All Questions
Tagged with elementary-number-theory summation
482
questions
2
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Proof of $\frac{\Phi_n'(x)}{\Phi_n(x)} = \frac{1}{x^n-1}\sum_{k=1}^{n}c_k(n)x^{k-1}$
Let $c_k(n)$ denote Ramanujan's sum, and $\Phi_n(x)$ be the $n$th cyclotomic polynomial. Prove that
$$\frac{\Phi_n'(x)}{\Phi_n(x)} = \frac{1}{x^n-1}\sum_{k=1}^{n}c_k(n)x^{k-1}$$
My attempt was to ...
4
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3
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243
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Prove $z^n + 1/{z^n}$ is rational if $z+1/z$ is rational
Knowing that $z + \frac{1}{z} = 3$ ($z \in \mathbb{R}$), prove that $z^n + \frac{1}{z^n} \in \mathbb{Q}$ ($n \in \mathbb{N}$ and $n \geq 1$).
I've tried to find a general formula for $z^n + \frac{1}{...
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0
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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
$$\sum_{j = 1}^{A \left({n}\right)} \left\{{\frac{n - {j}^{2}}{2\, j}}\right\} = \sum_{k = 1}^{\lfloor{A \left(...
7
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1
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Sums of consecutive powers of 3 being perfect squares
I was recently considering the puzzle of finding consecutive (integer) powers of 3 that sum to a square. It's not hard to show that this can be reduced to finding values of $n$ such that
$S_3(n) = \...
0
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0
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Looking for a solution of $\sum_{i = 1}^{k} \sum_{{d}_{1}\, {d}_{2} = i (2k - i), {d}_{1} \le N, {d}_{2} \le N} [GCD(2 k, {d}_{1}, {d}_{2}) = 1]$
The double sum is
$$\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]$$
where [.....
1
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1
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91
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Prove that $\sum_{k=1}^{\infty }\left \lfloor \frac{n}{p^k} \right \rfloor=\frac{n-S_n}{p-1}$
If $p$ is a prime number, $n$ is a natural number, and $S_n$ is the sum of the digits of $n$ when expressed in base $p$.
$$\text{Prove that }\sum_{k=1}^{\infty }\left \lfloor \frac{n}{p^k} \right \...
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proof - Show that $1! +2! +3!+\cdots+n!$ is a perfect power if and only if $n=3$
Show that $1! +2! +3!+\cdots+n!$ is a perfect power if and only if $n =3$
For $n=3$, $1!+2!+3!=9=3^2$. I also feel that the word 'power' makes it a whole lot hard to prove. How do we prove this? What ...
0
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1
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58
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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:
$$2,2,2,3,3,3,4,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,...
3
votes
0
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79
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The number $1^1 + 3^3 + 5^5 + .. + (2^n - 1)^{2^n-1}$ is a multiple of $2^n$ but not a multiple of $2^{n+1}$
Prove that for all $n>1$ the number $1^1 + 3^3 + 5^5 + .. + (2^n - 1)^{2^n-1}$ is a multiple of $2^n$ but not a multiple of $2^{n+1}$.
Proof: Let $$S_n=1^1 + 3^3 + 5^5 + .. + (2^n - 1)^{2^n-1}$$
We ...
0
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1
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66
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Asymptotic for $\sum_{d\mid N}\frac{d^{2}}{\sigma\left(d\right)}$ [closed]
Let $N\in\mathbb{N}$. I'm looking for an asymptotic formula for $\sum_{d\mid N}\frac{d^{2}}{\sigma\left(d\right)}$ as $N\rightarrow\infty$. I tried to use known asymptotic formulas for similar ...
3
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0
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Is it possible to construct a sequence using the first $n$ prime numbers such that each segment has a unique sum?
For example, consider the sequence $2,7,3,5$. The sums of the segments of this sequence are as follows, and they are all unique:
$$2, 2+7, 2+7+3, 2+7+3+5, 7, 7+3, 7+3+5, 3, 3+5, 5$$
Can we generate ...
3
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Can we show that $\frac{\sum_{j=1}^n j^2\cdot j!}{99}$ generates only finite many primes?
Define $$f(n):=\frac{\sum_{j=1}^n j^2\cdot j!}{99}$$
Is $f(n)$ prime for only finite many positive integers $n\ge 10$ ?
Approach : If we find a prime number $q>11$ with $q\mid f(q-1)$ , then for ...
0
votes
2
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62
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Comparing integral with a sum
Show that \begin{equation}\sum_{m=1}^k\frac{1}{m}>\log k.\end{equation}
My intuition here is that the LHS looks a lot like $\int_1^k\frac{1}{x}\textrm{d}x$, and this evaluates to $\log k$. To ...
0
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1
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Differences between sums of reciprocals of primes and products thereof.
I am not a number theorist, but I know that number theorists are notorious for estimating sums including primes. So, I have the following question, which I may not have seen addressed, but I am sure ...
2
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3
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Sum of floored fractions $\lfloor \frac{1^3}{2009} \rfloor + \lfloor \frac{2^3}{2009} \rfloor + \cdots + \lfloor \frac{2008^3}{2009} \rfloor$
I want to compute the remainder of $A$ divided by 1000, where $A$ is
$$A = \lfloor \frac{1^3}{2009} \rfloor + \lfloor \frac{2^3}{2009} \rfloor + \cdots + \lfloor \frac{2008^3}{2009} \rfloor.$$
I tried ...