## How many pairs $(x,y)$ such that $x+2=y$ have an $x$ or $y$ divisible by 5 and not 3?

Let the set $$S_{n}$$ = {$$(x,y):x,y \in \mathbb{O}$$} such that $$x+2=y$$ and $$y$$ is less than or equal to odd integer $$n$$ and $$\mathbb{O}$$ is set of odd integers > 1.

Let us define the function $$f(n) = |S_{n}|$$ that counts the number of pairs in $$S_{n}$$.

Examples: $$S_{9} =\{(3,5),(5,7),(7,9)\}$$ and $$f(9) = 3$$.

$$S_{13} = \{(3,5), (5,7), (7,9), (9,11),(11,13)\}$$ and $$f(13) = 5$$

$$S_{35} = \{(3,5), (5,7), (7,9), (9,11), (11,13), (13,15), (15,17), (17,19), (19,21), (21,23), (23,25), (25,27), (27,29), (29,31),(31,33),(33,35)\}$$

and $$f(35) = 16$$

It turns out that $$f(n) = ((n-3)/2)$$ for odd integers $$n$$.

Let us define another function $$g(n)$$ that counts the number of pairs $$(x,y)$$ such that either $$x$$ or $$y$$ is divisible by 5 and not divisible by 3, and $$x \neq 5$$ and $$y\neq 5$$.

Example: $$g(35) = 3$$ because there are 3 pairs where $$x$$ or $$y$$ is divisible by 5 and not 3, and $$x$$ or $$y$$ is not equal to 5. They are (23,25), (25,27) and (33,35).

Note, the pairs (3,5) and (5,7) are not included since they contain 5, and the pairs (13,15) and (15,17) are not included because 15 is also divisible by 3.

What is the formula for $$g(n)$$ in terms of $$f(n)$$ ?

What is the formula for $$g(n)$$ in terms of $$f(n)$$ for limit $$n \to\infty$$?

## Random variables $X,Y$ such that $E(X|Y)=E(Y|X)$ a.s.

Let $$X$$ and $$Y$$ be random variables such that $$E(|X|), E(|Y|)<\infty$$ and $$E(X|Y)=E(Y|X)$$ a.s. Then is it true that $$X=Y$$ a.s. ? If this is not true in general, what happens if we also assume $$X,Y$$ are identically distributed ?

## $X,Y$ spaces such that are cones over $X\cap Y$ generate $X\cup Y=\sum (X\cap Y)$

I have this question from this Kratzer and Thevenaz article :

In page 89, they explain that

If $$Y=Y_1\cup Y_2$$ are such that $$X=Y_1\cap Y_2$$ and if each $$Y_i$$ is a cone over $$X$$, then $$Y$$ is the suspension of $$X$$

But in page 90, in the proof of the Proposition 2.5, they use this in a way I don’t understand. They prove that $$\mid X\mid \simeq C\mid E\mid$$ and $$\mid Y\mid \simeq C\mid F\mid$$, but those aren’t cones over $$\mid X\cap Y\mid$$, so why I can continue the proof of this proposition saying that this implications make $$\mid G\mid \simeq \sum(\mid X\cap Y\mid)$$ ?

My concept of a “cone over $$\mid X\cap Y\mid$$” is:

$$\mid X\cap Y\mid\times [0,1]/ (a,i)\thicksim(b,1)$$

with $$a,b\in X\cap Y$$.

Am I misunderstanding what is a cone over $$\mid X\cap Y\mid$$?