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0 votes
0 answers
39 views

How to show that $\sum_{j = 1}^{A \left({n}\right)} \left\{{\frac{n - {j}^{2}}{2\, j}}\right\}$ satisfies Weyl's criterion.

We can write this sum in terms of even ($j = 2k$) and odd ($j = 2k-1$) summation index as $$\sum_{j = 1}^{A \left({n}\right)} \left\{{\frac{n - {j}^{2}}{2\, j}}\right\} = \sum_{k = 1}^{\lfloor{A \left(...
Lorenz H Menke's user avatar
0 votes
0 answers
81 views

Looking for a solution of $\sum_{i = 1}^{k} \sum_{{d}_{1}\, {d}_{2} = i (2k - i), {d}_{1} \le N, {d}_{2} \le N} [GCD(2 k, {d}_{1}, {d}_{2}) = 1]$

The double sum is $$\sum_{i = 1}^{k} \sum_{\substack{{d}_{1}\, {d}_{2} = i \left({2k - i}\right), \\ {d}_{1} \le N, {d}_{2} \le N}} \left[{\left({2\, k, {d}_{1}, {d}_{2}}\right) = 1}\right]$$ where [.....
Lorenz H Menke's user avatar
0 votes
1 answer
57 views

How to define a function that satifies this condition?

I would like to define a function $f(n)$. It must be such that it should produce the sum of all elements till the nth term of the series mentioned below: $$2,2,2,3,3,3,4,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,...
Teflon's user avatar
  • 73
3 votes
0 answers
75 views

The number $1^1 + 3^3 + 5^5 + .. + (2^n - 1)^{2^n-1}$ is a multiple of $2^n$ but not a multiple of $2^{n+1}$

Prove that for all $n>1$ the number $1^1 + 3^3 + 5^5 + .. + (2^n - 1)^{2^n-1}$ is a multiple of $2^n$ but not a multiple of $2^{n+1}$. Proof: Let $$S_n=1^1 + 3^3 + 5^5 + .. + (2^n - 1)^{2^n-1}$$ We ...
Student's user avatar
  • 101
0 votes
1 answer
66 views

Asymptotic for $\sum_{d\mid N}\frac{d^{2}}{\sigma\left(d\right)}$ [closed]

Let $N\in\mathbb{N}$. I'm looking for an asymptotic formula for $\sum_{d\mid N}\frac{d^{2}}{\sigma\left(d\right)}$ as $N\rightarrow\infty$. I tried to use known asymptotic formulas for similar ...
user23571119's user avatar
3 votes
0 answers
58 views

Is it possible to construct a sequence using the first $n$ prime numbers such that each segment has a unique sum?

For example, consider the sequence $2,7,3,5$. The sums of the segments of this sequence are as follows, and they are all unique: $$2, 2+7, 2+7+3, 2+7+3+5, 7, 7+3, 7+3+5, 3, 3+5, 5$$ Can we generate ...
dodicta's user avatar
  • 1,451
3 votes
0 answers
75 views

Can we show that $\frac{\sum_{j=1}^n j^2\cdot j!}{99}$ generates only finite many primes?

Define $$f(n):=\frac{\sum_{j=1}^n j^2\cdot j!}{99}$$ Is $f(n)$ prime for only finite many positive integers $n\ge 10$ ? Approach : If we find a prime number $q>11$ with $q\mid f(q-1)$ , then for ...
Peter's user avatar
  • 85.1k
0 votes
2 answers
58 views

Comparing integral with a sum

Show that \begin{equation}\sum_{m=1}^k\frac{1}{m}>\log k.\end{equation} My intuition here is that the LHS looks a lot like $\int_1^k\frac{1}{x}\textrm{d}x$, and this evaluates to $\log k$. To ...
turkey131's user avatar
  • 135
0 votes
1 answer
39 views

Differences between sums of reciprocals of primes and products thereof.

I am not a number theorist, but I know that number theorists are notorious for estimating sums including primes. So, I have the following question, which I may not have seen addressed, but I am sure ...
User's user avatar
  • 717
2 votes
3 answers
199 views

Sum of floored fractions $\lfloor \frac{1^3}{2009} \rfloor + \lfloor \frac{2^3}{2009} \rfloor + \cdots + \lfloor \frac{2008^3}{2009} \rfloor$

I want to compute the remainder of $A$ divided by 1000, where $A$ is $$A = \lfloor \frac{1^3}{2009} \rfloor + \lfloor \frac{2^3}{2009} \rfloor + \cdots + \lfloor \frac{2008^3}{2009} \rfloor.$$ I tried ...
mathhello's user avatar
  • 918
-2 votes
1 answer
61 views

Summation of $\frac{\sum_{n=1}^{\infty}\frac{T\left(n\right)}{n^{3}}}{\sum_{n=1}^{\infty}\frac{\phi\left(n\right)}{n^{3}}}$ [closed]

Define $T(x)$ as $T(x) = \operatorname{gcd}(1,x)+\operatorname{gcd}(2,x)+ \dots+\operatorname{gcd}(x,x)$. How do I find the value of $$\frac{\sum_{n=1}^{\infty}\frac{T\left(n\right)}{n^{3}}}{\sum_{n=1}...
FreshestCheese's user avatar
2 votes
1 answer
79 views

$m_i, n_j$ integers and $\{m_i\}_{i=1}^{k}\neq\{n_j\}_{j=1}^{k'}.$ Does $\sum\frac{1}{m_i}=\sum\frac{1}{n_j}\implies\sum m_i\neq\sum n_j?$

Suppose $\{m_i\}_{i=1}^{k}$ and $\{n_j\}_{j=1}^{k'}$ are each finite subsets of $\mathbb{N},$ $\{m_i\}_{i=1}^{k}\neq\{n_j\}_{j=1}^{k'},$ and $\displaystyle\sum_{i=1}^{i=k}\frac{1}{m_i} = \sum_{j=1}^{...
Adam Rubinson's user avatar
1 vote
1 answer
69 views

Summation notation over divisors confusion

What does the following summation notation represent? $\sum\limits_{d_1 \mid a, \; d_2\mid b}f(d_1d_2)=\sum\limits_{d_1\mid a }\sum\limits_{d_2 \mid b}f(d_1)f(d_1)=\sum\limits_{d_1\mid a}f(d_1)\sum\...
Jason Xu's user avatar
  • 637
5 votes
2 answers
260 views

Determine all positive integers $n$ such that: $n+d(n)+d(d(n))+\dotsb=2023$.

For a positive integer number $n>1$, we say that $d(n)$ is its superdivisor if $d(n)$ is the largest divisor of $n$ such that $d(n)<n$. Additionally, we define $d(0)=d(1)=0$. Determine all ...
Kokos's user avatar
  • 418
13 votes
0 answers
189 views

Can the sequence $\{\lfloor \alpha n \rfloor\}$ be divided into two parts with equal sums, for all $\alpha \in \mathbb{R}$?

Define the sequence $a_n = \lfloor \alpha n \rfloor$ for a real number $\alpha$. Is there any pair of natural numbers $k, l$ satisfying the following condition?: $$\sum_{n=1}^k a_n = \sum_{n=k+1}^l ...
dodicta's user avatar
  • 1,451

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