N37, 11Y60, 40A25. Keywords: palindrome, palindromic number, reciprocal, asymptotic, log-concave.

(Concerned with sequences \underline{A002113} and \underline{A244162}.)

Received May 27 2019; revised versions received October 7 2019; December 10 2019. Published in Journal of Integer Sequences, December 27 2019.

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