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A sharp upper bound for the sum of reciprocals of least common multiples

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Let $$n$$ n and $$k$$ k be positive integers such that $$n\ge k+1$$ n ≥ k + 1 and let $$\{a_i\}_{i=1}^n$$ { a i } i = 1 n be… Click to show full abstract

Let $$n$$ n and $$k$$ k be positive integers such that $$n\ge k+1$$ n ≥ k + 1 and let $$\{a_i\}_{i=1}^n$$ { a i } i = 1 n be an arbitrary given strictly increasing sequence of positive integers. Let $$S_{n, k}:=\sum _{i=1}^{n-k} \frac{1}{ 1{\rm cm} (a_{i},a_{i+k})}$$ S n , k : = ∑ i = 1 n - k 1 1 cm ( a i , a i + k ) . Borwein [3] proved a conjecture of Erdős stating that if $$n\ge 2$$ n ≥ 2 , then $$S_{n,1}\le 1-\frac{1}{2^{n-1}}$$ S n , 1 ≤ 1 - 1 2 n - 1 , with the equality holding if and only if $$a_{i}=2^{i-1}$$ a i = 2 i - 1 for $$1\le i \le n$$ 1 ≤ i ≤ n . In this paper, we first improve Borwein's upper bound by showing that $$S_{n,1}\le \frac{1}{a_{1}}(1-\frac{1}{2^{n-1}})$$ S n , 1 ≤ 1 a 1 ( 1 - 1 2 n - 1 ) with the equality occurring if and only if $$a_{i}=2^{i-1}a_{1}$$ a i = 2 i - 1 a 1 for all integers $$1 \le i \le n$$ 1 ≤ i ≤ n . Then we use this improved upper bound to show that if $$n\ge 3$$ n ≥ 3 , then $$S_{n, 2}\le \frac{7}{6}+\frac{1}{2^{\lfloor \frac{n}{2}\rfloor }} (\frac{2}{3}\delta _{n}-\frac{7}{3})$$ S n , 2 ≤ 7 6 + 1 2 ⌊ n 2 ⌋ ( 2 3 δ n - 7 3 ) , with the equality holding if and only if $$a_1=1, a_{2i}=2^i$$ a 1 = 1 , a 2 i = 2 i and $$a_{2i+1}=3\times 2^{i-1}$$ a 2 i + 1 = 3 × 2 i - 1 for all integers $$1\le i\le \lfloor \frac{n}{2}\rfloor $$ 1 ≤ i ≤ ⌊ n 2 ⌋ , where $$\delta _{n}:=0$$ δ n : = 0 if $$n$$ n is even, and 1 if $$n$$ n is odd. Furthermore, we show that if $$n\ge 7$$ n ≥ 7 , then $$S_{n, 3}\le \frac{17}{15}-\frac{37}{15}\cdot \frac{1}{2^{\lfloor \frac{n}{3}\rfloor }} +\frac{\epsilon _{n}}{2^{\lceil \frac{n}{3}\rceil }}$$ S n , 3 ≤ 17 15 - 37 15 · 1 2 ⌊ n 3 ⌋ + ϵ n 2 ⌈ n 3 ⌉ , with equality occurring if and only if $$a_i=i$$ a i = i for all $$i\in \{1, 2, 3\}$$ i ∈ { 1 , 2 , 3 } and $$a_{3i+1}=2^{i+1} (1\le i\le \lfloor \frac{n-1}{3}\rfloor ), a_{3i+2}= 5\times 2^{i-1} (1\le i\le \lfloor \frac{n-2}{3}\rfloor )$$ a 3 i + 1 = 2 i + 1 ( 1 ≤ i ≤ ⌊ n - 1 3 ⌋ ) , a 3 i + 2 = 5 × 2 i - 1 ( 1 ≤ i ≤ ⌊ n - 2 3 ⌋ ) and $$a_{3i+3}=3\times 2^i (1\le i\le \lfloor \frac{n}{3}\rfloor -1)$$ a 3 i + 3 = 3 × 2 i ( 1 ≤ i ≤ ⌊ n 3 ⌋ - 1 ) , where $$\epsilon _n=0$$ ϵ n = 0 if $$3 \mid n$$ 3 ∣ n , 1 if $$n\equiv 1~({\rm mod} \; 3)$$ n ≡ 1 ( mod 3 ) and $$\frac{9}{5}$$ 9 5 if $$n\equiv 2~({\rm mod} \; 3)$$ n ≡ 2 ( mod 3 ) . We also present a tight upper bound for $$S_{n, 3}$$ S n , 3 if $$n\in \{4,5,6\}$$ n ∈ { 4 , 5 , 6 } .

Keywords: upper bound; frac; frac rfloor; lfloor frac

Journal Title: Acta Mathematica Hungarica
Year Published: 2019

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