Mathematicianadda

Suppose $A=(a_{ij})$ is a n×n matrix by $a_{ij}=\sqrt{i^2+j^2}$. I have tried to check its sign by matlab. l find that the determinant is positive when n is odd and negative when n is even. How to prove it？

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Jimmy

Question: How to show the relation $$ J:=\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\frac 1{64}\pi^4 $$ (using a "minimal industry" of relations, possibly remaining inside the real analysis)?

So i have found a solution to the problem, it is part of my solution for math.stackexchange.com - questions - 3854736, but not a satisfactory solution. "There should be more", explaining why there is a "clean result" for the integral.

Here, i am not strictly interested in a computational approach. I just want to share this with the community in these days of isolation. Any idea to attack this, or a related integral involving "three log factors" is welcome. (Well, the $\arctan$ is a sort of $\log$ in a sense that i don't want to define closer, see below.) Computations may be safely done "modulo integrals involving two or one log factor". But an illuminating, short way to show the above formula for $J$ would be wonderful.

---

Motivation: The above relation appeared as i tried to solve the integral posted at the above link:

Calculate $\displaystyle\int_0^{2\pi} x^2\; \cos x \cdot\operatorname{Li}_2(\cos x)\; dx$ .

After several simplifications and substitutions, it turns out that the above integral is related to integrals of the shape

• $\int_0^1\log t\; R(t)\; dt$ , and
• $\int_0^1\arctan t\cdot \log t\; R(t)\; dt$ , and
• $\int_0^1\arctan^2 t\cdot \log t\; R(t)\; dt$ ,
• and "similar" expressions.

Here $R$ is in each case a (rather simple) rational function. (The more log and/or arctangent factors, the higher the computational complexity.)

I could compute more or less algorithmically most of the the needed integrals to solve the linked problem, all of them but the integral $$ K=\int_0^1\arctan^2 t\cdot\log t\cdot\frac2{1-t^2}\; dt\ , $$ which turned out to be very hard to attack with the methods of real analysis. Computing this integral is more or less equivalent to computing $J$, and the question wants $J$ instead, since we have a "clean formula", so that some speculation about a "clever substitution" may be accepted.

My solution (for $K$) works in complex analysis, the first step is to write $$ \int_0^1 =\int_0^i+\int_i^1\ , $$ then parametrize the first integral using a linear path, the second one using a path on the unit circle.

---

Some comments: I will say some more words, because the situation is rich in coincidences. Since a numerical evidence is the simplest and shortes way to present (instead of showing how to show), i will use this method to at least list the coincidences. Many equalities below are "equivalent" (modulo computation of integrals of lower complexity) to the formula for $J$.

• First of all, a numerical experiment using pari/gp delivers some connection between $K$ and a "cousin" of $J$:

    ? 2 * intnum( t=0, 1, atan(t)^2 * log(t) / (1-t^2) )
    %88 = -0.357038604620289042902893412499686912781214141574556097366337
    ? real(intnum( t=0, I, (pi/4 - atan(t))^2 * log(t) / (1-t^2) ))
    %89 = -0.357038604620289042902893412499686912781214141574556097366337
    ? intnum( t=0, 1, (pi/4 - atan(t))^2 * log(t) / (1-t^2) )
    %90 = -0.357038604620289042902893412499686912781214141574556097366337
  

In words: $$ \begin{aligned} K &= \int_0^1\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt \\ &= \Re \int_0^i\left(\frac \pi 4-\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt \\ &= \int_0^1\left(\frac \pi 4-\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt \ . \end{aligned} $$ Note the integration margins. What happens if we take the integral on $[0,i]$ instead of $[0,1]$ in the $K$-integral? Numerically:

    ? 2 * real(intnum( t=0, i, atan(t)^2 * log(t) / (1-t^2) ))
    %98 = 1.52201704740628808181938019826101736327699352613570971392919
    ? pi^4/64
    %99 = 1.52201704740628808181938019826101736327699352613570971392919

In words: $$ \begin{aligned} K^* &:= \Re\int_0^i\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt \\ &=\frac 1{64}\pi^4 \\ &= -\int_0^1\left(\frac \pi 4+\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt \\ &=-J\ . \end{aligned} $$

(These observations were leading to the formula for $K$ in loc. cit. .)

• One idea is to use partial integration in $J$ or $K$. Well, we have for $K$: $$ \begin{aligned} K &= \int_0^1\arctan^2 t\cdot\log t\;\left(-\log\frac {1-t}{1+t}\right)'\; dt \\ &= \underbrace{\int_0^1\arctan^2 t\cdot\frac 1t\cdot \log\frac {1-t}{1+t}\; dt}_{=2K\text{ (why?)}} \\ &\qquad\qquad+ \underbrace{\int_0^12\arctan^2 t\cdot\frac 1{t+t^2}\cdot \log t\cdot \log\frac {1-t}{1+t}\; dt}_{=-K\text{ (why?)}} \ . \end{aligned} $$

• Note that $\arctan$ is related to the logarithm (over $\Bbb C$), we have the relation (around $0$) $$ \arctan t=\frac 1{2i}\log\frac {1+it}{1-it}\ . $$ The substitution $t=\frac{1-u}{1+u}$ and the formula for $\tan(\arctan 1-\arctan u)$ are giving: $$ \begin{aligned} K &= \int_0^1\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt \\ &=\int_0^1 \left(\frac\pi2-\arctan u\right)^2\cdot\log\frac {1-u}{1+u}\cdot \frac {du}u\ . \\ &=\int_1^\infty \left(\frac\pi2-\arctan u\right)^2\cdot\log\frac {u-1}{1+u}\cdot \frac {du}u\ . \end{aligned} $$ (Write $\log t=\frac 12\log t^2$ to have the same expression under the integral on $(0,1)$ and on $(1,\infty)$.)

