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for questions about sequences and series, e.g. convergence, closed form expressions, etc. Note that there is a different tag for spectral sequences, and also note that MathOverflow is not for homework. Please consider consulting the online encyclopedia for integer sequences, if you are trying to identify a given sequence that you have found in your research.
9
votes
1
answer
229
views
Riccati-type recurrence: infinitely many sign changes?
Suppose $b_1, b_2, b_3, \dots \in \Bbb{R}$ satisfy the Riccati-type recurrence
$$b_{k+1}=\frac{1+kb_k}{k-b_k},\quad k\ge 1.$$
Is it true that such a sequence reaches infinitely many positive as we …
10
votes
4
answers
1k
views
Adventure with infinite series, a curiosity
It is easily verifiable that
$$\sum_{k\geq0}\binom{2k}k\frac1{2^{3k}}=\sqrt{2}.$$
It is not that difficult to get
$$\sum_{k\geq0}\binom{4k}{2k}\frac1{2^{5k}}=\frac{\sqrt{2-\sqrt2}+\sqrt{2+\sqrt2}}2.$$ …
9
votes
0
answers
179
views
Infinite series identities in search of a proof
This comes in relation to the Fishburn numbers.
I stumbled on the following relation for which I ask a proof if true.
Let $Q_i(z):=1-(1-t)^{i-1}(1-zt)$. Then
$$\sum_{n=0}^{\infty}\frac{(n+1)zt …
10
votes
2
answers
308
views
Denominators of certain Laurent polynomials
Consider the following somos-like sequence
$$x_n=\frac{x_{n-1}^2+x_{n-2}^2}{x_{n-3}}.$$
It's known that $x_n$ is a Laurent polynomial in $x_0, x_1$ and $x_2$. I got interested in the denominators of t …
7
votes
1
answer
345
views
Descartes' rule of signs for infinite series
Consider the function given by
$$f(x)=1-a_1x-a_2x^2-a_3x^3-\cdots$$
where each $a_k\geq0$ and some $a_j>0$. If $f(x)$ is a polynomial then Descartes' Rule of signs tells us there is exactly one positi …
2
votes
1
answer
69
views
Decaying of a certain ratio of binomial sums
Consider the two sequences
$$a(n)=\sum_{k=1}^n\binom{n}k\sum_{j=1}^{k/2}\binom{k}{2j}\frac{(2j)!}{j!}$$
and
$$b(n)=\sum_{k=0}^n\binom{n}k^2k!$$
QUESTION. Is this true?
$$\frac{a(n)}{b(n)}\longrightar …
-1
votes
1
answer
128
views
Collapsed partitions and ordinary partitions
Adopt the standard notation for integer partitions, writing $\lambda_1^{a_1} \cdots \lambda_k^{a_k}$ as shorthand for the partition $a_1 \lambda_1 + \cdots + a_k \lambda_k$ with parts $\lambda_1 > \cd …
1
vote
1
answer
122
views
A $1$-step convolution identity involving the Motzkin triangle
The Motzkin triangle $T(n,k)$ is built according to the rules:
(1) $T(n,0)=1$;
(2) $T(n,k)=0$ if $k<0$ or $k>n$;
(3) $T(n,k)=T(n-1,k-2)+T(n-1,k-1)+T(n-1,k)$.
After some numerical evidence I ask:
…
11
votes
3
answers
1k
views
Integrality of a binomial sum
The following sequence appears to be always an integer, experimentally.
QUESTION. Let $n\in\mathbb{Z}^{+}$. Are these indeed integers?
$$\sum_{k=1}^n\frac{(4k - 1)4^{2k - 1}\binom{2n}n^2}{k^2\binom{2 …
4
votes
1
answer
187
views
Is there a generalization of these q-series identities?
Denote $(q;q)_n=(1-q)(1-q^2)\cdots(1-q^n)$.
The below three identities are known.
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(q;q)_n}
&=1-\sum_{n\in\mathbb{Z}}(-1)^nq^{\frac{n …
3
votes
1
answer
168
views
Divisibility question while enumerating endomorphisms: PART 2
From endomorphisms of rank 1 of the full transformation semigroup $[n]^{[n]}$ or idempotents in $[n]^{[n]}$, we have
$$c_n:=\sum_{m=1}^n\binom{n}mm^{n-m}.$$
Denote the $2$-adic valuation of $x$ by $\n …
4
votes
1
answer
146
views
Integrality of ratios of $\ell$-sequences
The so-called $\ell$-sequences are defined by $a_0=0, a_1=1$ and $a_n=\ell\,a_{n-1}-a_{n-2}$. The Generalized Lecture Hall Theorem (due to Mireille BousquetMelou and Kimmo Eriksson) depends on a polyn …
2
votes
0
answers
96
views
Log-convexity of Lassalle's sequence
Lassalle's sequence is defined by the recurrence $A_1:=1$ and for $n\geq2$,
$$A_n=(-1)^{n-1}C_n + (-1)^{n- 1}\sum_{j=1}^{n-1}(-1)^j\binom{2n - 1}{2j - 1}A_jC_{n - j}$$
where $C_k=\frac1{k+1}\binom{2k} …
1
vote
2
answers
266
views
An elementary proof for a limit? [closed]
This question is motivated by pedagogical reason, not research. I will provide a simple proof for contrast, but I would like to see another approach that does not involve integrals, instead even more …
1
vote
0
answers
98
views
Divisibility question while enumerating endomorphisms: PART 1
From endomorphisms of rank 1 of the full transformation semigroup $[n]^{[n]}$ or idempotents in $[n]^{[n]}$, we have
$$c_n:=\sum_{m=1}^n\binom{n}mm^{n-m}.$$
QUESTION. Is it true that $n$ divides $c_ …