17
votes

Accepted

### The average of an infinite sequence

By induction, $a_i=(p+u)\{i\frac{u}{p+u}\}$, where $\{x\}$ is a fractional part of $x$. Thus, by Weyl equidistribution theorem, if $p/u$ is irrational, the average equals $(p+u)/2$, otherwise $u/(p+u)=...

14
votes

Accepted

### Closed form for $\sum_{n=0}^\infty \frac1{2^{2^n}}$?

If you allow for a named number to be a closed form representation, the answer is "yes".
$\sum_{n=0}^\infty (1/2)^{2^n}$ is known as the Kempner number [1], a transcendental number [2].
More ...

12
votes

### Closed form expression for $\sum_{n=0}^{\infty} J_n^2(x) \cos(ny)$, where $J_n(x)$ is the Bessel function of order $n$

Just solve for the sum in the addition formula (31) of Neumann 1867, p. 65 (also in Watson p. 128, or more conceptually Vilenkin 1968, formula (4) p. 209):
$$
J_0\left(2r\sqrt{\frac{1-\cos\theta}2}\...

10
votes

Accepted

### Permutations of the natural numbers with a common conditionally convergent series

Let me rephrase your Question 2: How many permutations of the natural numbers are needed so that every conditionally convergent series of real numbers can be rearranged to no longer converge? This ...

9
votes

Accepted

### Summation of binomial coefficients with alternating signs

It's not hard to obtain explicit formulas:
$$S_1(n,\alpha) = \alpha^{-1}\binom{n+1-\frac1{\alpha}}n^{-1}$$
and
$$S_2(n,\alpha) = (1-2\alpha)\alpha^{-2}\binom{n+2-\frac1{\alpha}}n^{-1}.$$

6
votes

### Possible new series for $\pi$

A related, and perhaps easier, question is whether there are other known series for 𝜋 that involve a complex parameter 𝜆 in the summand, but where the sum of the series is independent of the value ...

5
votes

### Is it known that the Collatz-like sequence with 7n+1 diverges to infinity starting with 7?

Since others are contributing knowledge about other factors than $7$, I figured that I would chip in. It's been known since 1987 that the orbit of $t^3 + 1$ diverges in the ring $\mathbb{F}_2[t]$ ...

5
votes

Accepted

### Do disjunctive sequences eventually get palindromic at some point?

Any disjunctive sequence of $0$'s and $1$'s will always have a palindromic initial segment: say it starts with $1$, the next digit must be $0$ to avoid an immediate palindrome, and then the next $1$ ...

4
votes

Accepted

### 5 different ways to define the same family of integer sequences

Let's prove that $a_3=a_1$.
Note that the recurrence for $R$ translates to the following PDE for the generating $F(x,y):=\sum_{n,m\geq0} R(n,m,p,q) \frac{x^n}{n!}\frac{y^m}{m!}$:
$$\frac{\partial}{\...

4
votes

### 5 different ways to define the same family of integer sequences

Here is a proof that $a_1(n, p, q) = a_2(n,p,q)$.
The generating function for the Stirling numbers of the second kind is
$$\sum_{n=i}^\infty {n\brace i}\frac{x^n}{n!}= \frac{(e^x-1)^i}{i!}.$$
Also
$$\...

4
votes

Accepted

### An inequality about binomial distribution

Another try, now I claim that the inequality holds.
For a random variable $X$, denote $\|X\|=(\mathbb E |X|^\sigma)^{1/\sigma}$. It is indeed a norm.
Then
$$
\|X_1+\ldots+X_n\|\leqslant \|X_1+\ldots+...

3
votes

### Is there another representation for the summation: $\sum_{j=1}^{N}\frac{a_j}{(c+a_j)(c+a_j+1)} $, how to reformulate that to keep $c$ out of the sum

A few words on existence and approximation. Write your equation as
$$ \sum_{n=1}^N \frac{c^2a_n}{(c+b_n)(c+b_n+1)}=1.$$
There is a nice partial fractions decomposition,
$$\frac{c^2 }{(c+b )(c+b +1)}=1+...

3
votes

Accepted

### $R$-recursion for Fibonacci numbers using signed Catalan numbers

Let $R(x,y) = \sum_{n,q} R(n,q)x^ny^q$. Then, you can show that
\begin{eqnarray*}
R(x,y)\left(x - y - xyC(-y)\right) &=& x R(x,0) - yR(0,y) \\
&=& xR(x,0) - \frac{y}{1-y}
\end{eqnarray*...

2
votes

### Possible new series for $\pi$

This answers the final question about other parametric similar series for $\pi$. Working on the Saha & Sinha paper, it is possible to get a bi-parametric generalization of a related formula. If $s\...

2
votes

### Conjectured Somos-like closed form of recurrences with polynomial coefficients

This is just an extended comment.
There is no need to invoke the algebraic dependency search. The recurrence can be found by directly constructing Groebner basis $B$ under any term order, in which $n$ ...

2
votes

### Lower bound in recurrence relation

From the definition $$v_{k-1}=\frac1{15}\prod_{j=1}^k(2^j+1)=$$$$= \frac1 {15}{2^{ {\frac12k(k+1)}}}\prod_{j=1}^k(1+2^{-j}),$$
and from the bound of the infinite product
$$\frac1 {15}\prod_{j=1}^\...

1
vote

Accepted

### Lower bound in recurrence relation

This recurrence looks fairly random, so let's try tackling it in parts. First, we will show that $\log v_k$ and $\log n_k$ grow like $k^2$. Indeed, it can not grow slower since even if we leave the ...

1
vote

Accepted

### $R$-recursion for unsigned Genocchi numbers (of first kind) of even index

From the recurrence it follows that the generating function $F(x,y) := \sum_{n,k\geq 0} R(n,k) \frac{x^{2n}}{(2n)!} \frac{y^k}{k!}$ satisfies a neat PDE:
$$\frac{\partial^2}{\partial^2 x} F(x,y) = \...

1
vote

### Simplest way to generate integer coefficients with row sums equal to the terms of an arbitrary given sequence

I don't have any answer, but here are a few terms to be sure it is clear and to show some regularity in the constant or linear terms. Writing $f_n := f(n)$ and taking $f_0=1$, it looks like
$$\begin{...

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