• Note the fact that the factor $\frac 2{1-t^2}$ is not "random". It is the right one to make things feasible. It is the derivative of $\displaystyle -\log\frac{1-t}{1+t}$, and plugging in $t=iu$ into $\displaystyle \log\frac{1-t}{1+t}$ leads to an expression related to $\arctan u$. And conversely, $\arctan(iu)$ is related to such a logarithmic expression in $u$.

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dan_fulea

4 Jobs You Can Do With a Major in Software Engineering

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In ordinary discourse, when we say that $A$ implies $B$, we shall formalize it by writing the following: $$A\rightarrow B$$ But when we say that $A$ does not imply $B$, we cannot formalize it as the following: $$\neg(A\rightarrow B)$$ Because by material implication, we have the following equivalences for the second formula: $$\neg(A\rightarrow B)\Longleftrightarrow\neg(\neg A\vee B)\Longleftrightarrow A\wedge\neg B$$ But the ordinary meaning of the sentence "$A$ does not imply $B$" is that $B$ does not follow from $A$: if $A$ is true, $B$ is either false or undecidable. I wonder how "$A$ does not imply $B$" is formally expressed in the object language (not at the meta-level). Thanks!

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Fred

Given the following recurrence relation :

$f(n) = 5f(n-1) - 2f(n-2)$ where $f(0) = 0, f(1) = 1$

I need to find out if an integer $F_n$ is present in the sequence in $O(1)$ time and space.

Solving the equation, there are two distinct real roots.

$\phi = \frac{5 + \sqrt17}2$

$\psi = \frac{5 - \sqrt17}2$

Therefore, $F_n = \frac{\phi^n - \psi^n}{\sqrt17}$

Similar to Binet's rearranged formula, I want to solve for $n$ in terms of $F_n$.

Since, $\psi = \frac{2}{\phi}$

$\sqrt17F_n = \phi^n - \frac{2^n}{\phi^n}$

$Or,$

$\phi^{2n} - \sqrt17F_n\phi^n-2^n = 0$

Here I'm not able to find out a solution to express $n$ purely in terms of $F_n$ so that I can calculate the perfect square just like in Binet's formula.

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Rohit Roy Chowdhury

I was looking into the proof that there are only five Platonic solids in Basic Concepts of Algebraic Topology by F.H. Croom at page 29, Theorem 2.7. To clarify,

• We define a Platonic solid as a simple, regular polyhedron.
• We define a simple polyhedron to be a polyhedron homeomorphic to the $2$-sphere.
• We define a regular polyhedron to be a polyhedron whose faces are regular polygons all congruent to each other and whose local regions near the vertices are all congruent to each other.

Using homology theory, one can prove that the Euler formula $V-E+F=2$ must hold for Platonic solids. Then by using Euler's formula and invoking a counting argument, we find that there are five possible tuples $(V, E, F)$. This is a beautiful proof, but I am unsatisfied with a question: How do we know there can't be two non-similar Platonic solids that have the same $(V, E, F)$-tuple?

Almost all sources I've looked at seem to assume it is obvious that two Platonic solids with the same $(V, E, F)$-tuple are similar, and it is not obvious to me.

Does anyone have any suggestions for how to prove this? Alternatively, does anyone know of a reference where this is proved rigorously?

---

Edit 1: It's not completely clear, but it seems like the definition I used for "regular polyhedra" is different than the one commonly used. Note that I am not assuming any global symmetry, so if any global symmetry is to be invoked, it needs to be proven.

Edit 2: I've been made aware of Cauchy's rigidity theorem, which is proven in, e.g., Proofs From the BOOK by Aigner & Zeigler. One can show that any two Platonic solids that have the same $(V, E, F)$-tuple must be combinatorically equivalent. However, in order for the theorem to apply, we need to show that our Platonic solids are convex. I can't seem to think of any rigorous argument for why the Platonic solids have to be convex.

And actually, you don't need to show that the entire polyhedron is convex. If I'm not mistaken, the proof for Cauchy's rigidity theorem only relies on the vertices of the polyhedron being locally convex. So really it suffices to show the vertices are convex.

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Maximal Ideal

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I know there is a result that $S=$ collection of all trace $0$ matrices, and that collection forms a vector space. But I want to prove it independently i.e.
For any $A,B,C,D\in M_n(K)$ we have to find $E,F\in M_n(K)$ such that $(AB-BA)+(CD-DC)=EF-FE$.
But I don't know how to solve this. But the thing I can observe that the above equation gives rise to $n^2$ equation (equating each entries of matrices of both the sides) with $2n^2$ variables (Total number of entries of $E,F$ is $2n^2$).Can this problem be simplified if we choose special kind of $E$ say diagonal matrix.
Edit-(This is valid only for $\Bbb{R}$ or $\Bbb{C}$)
$AB-BA=(A+aI)(B+bI)-(B+bI)(A+aI)$ for all $a,b\in\Bbb{R}$. And there is $a,b$ in $\Bbb{R}$ such that $A+aI,B+bI$ are invertible.
Hence, we can assume $A,B$ to be invertible matrices i.e. $S=\{AB-BA|A,B\in GL_n(K)\}$

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Biswarup Saha

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What is the biggest circle that is contained in the region bounded by the graph of the polynomial $f(x) = x(1-x)(2x+1)$ and the x-axis interval $[0, 1]$?

