Given a sequence of integers $A_1,A_2,A_3 ……A_n$ , find number triples giving equal xor?

For giving sequence $ A_1,A_2,A_3 ……A_n$ , find number of triples $ i,j,k$ such that $ 1<=i<j<=k<=N$ and $ A_i + A_{i+1} + … A_{j-1} = A_{j} + A_{j+1} ….. A_{k}$ .Where $ +$ is bitwise xor operation .

I tried to solve it using dynamic programming somewhat similar to https://www.geeksforgeeks.org/count-number-of-subsets-having-a-particular-xor-value/ , but it’s time complexity if $ O(n*m)$ , where m is maximum element in the array .Can we do better than $ O(n*n)$ or $ O(n*m)$ ?

A series such that $\sum {a_n}$ converges, but $\sum {a_{3n}}$ diverges.

  1. Give an example of a convergent series $ \sum {a_n}$ such that the series $ \sum {a_{3n}}$ is divergent.

  2. Give an example of a divergent series $ \sum {b_n}$ such that the series $ \sum {b_{3n}}$ is convergent.

Attempt:

  1. I am not sure if this is a valid forumla for a sequence : $ a_{3n-2} = \frac{1}{1+4(n-1)} ,a_{3n-1} = \frac{1}{3+4(n-1)}, a_{3n} = -\frac{1}{2n}$ . This series converges to $ \frac{3}{2} \log(2)$ . But, $ \sum {a_{3n}}$ diverges.

  2. We define $ b_{3n-2}=1 , b_{3n-1}=1, b_{3n} = \frac{1}{n^2}$ . The series diverges, but $ \sum{b_{3n}}$ converges to $ \frac{\pi^2}{6} $

The problem is, I am not sure if the this type of “formula” works [unlike the sequence defined by $ 1/n$ or something. Is this valid to define the sequence “term-by-term” (here, three different types of indices)?].

If $(a_n)$ monotonous and not zero convergent then $(2^na_{2^n})$ is also not zero convergent

There was an excercise About condensed sums. We should prove that:

If $ (a_n)$ is monotonous then $ \sum_{n}a_n$ converges/diverges iff $ \sum_{k}2^ka_{2^k}$ converges/diverges

It is sufficient to look at a decreasing Zero convergent sequence $ (a_n)$

$ s_n=\sum_{k=1}^{n}a_k,v_n=\sum_{k=1}^n2^ka_{2^k}$

If $ K,N\in\mathbb{N}$ with $ 2^{K-1}<N\leq 2^K$ then

I am sorry for the duplicate please answer one specific Question About the Problem I have in the comment and I will delete the answer afterwards

$ S_N\le S_{2^K}\leq a_1+v_{K-1}$ and

$ S_N\ge S_{2^{K-1}}\geq a_1/2 +v_{k-1}/2$

I have understood that the partialsums are Always positive since every $ a_n$ must be positive. And if the partialsums can be bounded then because the partialsums are monotonously rising the sequence of partialsums is converging to its Supremum. And $ s_k$ is bounded if and only if $ v_k$ is bounded.

Again because $ a_n$ is decreasing and zeroconvergent the partialsums are always positive. If one cannot be boundet then also the other one cannot be bounded.

My Question is that in the excercise the Task was to Show that the Statement holds if $ (a_n)$ is monotonous. But we have only looked at decreasing monotonous $ (a_n)$ and what is more important we have assumed that $ (a_n)$ is Zero convergent.

What if $ (a_n)$ is not Zero convergent then $ \sum_n a_n$ necessarily diverges.

But to prove the Statement we also have to Show that $ \sum_n 2^na_{2^n}$ diverges. It would be sufficient in this case to Show that $ (2^na_{2^n})$ is not Zero convergent.

How can I do that?

Show if $\{a_n^2\}$ is bounded then $\{a_n\}$ is bounded

If $ \{a_n^2\}$ is bounded then I know for a non-negative number $ M$ , $ \mid a_n^2 \mid \leq M, \forall n$ .

How do I show that this leads to $ \{a_n\}$ being bounded?

$ \mid a_n^2 \mid \leq M$

$ -M \leq a_n^2 \leq M$

but now I don’t think I can replace $ M$ with $ \sqrt{M}$ , can I?

$ -\sqrt{M} \leq a_n \leq \sqrt{M}$ is probably allowed, but how do I justify this?

Let $a_n $ complex sequence prove that if $ a_n\to \infty$ then $|a_n|\to\infty$. Note that $a_n = x_n + y_ni$

Let $ a_n $ complex sequence prove that if $ a_n\to \infty$ then $ |a_n|\to\infty$ . Note that $ a_n = x_n + y_ni$ i dont know how to write that mathmatically.

trial :

Can i say that for every $ M>0$ there exist $ N$ such that for every $ n>N$ ,

$ ~~|x_n|>M~~ OR ~~~|y_n|>M$ ( At least one of them goes to $ \infty$ )

because of that $ |an| = \sqrt{(x_n)^2+(y_n)^2} > M$ and so $ |a_n|\to\infty$ .

What are the prime divisors of $\det(A_n)$, where $A_n$ is the $n\times n$ matrix given by $(A_n)_{i,j}={n\choose|i-j|}$?

For $ n\in\mathbb{Z}_{\ge 1}$ , let $ A_n$ be the $ n\times n$ matrix given by $ (A_n)_{i,j}={n\choose |i-j|}$ . From this post it is clear that $ $ \det(A_n)=\prod_{k=0}^{n-1}\left[\left(\exp\left(\frac{2\pi k i}{n}\right)+1\right)^n-1\right]=\prod_{k=0}^{n-1}\left(2^n(-1)^k\cos^n\left(\frac{\pi k}n\right)-1\right).$ $ Also, $ \det(A_n)$ is obviously integer for all $ n$ .

Question: What is known about the (prime) divisors of $ \det(A_n)$ ?

If this is too broad, I am particularly interested in pairs $ (p,d)$ where $ p$ is prime with $ d\mid p-1$ and $ p\nmid\det(A_{(p-1)/d})$ .

Asymptotic estimation of $A_n$

Let $ A_n$ represent the number of integers that can be written as the product of two element of $ [[1,n]]$ .

I am looking for an asymptotic estimation of $ A_n$ .

First, I think it’s a good start to look at the exponent $ \alpha$ such that :

$ $ A_n = o(n^\alpha)$ $

I think we have : $ 2 < alpha $ . To prove this lower bound we use the fact that the number of primer numbers $ \leq n$ is about $ \frac{n}{\log n}$ . Hence we have the trivial lower bound (assuming $ n$ is big enough) :

$ $ \frac{n}{\log n} \cdot \binom{ E(\frac{n}{\log n})}{2} = o(n^3)$ $

Now is it possible to get a good asymptotic for $ A_n$ and not just this lower bound ? Is what I’ve done so far correct ?

Thank you !