All Questions
39
questions
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 [.....
- 967
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 ...
- 918
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?:
$$\sum_{n=1}^k a_n = \sum_{n=k+1}^l ...
- 1,451
4
votes
1
answer
106
views
Simplifying floor function summation $\sum^m_{k=0} \left(\left\lfloor{\frac{n}{m}k}\right\rfloor-\left\lceil{\frac{n}{m}(k-1)}\right\rceil\right)$
Is there a way this summation can be simplified/cast in a more revelatory form (non-summation representation, single-term only, etc.)?
$$\sum^m_{k=0} \left(\left\lfloor{\frac{n}{m}k}\right\rfloor-\...
- 315
1
vote
1
answer
129
views
Proving $\sum_{1}^{n} \left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil=2n -1 $
Show that $$\sum_{i=1}^{n} \left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil=2n -1 $$ where $ \lceil\cdot\rceil$ denotes the ceiling function.
My method: one way would be observe each part of the ...
- 977
-1
votes
1
answer
55
views
Some alternating sum of integer part of $\frac{kb}{1722}$.
For every integer $k$ coprime to 1722, how can one compute the sum
$\sigma(k)= \sum_{1\leq b\leq 1722, (1,b)=1722} \lfloor\frac{kb}{1722}\rfloor (-1)^{b-\lfloor \frac{b}{2}\rfloor-\lfloor \frac{b}{3}\...
- 119
4
votes
1
answer
405
views
Prove an Elementary sum of floor function
Prove:
If $a$ and $b$ are odd and relatively prime,
$$\sum_{\substack{0 \lt x \lt b/2\\x \in Z}} \left\lfloor \frac{ax}{b} \right\rfloor + \sum_{\substack{0 \lt y \lt a/2\\y\in Z}} \left\lfloor \frac{...
- 113
4
votes
0
answers
90
views
Prove that $\sum_{r=2}^{n} \left \lfloor n^{\frac{1}{r}} \right \rfloor = \sum_{r=2}^{n} \left \lfloor \log_{r}(n) \right \rfloor$.
Prove that
$$\sum_{r=2}^{n} \left \lfloor n^{\frac{1}{r}} \right \rfloor = \sum_{r=2}^{n} \left \lfloor \log_{r}(n) \right \rfloor\,.$$
I have tried to use substitutions of $n=p^k$ in order to try ...
- 331
2
votes
4
answers
62
views
For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$
For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$
By dividing the numbers between $1$ and $5^m$ as intervals of $5^k$, I was getting the following ...
user675768
4
votes
0
answers
116
views
Simplify $\sum_{k = 0}^{[p/3] - 2} \sum_{i = 1}^{[p/(k + 1)] - 3} \sum_{j = 0}^{p - (k + 1) (i + 2) - 1} {\delta}_{N, i\, p + j + k + 1}$
Where in the title [...] denotes the ceiling function which is used for a shorter title. This sum arises as corrections terms to the number of reducible quadratics of height $N$. Looking for ...
- 967
3
votes
2
answers
690
views
How many values of $n$ are there for which $n!$ ends in $1998$ zeros?
How many values of $n$ are there for which $n!$ ends in $1998$ zeros?
My Attempt:
Number of zeros at end of $n!$ is
$$\left\lfloor \frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{5^2}\right\rfloor+\...
- 9,599
4
votes
0
answers
146
views
A $p$-congruence involving binomial coefficients and the floor function.
Let $p$ be a prime number.
Let $k$ and $m$ be positive integers such that $k \ge m$ and $p-1$ does not divide $k-m$.
It seems to be true that $$ (-1)^{m-\lfloor
\frac{k-m}{p-1}\rfloor}\sum_{i\...
- 3,716
1
vote
0
answers
162
views
Find the asymptotic expansion as $N \rightarrow \infty$ of $\sum_{n=1}^{n_{\max}} \left\lceil{\sqrt{2nN+n^2}}\right\rceil$
We can write the fractional part $\left\{{\sqrt{N^2 + k^2}}\right\} = \sqrt{N^2 + k^2} - N - n$ for some $n \ge 0$. After solving for $n$ we get $$\sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \left\{...
- 967
1
vote
1
answer
78
views
Prove that $\sum_{j=1}^{\infty} \left[ \frac{n}{2^j}+\frac{1}{2} \right] = n$ for interger $n\ge 1$ [duplicate]
This is the full question:
Consider an integer $n\ge 1$ and the integers $i$,$1\le i \le n$. For each $k=0,1,2,\cdots$, find the number of $i$'s that are divisble by $2^k$ but not by $2^{k+1}$.Thus ...
- 3,360
4
votes
1
answer
485
views
On the sum of sum of divisors $\sum_{a=1}^{N} D \left({\left\lfloor{\frac{N}{a}}\right\rfloor}\right)$.
where $D \left({x}\right)$ is the sum of divisors. This sum comes from my work on the number of reducible monic cubics. This is a two part question. By writing out all the divisors $\tau \left({a}\...
- 967