(Here's the thing in Desmos)

---

Here's what I have tried

Let's denote the center of the circle by $(c_x, c_y)$ and the points of tangency by $(x_k, y_k)$, $k=1,2$. From distance to the x-axis, we know that the radius of the circle is then $c_y$. We can express the tangency by the dot product of the vectors $(x_k-c_x, y_k-c_y)$ and $(1, f'(x_k))$ being zero. So we get the group of equations (all for $k=1,2$)

$$\begin{cases} (x_k-c_x)^2 + (y_k-c_y)^2 &=& c_y^2 \\ x_k-c_x + (y_k-c_y)f'(x_k) &=& 0 \\ f(x_k) &=& y_k \end{cases} $$

Then I tried to solve this group by using Gröbner basis and elimination ideals. I wrote this SageMath code:

f(x) = x*(1-x)(2*x+1)
fd = f.derivative()

R.<x1, y1, x2, y2, cx, cy> = PolynomialRing(QQ, order='lex')
polys1 = [(cx-xk)**2+(cy-yk)**2 - cy**2 for (xk, yk) in [(x1,y1), (x2, y2)]]
polys2 = [(xk-cx)+(yk-cy)*fd(xk)  for (xk, yk) in [(x1,y1), (x2, y2)]]
polys3 = [f(xk)-yk for (xk, yk) in [(x1,y1), (x2, y2)]]
I = R.ideal(polys1+polys2+polys3)
gb = I.groebner_basis()
for poly in gb:
    print (poly)
print (I.dimension())

It gave me a long list of polynomials, the last one being

$$c_x^6 - 2c_x^4c_y^2 + \frac{5}{4}c_x^4c_y + \frac{1}{64}c_x^4 + c_x^2c_y^4 + \frac{3}{4}c_x^2c_y^3 + \frac{3}{16}c_x^2c_y^2 + \frac{1}{64}c_x^2c_y$$

and said the dimension is $1$. I now realize the reason we don't get dimension $0$ has probably to do with the fact that the points $(x_1, y_1)$ and $(x_2, y_2)$ could be the same point(?) So for a solution, the point $(c_x, c_y)$ should be a zero of that polynomial? But is even that correct, since the points could be collapsed and the other side go over the curve??

How to solve this even numerically? I tried with Sage by minimizing the square sum of all those polynomials that should be zero. I also tried adding constraints for the variables to be near the solution that can be seen from the picture but that didn't work either.

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minkbag

I need to compute a limit:

$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$

I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.

$$ \lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\ = \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\ = \exp (\lim_{x \to 0+} \frac {\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})} {\frac 1 x}) \\ = \exp \lim_{x \to 0+} \dfrac {\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x} - \dfrac {\cos \dfrac 1 x} {x^{3/2}}} {- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)} $$

I've calculated several values of this function, and it seems to have a limit of $1$.

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Andrey Surovtsev

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to […]

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John Quintanilla

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Does anyone know how to compute analytically the following integral:

$$\int\limits_{0}^{+\infty}\dfrac{x^2\mathrm{d}x}{e^{x}-1}$$

It should be equal to $2\zeta(3)$ according to Maple. I tried the following using the binomial theorem for negative integer exponents:

$$I = \int\limits^{+\infty}_{0} e^{-x}(1-e^{-x})^{-1}x^2\mathrm{d}x = \int\limits^{+\infty}_{0}\left[\sum_{k=0}^{+\infty}(-1)^k(-1)^ke^{-(k+1)x}\right]x^2\mathrm{d}x=\int\limits^{+\infty}_{0}\left[\sum_{k=0}^{+\infty}(-1)^{2k}e^{-(k+1)x}\right]x^2\mathrm{d}x$$

After another change of variables, $y=(k+1)x$:

$$I = \sum_{k=0}^{+\infty}(-1)^{2k} \frac{1}{(k+1)^3}\int\limits_0^{+\infty} y^2e^{-y}\mathrm{d}y$$

The keen eye might recognize $\int\limits_0^{+\infty} y^2e^{-y}\mathrm{d}y$ as the gamma function, $\Gamma(3)=(3-1)!=2$. This, together with a slight nudge to the bottom limit of the summation we can rewrite things as:

$$I = \Gamma(3)\sum_{k=1}^{+\infty} \dfrac{(-1)^{2k}}{k^3}$$

And i see immediately (since the beginning in fact...) an infinite sum that makes me troubles and i can't get rid of. I tried to found if i did any trivial error but i'm focusing since to many hours to found it. That's why I need an external view to point me out my obvious error.

Thanks in advance for your help

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Vincent ISOZ

Solving System of Linear Equations Using Rank Method

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A conversation about mathematics inspired by a ball of wool (yarn). Presented by Katie Steckles and Peter Rowlett, with special guest Pat Ashforth.

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Katie Steckles and Peter Rowlett

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Question: A not uncommon calculus mistake is to believe that the product rule for derivatives says that $(fg)'=f'g'$. If $f(x)=e^{x^2}$, determine, with proof, whether there exists an open interval $(a,b)$ and a nonzero differentiable function $g$ defined on $(a,b)$ such that this wrong product rule is true for $x$ in $(a,b)$.

Is my solution correct? Is there anything I need to improve?

My solution:

We're trying to find a function $g$ such that $(fg)'=f'g'.$ Using the product rule, we get $$f'g + fg' = f'g'.$$ Plugging in both $f(x)=e^{x^2}$ and $f'(x)=2xe^{x^2},$ we get $$2xe^{x^2}g + e^{x^2}g' = 2xe^{x^2}g'.$$ Simplifying and canceling $e^{x^2},$ we get $$\frac{g'}{g} = \frac{2x-1}{2x} = 1 + \frac{1}{2x-1}.$$ Taking the integral of both sides, $$\int \frac{dg}g = \int 1+ \frac{1}{2x-1} \, dx.$$ $$\log|g| = \frac{\log|2x-1|}{2} + x + C.$$ $$|g| = e^{\frac{\log|2x-1|}{2}}e^xe^C.$$ Letting $e^C=K,$ we get $$\boxed{g = Ke^x\sqrt{|2x-1|}},$$ Where K is a constant greater than 0. Therefore, on any interval $(a,b)$ that does not contain the value of $\frac{1}{2},$ there exists nonzero differentiable function $g$ defined on $(a,b)$ such that $(fg)'=f'g'$ is true for $x$ in $(a,b)$.

