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Results tagged with sequences-and-series
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user 46140
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.
10
votes
Accepted
Radio-playing sequence
I think you're just asking for a de Bruijn sequence of order $2$ on $n$ symbols, in which case the answer is $n^2$ because the non-simple digraph on $n$ vertices where $u \to v$ for every $u, v$ (incl …
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1
vote
A $1$-step convolution identity involving the Motzkin triangle
Using different bound variables on the two sides for clarity in the subsequent discussion, the goal is:
$$\sum_{k=0}^{n-1}T(n,k) \, T(n,k+1)=\sum_{j=0}^{n-1}\binom{2n}{2j+1}\binom{2j+1}{j}\frac1{j+2}
…
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1
vote
Sum of reciprocal of Pochhamer symbols through multiples of a natural L
The identity you link on the special functions wiki can be rewritten as
$$\sum_{k=1}^n \frac{\Gamma(k)}{\Gamma(k+r)} = \frac{1}{(r-1)\Gamma(r)} - \frac{n\Gamma(n)}{(r-1)\Gamma(n+r)}$$
This clearly has …
- 7,226
7
votes
Accepted
Numbers $m$ for which coefficients of the polynomial $p(m,x)$ are relatively prime
Counterexample: $463 \in b(n)$ (it's a prime and $464 = 2^4 \cdot 29$ is not squarefree), but $463 \not \in a(n)$ because it's a factor of the GCD of the coefficients of $p(463, x)$.
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15
votes
Accepted
Multiple roots of polynomials with coefficients $\pm 1$
Question P. Can a polynomial $P(x)=\sum_{n=0}^ma_nx^n$ with coefficients $a_n\in\{-1,1\}$ (and $P(1)=0$) have a multiple root in the interval $(\tfrac12,1)$?
Yes. The following four Littlewood polyn …
- 7,226
1
vote
Connection between central factorial numbers and the Stern–Brocot tree
As I noted earlier in a comment, you can substitute the Stirling numbers of the second kind for $U$, since the difference between the recurrence $U(n,k) = U(n-1,k-1) + k^2 U(n-1,k)$ and the recurrence …
- 7,226
4
votes
Accepted
Periodic sequences of integers generated by $a_{n+1}=\frac{\operatorname{rad}(pa_{n})}{p}+\f...
For any odd $p$, $q$ (not necessarily prime) the values modulo $2$ follow a cycle of order 3.
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3
votes
$a(16n+k)=b(16n+k)-c(16n)$ for $n\geqslant0$, $0 < k < 16$ where $c(n)=b(n)-a(n)$
In this answer, $\wedge$ denotes bitwise AND and $[\;]$ are Iverson brackets.
$$b(n) = \sum_{k \ge 0} \sum_{j \ge 0} (-1)^{k}(j+1) [n \wedge 2^{j+k}] = \sum_{e \ge 0}[n \wedge 2^e] \sum_{j=0}^e (-1)^{ …
- 7,226
3
votes
Operation preserving log-concavity of sequences
If my calculations are correct, a counterexample for question 2 is $$f= \frac{720 + 1684x + 1350x^2 + 585x^3 + 90x^4 + 11x^5}{120} \\
g = \frac{600 + 1434x + 1175x^2 + 535x^3 + 85x^4 + 11x^5}{120}$$ h …
- 7,226
8
votes
Accepted
Subsequence of the cubes
Experimenting with a CAS suggests an induction. In order to handle the induction, we need to consider the forms of the numbers involved. $\frac{4^m-1}{3} = 1 + 2^2 + 2^4 + \cdots + 2^{2m-2}$ alternate …
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3
votes
Subwords of the infinite Fibonacci word
Such a "splitter" turns out to be simply a reversal of an initial word of $W$, so that the first few splitters are $,0,10,010,0010,10010,\ldots$. The corresponding Wythoff composites are
$$A,B,AA,AB, …
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3
votes
Accepted
Stern-Brocot tree and subtree
The second half is already given in the question, so really what you're asking is whether $$b(2n-1)=b(2n-3)+b(n-1)-2(b(2n-3)\bmod b(n-1))$$
But as noted in OEIS (quoted with relabelling),
Moshe Newma …
- 7,226
4
votes
Accepted
Another combinatorial identity
Subst $k = p - 2n \ge 0$ and $s = r - i$ to get the symmetric
$$\sum_{s \ge 0,i \ge 0} [s + i \le 2n + k] \frac{(-1)^{s+i} (3n+k-s-i-1)! (2n^2 + nk - is)}{i!(n-i)!(2n + k-i)! s!(n-s)!(2n + k-s)!}$$
Bu …
- 7,226
1
vote
Accepted
Recurrence for the number of steps required to get one ball in each box
Generalise $a$: $a(n, k)$ is the number of steps to perform this process with $n+k$ boxes and balls starting with $n \ge 1$ balls in the first box and one ball each in the next $k$ boxes. Then the ori …
- 7,226
1
vote
Accepted
Sequences that sum up to Dowling numbers
Cleaning up the notation a bit,
$$b_{m,k}(n) = m\, b_{m,k}(n-2^{\ell(n)}) + k \sum_{j=0}^{\ell(n)-1} [n \,\&\, 2^j = 0] \,b_{m,k}(n - 2^{\ell(n)} + 2^j)$$ where $\&$ is bitwise AND.
$$s_{m,k}(n) = \su …
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