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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.
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
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 …
- 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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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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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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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 …
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4
votes
Accepted
Property of some permutations of non-negative integers such that $a(n)<2^k$ iff $n<2^k$
I'm going to use $\operatorname{msb}$ (for most significant bit) as an alias of $f$.
Since $q_i$ is a permutation, the property that $q_i(n)<2^k$ iff $n < 2^k$ is equivalent to $\operatorname{msb}(q_i …
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4
votes
Accepted
Correctness of the algorithm for the A329369, A347205 and related sequences
Generalise to $$b(2^m(2k+1)) = \sum\limits_{j=0}^{m}C_{m+1,j} \, b(2^jk), \\
b(0) = 1$$
Consider the infinite matrices: $$M_0 = \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots \\
0 & 0 & 1 & 0 & \cdots \\
0 & …
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3
votes
Subwords of the infinite Fibonacci word
Such a "splitter" turns out to be simply a reversal of an initial word of $W$, so that the first few splitters are $,0,10,010,0010,10010,\ldots$. The corresponding Wythoff composites are
$$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 …
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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.
$$b(n) = \sum_{k \ge 0} \sum_{j \ge 0} (-1)^{k}(j+1) [n \wedge 2^{j+k}] = \sum_{e \ge 0}[n \wedge 2^e] \sum_{j=0}^e (-1)^{ …
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3
votes
Operation preserving log-concavity of sequences
If my calculations are correct, a counterexample for question 2 is $$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}$$ h …
- 7,226
3
votes
Accepted
Permutation of the natural numbers from operation related to binary expansion of $n$
$\ell(0)$ is problematic, so I will assume that you actually mean to restrict to positive integers rather than natural numbers.
We can rephrase the construction of $a(n)$ to emphasise the increment: l …
- 7,226
2
votes
Accepted
$q$-series and Stirling of the 1st kind
QUESTION. Is this true? Or, can you provide a reference to it.
$$\mathbf{F}_a(q)=\frac1{(2a-1)!}\mathbf{G}(\mathbf{G}^2-1^2)(\mathbf{G}^2-2^2)(\mathbf{G}^2-3^2)\cdots(\mathbf{G}^2-(a-1)^2);$$
where w …
- 7,226
2
votes
Sequence derived from transform of a given vector (with Fibonacci as partial sums)
Not a complete answer, but too long for a comment and addressing the conjecture which I take to be the most important part of the question.
The double-loop transformation process seems familiar to me …
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