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CSS Jowoo

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Question - Given a randomly generated set of binary numbers, what is the probability that all of them, when XORed with each other, yield $0$?

Additional information -

1. All of the binary numbers in the set are different from each other. As an example, $\{0001, 0010, 0011\}$ is a valid set, but not $\{0001, 0010, 0010\}$ or $\{0010, 0010, 0010\}$.

2. All of the binary numbers have equal length which is some given. For example, one could have a given length of 4, meaning numbers such as $00001$ and $010$ are not allowed.

3. The length of the set is a given, its members are randomly generated.

4. Each possible valid binary number has an equal likelihood of being generated.

Context - For a school math project, I decided to analytically investigate the average percentage of errors an error correction code can correct given a data set and error rate. This question I am asking it a part of said project that I am unable to solve. If necessary, I can provide additional context.

I already know how to do this problem if restriction 1 is lifted, but did not really know how to proceed. I visualized the randomly generated data set as being one number stacked on top of another, kind of how we write when adding large numbers on paper by hand, and then taking my answer to be the probability that each column of bits had an even number of $1$s.

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Nitin Singhal

How to find equation of tangent to the curve without derivative

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Median Practice Questions

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How to Find the Function From the Derivative

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Find the derivative of polynomial

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Prove that Problems in Derivatives

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Finding Median for Grouped Data

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Quotient Rule of Derivative

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How to Find Derivatives Using First Principle

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Using Chain Rule to Find Derivative

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Using Chain Rule to Differentiate

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Empirical Relationship between Mean Median and Mode

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Mode of a Grouped Data - Formula - Example

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When to Use Chain Rule

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Consider the matrix $A$ given by $A=I-\alpha vv^{T}$ with $v\neq0$ and $v\in\mathbb{R}^{n}$ and $\alpha\neq0$. we want to show that there are two distinct eigenvalues $\lambda_{1},\lambda_{2}$ to be found with their corresponding eigenvectors $x_{\lambda_{1}}$ and $x_{\lambda_{2}}$.

My attempt : By definition, we have that $Ax=\lambda x$ thus : $$ (I-\alpha vv^{T})x=x-\alpha vv^{T}x=x-(\alpha v^{T}x)v $$ One can easily notice that $v$ is nothing but a scalar multiple of $x$ that is to say $x=\beta v$ and thus we have that for $x=v$ we get : $$ Av=(1-\alpha v^{T}v)v $$ Thus, an eigenvalue of $A$ is $\lambda_{1}=1-\alpha v^{T}v$. I am unable to find the second eigenvalue nor the corresponding eigenvectors. I would truly appreciate help as I am lost in the process.

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WiWo

We have had many questions here about the divisibility of $\binom{n}{k}$, many of them dealing with divisibility by powers of primes, or expressions involving the $\textrm{gcd}(n,k)$ (I originally gave several more examples but took TheSimpliFire's advice to shorten the list, and many other examples can still be found by looking at this question's edit history):

• How can we prove that a binomial coefficient (n choose k) is divisible by the ratio of n and the gcd(n,m)?
• Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$

The topic has also lead to some interesting research papers recently:

• In 2014: Some divisibility properties of binomial and $q$-binomial coefficients.
• In 2017: Divisibility of binomial coefficients and generation of alternating groups.
• In 2019: On the divisibility of binomial coefficients.

There's also many associated theorems:

• Kummer's theorem
• Lucas's theorem
• Granville's theorem
• Désarménien's theorem

However I've come across a related problem which is expressed completely in the title, and is remarkably not covered by the above extensive body of literature. In MathJax, if $s>0$ is an integer (let's also make it the largest one for which $2^s$ divides $n$), under what conditions of $k$ do we have the following:

$$ 2^s \mid n \implies 2^s \left\vert \binom{n}{k} \right. . \tag{1} $$

Since $\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$ and $2^s \mid n$ we are left with determining when $\frac{q}{k} \binom{n-1}{k-1}$ is an integer ($q$ being the result when $n$ is divided by $2^s$).

The implication works for a remarkable number of cases, but not always (for example if $\binom{n}{k}$ is odd).

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user1271772

Suppose that when you submit a job application you have a probability of $0.1$ to receive an interview, and a job interview results in a job offer with probability $0.2$. Also assume you only submit one application at a time (i.e. you wait to know if you have been rejected from the job, either at the application stage or the interview stage before applying to another job).

What is the probability of getting your first job offer after submitting at most 3 job applications?

---

Attempt: Either you get a job offer after your first, second, or third application.

• The probability of getting an offer after your 1st application is $(0.1)(0.2)=0.02$.
• The probability of getting an offer after your 2nd application is $(1-0.02)(0.1)(0.2)=0.0196$
• The probability of getting an offer after your 3rd application is $(1-0.02)(1-0.02)(0.1)(0.2)=0.0192$

Thus, $0.02+0.0196+0.0192=0.0588$

I don't think my approach is correct, it doesn't make sense to me that you have a higher probability of getting an offer on your first application than you do on your 2nd or 3rd. I'm stumped as to what the mistake is that I am making.

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diteni1927

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Finding Mode of Ungrouped Data Worksheet

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How to Find Mode of Ungrouped Data - Concept - Examples

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I have the following elementary problem/question that I do not know how to tackle. It comes with a "math-olympiad-flavor" but I suspect it may be much more difficult than an high-school olympiads problem.

Let $\mathscr{P}$ be a simple polygon (not necessarily convex) in the plane, and such that the number of sides of $\mathscr{P}$ is an odd number $2k+1$.

For each vertex $v$ of $P$ there is an intuitive notion of opposite side (because the number of sides is odd), just label the sides counterclockwise at $v$ from $1$ to $2k+1$, and it would be the one with the label $k+1$.

Now suppose that for each vertex and its corresponding opposite side I consider the triangle the determine in the plane. Let us call $\Delta_v$ this triangle and call it a central triangle.

Is the following:

$$ \mathscr{P} \subseteq \bigcup_{v\in \mathscr{P}} \Delta_v$$

always true? In other words, if one paints all central triangles is it true that the whole polygon is then painted?

from Hot Weekly Questions - Mathematics Stack Exchange
Luis Ferroni

We will denote by $\mathbb{K}$ one of the fields $\mathbb{Q}, \mathbb{R}$ or $\mathbb{C}$. On $\mathbb{K}\times \mathbb{K}$ we define the following equivalence relation: $$(a,b)\equiv (a', b') \iff \exists (q,\alpha)\in \mathbb{K}^{*}\times \mathbb{K} \text{ such that } \begin{cases} a=q^2a'+\alpha^2-b\alpha \\ b=qb'+2\alpha \end{cases}.$$ We wish to determine the quotient set $\mathbb{K} \times \mathbb{K}/\equiv$ $\space$ for all $\mathbb{K}$s.
This problem was an extra problem in my abstract algebra class (don't worry, this isn't an attempt to cheat, I am posting this a week after the solution was due to be sent) and I kind of got stuck when it comes to $\mathbb{K}=\mathbb{Q}$.
For $\mathbb{K}=\mathbb{C}$, the things were nice and easy, because the original question asked me to prove that $\mathbb{C}\times \mathbb{C}/\equiv$ is equal to $\{\hat{(0,0)}, \hat{(0,1)}\}$ and this can be checked through (tedious) direct computations.
For $\mathbb{K}=\mathbb{R}$, a friend came up with the idea of expressing $\alpha$ from the second equation and then substituing it in the first one. This gives us the following equivalent characterisation of the equivalence relation: $$(a,b)\equiv (a', b') \iff \exists q\in \mathbb{K}^{*} \text{ such that } 4a+b^2=q^2(4a'+b'^2) \space (*).$$ (notice that this works for all $\mathbb{K}$s, the case $\mathbb{K}=\mathbb{C}$ can be solved much easier by using this, but I didn't really need to think that much for that one since direct computations worked in my context)
For real numbers, this rewrites as $(a,b)\equiv (a', b') \iff \operatorname{sgn}(4a+b^2)=\operatorname{sgn}(4a'+b'^2)$.
As a result, there will be three equivalence classes: the parabola $4x+y^2=0$, its interior and its exterior. A representative for each of these are, respectively, $(0,0), (-1,0)$ and $(0,1)$, so $\mathbb{R}\times \mathbb{R}/\equiv \space = \{\hat{(0,0)}, \hat{(-1,0)}, \hat{(0,1)}\}$.
For $\mathbb{K}=\mathbb{Q}$, the things get pretty nasty by this approach. In this case, $(*)$ rewrites as $(a,b)\equiv (a',b') \iff \sqrt{\frac{4a+b^2}{4a'+b'^2}}\in \mathbb{Q}$ and I haven't been able to make any further progress.

from Hot Weekly Questions - Mathematics Stack Exchange
Alexdanut

QUESTION: Let $f:A \rightarrow \mathbb{R}$ be a limited function and let $P$ be a partition of the block $A$ ($A$ is a block in $\mathbb{R}^m$). Then $f$ is integrable $\iff$ for every sub-block $B$ we have that the function $f|_{B}$ is integrable and in this case, $$\int_{A}f=\sum_{B}\int_{B}f|_{B}$$.

REMARK: The professor allowed us to use the following concepts:

1. Proposition: Let $P_0$ be an arbitrary partition of the block $A$. In order to consider the upper and lower integrals of the limited function $f:A \rightarrow \mathbb{R}$, we just need to consider partition refinements of $P_0$. That is, we have $$\underline\int_{A} f(x) dx= \underset{P\supset P_0}{sup} s(f; P)$$ and $$\overline\int_{A} f(x) dx= \underset{P\supset P_0}{inf} S(f; P)$$
2. Theorem: The limited function $f: A \rightarrow \mathbb{R}$ is integrable $\iff$ for every $\epsilon>0$ it is possible to find a partition $P$ of the block $A$ such that $$\displaystyle\sum_{B\in P} \omega_{B}\cdot vol B<\epsilon$$ Where $\omega_{B}$ is the set of the oscillations, i. e., $$\omega_{B}:= sup\{|f(x)-f(y)|; x, y \in B\}$$

MY ATTEMPTY:

$(\Longrightarrow)$ Let $f: A \rightarrow \mathbb{R}$ be a limited function and let $P$ be a partition of the block $A$. Suppose that $f$ is integrable then $\forall \epsilon >0$ it is possible to obtain an partition $P=P_1 \times \cdots \times P_n$ of $A$ such that $\displaystyle\sum_{B\in P} \omega_B \cdot \text{vol}B <\epsilon$, where $B$ are blocks in $P$. Once $B$ are sub-blocks of $A$, let $P_0$ be an partition of $B$. Therefore for every limited function $f|_{B}$ we just need to consider the refinement partitions of $P_0$. Indeed, let $B=\displaystyle\Pi_{i=1}^{n}[b_i, c_i] \subset A$ then for every $i=1, \cdots, n$ lets define $Q_i= P_i\cap[b_i, c_i]$ from this we have a new partition $Q = Q_1 \times \cdots \times Q_n$ of $A$ that is a refinement of $P$ and, furthermore, the blocks of $Q$ are contained in $B$ makes a partition $P_0$ of $B$. Thus $$\underbrace{\displaystyle\sum_{B'\in P_0}\omega_{B'}\cdot \text{vol} B'}_{(I)}\leq\displaystyle\underbrace{\sum_{B\in P}\omega_{B}\cdot \text{vol} B<\epsilon}_{(II)}$$ $(I) \subset (II)$ therefore $f|_{B}$ is intagrable.

$(\Longleftarrow)$ We just need to consider $P=P_1 \times \cdots \times P_n$ as a partition of the block $A$ and we also need to consider that this partition is a composition of the block $A$ in sub-blocks like $B=I_1 \times \cdots \times I_n$ where every $I_j$ is an interval of the partition $P_j$, where every sub-block $B$ is the block of partition $P$, i.e., $B\in P$. So, writting $A=\displaystyle\bigcup_{i=1}^{n}B_i$ and remembering that every $f|_{B}$ is integrable. Note that if $P_i$ is a partition of $B_i$ we can consider $Q=\displaystyle\sum_{i=1}^{n}P_i$ as an refinement partition of $P$ thus $f:A \rightarrow\mathbb{R}$ is integrable.

Now we just need to show that: $$\int_{A} f \leq \displaystyle\sum_{B \in P} \int_{B} f|_{B}$$.

In $f:A \rightarrow \mathbb{R}$ considering the partition $P$ of the block $A$ we just need to consider refinement partitions of $P$, let $Q$ be an arbitrary partition of the block $A$ we can consider, for instance $P_0= P+Q$. It follows from upper integration definition that $$s(f, P)=\displaystyle\sum_{B \in P} m_B(f)\cdot \textbf{vol}B= \displaystyle\sum_{B \in P} m_{B}(f|_{B}) \cdot \textbf{vol} B$$. Then, for every $B$ we consider $B' \subset B$, the sub-blocks of $B$ resultants of the refinement of $P$, and $B=\bigcup B'$. Therefore, \begin{align*} \int_{A} f = \displaystyle sup_{P_0\supset P} s(f, P_0)& = sup \left(\displaystyle\sum_{B\in P}m_{B}(f|_{B}) \cdot \textbf{vol} B\right)\\ & = sup \left(\displaystyle\sum_{B\in P}m_{B}(f|_{B}) \displaystyle\sum_{B'\subset B} \textbf{vol} B'\right)\\ & = sup \left(\displaystyle\sum_{B\in P}\displaystyle\sum_{B'\subset B}m_{B}(f|_{B}) \textbf{vol} B'\right)\\ & \leq sup \left(\displaystyle\sum_{B\in P}\displaystyle\sum_{B'\subset B}m_{B'}(f|_{B}) \textbf{vol} B'\right)\\ & = \displaystyle\sum_{B\in P} sup \left(\displaystyle\sum_{B'\subset B}m_{B'}(f|_{B}) \textbf{vol} B'\right)\\ & = \displaystyle\sum_{B\in P} \underline{\int_{B}} f|_{B}\\ & = \displaystyle\sum_{B\in P} \int_{B} f|_{B} \end{align*} Thus, $$\int_{A} f \leq \displaystyle\sum_{B\in P} \int_{B} f|_{B}$$

Similarly, we can show for upper sum, and obtain $$\int_{A} f \geq \displaystyle\sum_{B\in P} \int_{B} f|_{B}$$ And finally, conclude $$\int_{A} f = \displaystyle\sum_{B\in P} \int_{B} f|_{B}$$

MY DOUBT: Would you help me to improve my answer? Specialy in this $(\Longleftarrow)$ way.

from Hot Weekly Questions - Mathematics Stack Exchange
Silvinha

AWM Social

Friday, November 13th

5 PM EST

 

AWM (Association for Women in Mathematics) is hosting our own virtual social on Friday, November 13th at 5 PM EST. We will chat about the Spring 2021 Course Catalog, and we encourage all students interested in taking any course in the department to come. This is a great and casual opportunity to meet and hear from current students about the Mathematics and Statistics courses that will be offered in the spring and to get answers for any questions you may have!

 

Please join us through this Zoom link: https://williams.zoom.us/j/99579131155

from Blog – Mathematics & Statistics
Mihai Stoiciu

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      During these days of classifying people and points of view, my thoughts turn again and again to Venn Diagrams and I am then reminded of a thoughtful and imaginative poem (by Pennsylvania poet and professor Marjorie Maddox) that I first read long ago -- and I offer it here:  

Learn about Venn Diagrams here

Venn Diagrams     

          by Marjorie Maddox   

There, stuck in that class,
chalking circles on a board 
       so high your toes ached,
an inch of sock exposed,
all for the sake of subsets,
        intersection.
That teacher with the tie too bright for day,
wide as your fingers spread  

Read more »

from Intersections -- Poetry with Mathematics
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noreply@blogger.com (JoAnne Growney)

Aside from $1!\cdot n!=n!$ and $(n!-1)!\cdot n! = (n!)!$, the only nontrivial product of factorials known is $6!\cdot 7!=10!$.

One might naturally associate these numbers with the permutations on $6, 7,$ and $10$ objects, respectively, and hope that this result has some kind of connection to a sporadic relation between such permutations - numerical "coincidences" often have deep math behind them, like how $1^2+2^2+\ldots+24^2=70^2$ can be viewed as an ingredient that makes the Leech lattice work.

The most natural thing to hope for would be a product structure on the groups $S_6$ and $S_7$ mapping to $S_{10}$, but as this MathOverflow thread shows, one cannot find disjoint copies of $S_6$ and $S_7$ living in $S_{10}$, so a product structure seems unlikely.

However, I'm holding out hope that some weaker kind of bijection can be found in a "natural" way. Obviously one can exhibit a bijection. For instance, identify the relative ordering of $1,2,\ldots 7$ in a permutation of size $10$, and then biject $_{10}P_{3}=720$ with $S_6$ in some way. But I'd like to know if there is a way to define such a bijection which arises naturally from the permutation structures on these sets, and makes it clear why the construction does not extend to other orders.

I tried doing something with orderings on polar axes of the dodecahedron ($10!$) and orderings on polar axes of the icosahedron ($6!$), in the hopes that the sporadic structure and symmetry of these Platonic solids would allow for interesting constructions that don't generalize, but ran into issues with the dodecahedron (sequences of dodecahedral axes aren't particularly nice objects) and the question of how to extract a permutation of length $7$.

I'm curious if someone can either devise a natural bijection between these sets or link to previous work on this question.

from Hot Weekly Questions - Mathematics Stack Exchange
RavenclawPrefect

Context
There are two different Wiener processes $W_t$ and $V_t$. It's known that they are independent. Additionally, we are given with third Wiener process $B_t$ that is given by the formula $$B_t = aW_t+bV_t, \quad \quad a^2 + b^2= 1.$$

Problem
Find the limit in $L^2$ of $$S_n = \sum_{i=1}^n\left[B_{it/n} - B_{(i-1)t/n}\right]\left[V_{it/n} - V_{(i-1)t/n}\right]$$
as $n$ tends to infinity.

My ideas
I assume that this is the type of task where we need to calculate the expected value and the variance. As the latter tends to $0$ (it should), we can say that the desired limit is the expected value. The issue is that it's very overextended work to calculate the $E(S_n)$.

Let $W_{it/n} = X_i$ and $V_{it/n} = Y_i$. We have
$$\Bbb E(S_n) = \Bbb E\sum_{i=1}^n [aX_i + bY_i - aX_{i-1} - bY_{i-1}][Y_{i} - Y_{i-1}]$$ which can be written as $$\sum \Bbb E\bigg( aX_iY_i + bY_i^2 - aX_{i-1}Y_i - bY_{i-1}Y_i - aX_iY_{i-1} - bY_iY_{i-1} + aX_{i-1}Y_{i-1} + bY_{i-1}^2\bigg).$$ Next calculations confuse me (what is $\Bbb E(X_i Y_{i-1})$?) Is it zero? And the main question how to calculate the variance?

If I'm not mistaken, $\Bbb E(S_n) = nb \to \infty$, so we don't need variance. Am I right?

from Hot Weekly Questions - Mathematics Stack Exchange
student

from Maths & Physics News
Evgeny Khukhro

Act One:  Take a look at the pic. What questions come to mind?

Just before Thanksgiving there are competitions all over the world to celebrate cool design, tricky engineering, and to donate a whole lot of food. At the end of the exhibit, all the creations are dismantled and the construction materials (thousands of cans of food) are given directly to soup kitchens, food banks, shelters and senior homes. Canstruction, Inc. now has events in over 200 cities worldwide.

Above is a picture of an exhibit called "Downside Up" which was the winning exhibit from Canstruction, 2010, in New York City.

Act Two:  How many cans does it take to build this structure?  What information do you need to determine this?  How did you determine your solution? What else did you notice that is mathematical?

The Activity: Canstruction.pdf

CCSS: 4.MD.3, 6.EE.1, 5.MD.5, 7.G.6, HSF.LE.2, MP2, MP3, MP7

Act Three: 

• We have a detailed solution and teacher tip page: Canstruction-solution.pdf and the Word docx if you would like to change the activity: Canstruction.docx
• You can also find a different solution at the website that covered the event:
  http://www.glenwoodnyc.com/manhattan-living/canstruction-nyc-2010-at-the/.  How they got their solution isn't explained and our count was different but there are some other lovely creations to view.    It might be an interesting discussion as to why in the article above they came to a different solution.
• Or let students continue their investigation with: Have a Heart.
• Or let students choose any of the creations in our movie to analyze.

from Yummy Math
Leslie

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Given $a_{n}\geq0$, let $b_{n}$ be rearrangement of $a_{n}$. Show that $\sum b_{n}=\sum a_{n}$

My Proof : Let $A_{n}=a_{0}+a_{1}+\cdots+a_{n}$ and let $B_{n}=b_{0}+b_{1}+\cdots+b_{n}$ be a rearrangements of $a_{n}$. Now we note that the sequence of the partial sums $A_{n}$ and $B_{n}$ are either finite, or either $A_{n}$ or $B_{n}$ approach infinity. Thus, it may either be the case that $A_{n}\leq\displaystyle\lim_{n\to+\infty}B_{n}$ or $B_{n}\leq\displaystyle\lim_{n\to+\infty}A_{n}$. We have that $\displaystyle\lim_{n\to+\infty}B_{n}$ involves all the the sum over all $b_{n}$ and so it involves the sum over $a_{0},a_{1},...,a_{n}$ and same for $\displaystyle\lim_{n\to+\infty}A_{n}$ . Taking the limit $n\to+\infty$ we find that $\displaystyle\lim_{n\to+\infty}B_{n}\leq \displaystyle\lim_{n\to+\infty}A_{n}$ and $\displaystyle\lim_{n\to+\infty}A_{n}\leq\displaystyle\lim_{n\to+\infty}B_{n}$ and thus we have that $\displaystyle\lim_{n\to+\infty}A_{n}=\displaystyle\lim_{n\to+\infty}B_{n}$

Is this a complete proof or is my proof missing some details?

from Hot Weekly Questions - Mathematics Stack Exchange
WiWo

Let $\mathcal{E}$ be a topos with subobject classifier $1\overset{t}{\rightarrow}\Omega$, then I want to show that $\Omega$ is injective.

So let $f:A\rightarrow\Omega$ and $g:A\rightarrow B$ be two maps in $\mathcal{E}$ with $g$ monic. Then I want to show that there exists a map $h:B\rightarrow\Omega$ such that $h\circ g = f$. Notice that by the property of a subobject classifier we have that there exists a unique map $\phi:B\rightarrow\Omega$ such that there is a pullback square $\require{AMScd}$ \begin{CD} A @>{\psi}>> 1\\ @VV{g}V @VV{t}V\\ B @>{\phi}>> \Omega \end{CD} Now I claim that $\phi$ does the job for $h$. But I don't see why $\phi\circ g= f$ (so maybe my claim is wrong). Any help would be appreciated!

from Hot Weekly Questions - Mathematics Stack Exchange
Peter

In this series of posts, we’ll be featuring mathematical podcasts from all over the internet, by speaking to the creators of the podcast and asking them about what they do.

We spoke to Rob Eastaway and Andrew Jeffrey about their new podcast, Puzzling Maths.

What is your podcast about, and when did it start? 

This podcast is about puzzles and the maths of everyday life, away from the context of school. It was partly inspired by the success of Andrew Jeffrey’s daily ‘Learner Drivers’ puzzle slot on BBC 5Live during Lockdown. The first episode went out in August 2020.

Tell us about yourselves – who are you?

We are Rob Eastaway and Andrew Jeffrey – two maths popularisers who want to spread the joy of maths beyond those who are already addicted to the subject.

Who is the intended audience for the podcast? 

This is aimed at that broad cross-section of the general public who are curious about maths and puzzles but would not necessarily regard themselves as mathematicians. The content will also appeal to teachers who are interested in hearing how those outside the classroom perceive and use the subject.

What is a typical episode like?

Our typical episode includes: a couple of quickie puzzles that we set each other; a discussion of unusual ways in which maths has cropped up in our everyday life since the previous podcast; an interview with a special guest to ask how maths features in their profession; and then a more involved puzzle that we leave with the listener to be answered in the next episode.

Why should people listen? Why is it different to other mathematical podcasts?

The podcast is very conversational, and an easy and inclusive way for ‘non-mathematicians’ to discover new mathematical ideas and discover how mathematicians think.

What are some highlights of the podcast so far?

Interviewing BBC radio broadcaster Anna Foster in Episode 3 and asking her why the media often appears to boast about not being good at maths. It’s also fascinating to discover what maths our guests would teach in schools if they were given control of the curriculum for a day.

What exciting plans do you have for the future? 

We want to include as diverse a range of guests as we can: historians, musicians, doctors – everyone has a connection with maths.

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Katie Steckles

from Maths & Physics News
Evgeny Khukhro

A Ramanujan summation is a

technique invented by the mathematician Srinivasa Ramanujan for assigning a value to divergent infinite series

In my case, I'm interested in assigning a value to the divergent series

$$\sum_{n=1}^\infty f(n) \ \ \ \ \ \ \ \text{where}\ \ \ \ f(n)=\sqrt[n]{2}$$

According to the Wikipedia page (and my understanding), the Ramanujan summation is

$$\sum_{n=1}^\mathfrak{R} f(n)=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N f(n)-\int_{1}^N f(t)dt\Bigg]$$

Thus

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\int_{1}^N \sqrt[t]{2}dt\Bigg]$$

Taking the antiderivative

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\Bigg(\ln2\Big(\text{li}\ 2-\text{Ei}\frac{\ln2}{N}\Big)+N\sqrt[N]{2}-2\Bigg)\Bigg]$$

Moving some constants outside the limit

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=2-\ln2\cdot\text{li}\ 2+\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\Bigg(N\sqrt[N]{2}-\ln2\cdot\text{Ei}\frac{\ln2}{N}\Bigg)\Bigg]$$

It's at this point I'm unsure of how to proceed. I'm not terribly confident what the limit converges to. From my computational estimates up to $N=10^8$, I find that

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}\approx1.6$$

But due to floating point errors or slow convergence, it deviates substantially enough for me to not be confident about any more digits.

I'd like to know if this converges at all, and if it does, is there a (reasonably) closed form / relation to other constants?

from Hot Weekly Questions - Mathematics Stack Exchange
Graviton

Porous numbers are numbers $k$ which are not multiples of 10 such that every m with sum of digits = $k$ and $k$ a divisor of both $m$ and rev($m$) has a zero in its digits. rev($m$) is the digit reversal of $m$ (e.g. rev(123) = 321).

Below 1000 there are only 11, 37, 74, 101 and 121 which fulfill these requirements (see OEIS sequence A337832).

Since $11$ and $11^2$ are porous, I wonder if $11^3$ and eventually all $11^n$ might be porous as well.

from Hot Weekly Questions - Mathematics Stack Exchange
Rüdi Jehn

Recently I came up with a problem regarding Fibonacci numbers:

For which $N$ is it possible to arrange all whole numbers from $1$ to $N$ in such a way that every adjacent pair sums up to a Fibonacci number?

I have manually tested a bunch of cases and I was also able to prove almost every case. The results that I was able to prove are the following:

• If $N$ is a Fibonacci number or exactly one less than a Fibonacci number, I was able to prove that an arrangement exists. I used an argument by induction to achieve this.
• If $F_k+2 \leq N \leq F_{k+1} -3$ , it is completely impossible, because the numbers $F_k$, $F_k + 1$ and $F_k + 2$ have only one possible pair and therefore have to be at the end of the arragement. This obviously gives a contradiction, because arrangements only have two ends (the first number and the last number).

The cases I was not able to solve are $N=F_k+1$ and $N=F_k-2$. My theory is that $N=9$ is the only working case of the form $N=F_k+1$, and $N=11$ the only working case of the form $F_k-2$. I expect every other $N$ of these two forms to be impossible.

Does anybody know a full proof to this problem or maybe the name of the official theorem (if this exists)?

from Hot Weekly Questions - Mathematics Stack Exchange
Rubenscube