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Results tagged with co.combinatorics
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user 7076
Enumerative combinatorics, graph theory, order theory, posets, matroids, designs and other discrete structures. It also includes algebraic, analytic and probabilistic combinatorics.
0
votes
Calculating the number of solutions of integer linear equations
Let $r:=\sum_j c_{1j}$, $c:=\sum_i c_{i,1}$, and $s:=c_{11} + c_{12} + c_{21} + c_{22}$. Clearly, $s\leq N$, $2r\geq s$, $2c\geq s$, and $2r+2c-s \leq N$.
Next, let $u:=c_{11}$, $v:=c{12}$, $w:=c_{21 …
- 29.5k
3
votes
Order of magnitude for trigonometric sum
I believe the problem can be approached from the generating-functions perspective.
First we notice that at least two of the $k_i$'s are positive. Based on the symmetry, let's consider the sum with $ …
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2
votes
Accepted
Show that $\sum_{i=0}^{2k} [ {n\choose i+1} + (-1)^{i+1}{n+i+1\choose i+1} ] \sum_{j=0}^i {i...
As pointed out by Ira Gessel, $u(k,j)=j!S(k+1,j+1)$. Correspondingly, the sum in question reduces to
$$f_{2k}(n) + f_{2k}(-n-1),$$
where
$$f_k(t):=\sum_{i=1}^{k+1} S(k+1,i)\frac{(t)_i}i,$$
where $(t)_ …
- 29.5k
2
votes
Accepted
Sequence that sums up to INVERTi transform applied to the ordered Bell numbers
First, we let $P(j,k):=1+f(\lfloor\tfrac{j}{2^k}\rfloor+1)$ and sum over $n$ of fixed weight $\ell:=\mathrm{wt}(n)$ (like in this answer):
\begin{split}
s(n) &= \sum_{\ell=0}^n \sum_{t_1 + \dots + t_\ …
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1
vote
Multivariate generating function
The given g.f. directly follows from the sequence definition.
Indeed, the sequence gives the number of partitions of vector $(n,m)$ into the sum of vectors with nonnegative integer components. Each p …
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2
votes
Accepted
probability of zero subset sum
First off, let me prove that all $P(n,k)=0$ for $n\geq k$, which follows from a simple lemma:
Lemma. In any sequence of $k\geq 1$ integers $m_1, m_2, \dots, m_k$, there exists a subsequence summing t …
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3
votes
Accepted
Compare AM and GM
The inequality $(2)$ (even with factor $\frac12$ in the r.h.s.) follows from the inequality quoted in this answer:
$$M_a - M_g \leq \frac1{2n\min_k x_k} \sum_{i=1}^n (x_i - M_a)^2.$$
First, we notice …
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6
votes
sum of odious numbers to the power of k
I think there is no simple formula here, although we can get some recurrence relations and related identities for generating functions as explained below.
Similarly to odious numbers, we have evil nu …
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2
votes
For which pairs $k$ and $n$, $n\mid{{n-2} \choose {k}}$
I doubt there exists a simple description for the general solution, however it's possible to give a characterization in some cases.
For example, in the case of square-free odd $n$, Lucas' theorem impl …
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3
votes
What is this sequence counting?
Notice that $P(n-1)$ counts the number of partition of $n$ that contain $1$, while $P(n-2)$ counts the number of partition of $n$ that contain $2$.
It follows that $P(n-1)+P(n-2)-P(n)$ equals the di …
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6
votes
1
answer
282
views
Explicit expression for recursive sums - II
A twist on just unfolded recursive summation formula. Let polynomials in nonnegative integer variables $t_1,t_2,\dots$ be defined by the recurrence:
\begin{split}
g_0 &= 1, \\
g_k(t_1,t_2,\dots,t_k) & …
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3
votes
Accepted
Counting numerical semigroups by largest element of minimal generating set
A key observation is that two sets of generators $g, g'\subseteq [n]$ produce the same semigroup if and only if $\langle g\rangle \cap [n] = \langle g'\rangle \cap [n]$. Hence, the number of different …
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2
votes
Decomposition of even symmetric polynomials and Euler numbers
It is easy to express $E[(x_1+\dots+x_n)^{2k}]$ in terms of monomial symmetric polynomials of $x_1^2,\dots,x_n^2$:
$$E[(x_1+\dots+x_n)^{2k}] = \sum_{\lambda \vdash k} \frac{(2k)!}{(2\lambda_1)!(2\lam …
- 29.5k
10
votes
2
answers
468
views
Explicit expression for recursive sums
Let $t_1,t_2,\dots,t_k$ be non-negative integers. Can the following sum
$$f_k(t_1,t_2,\dots,t_k):=\sum_{j_1=0}^{t_1} \sum_{j_2=0}^{t_2+j_1} \sum_{j_3=0}^{t_2+j_2} \dots \sum_{j_k=0}^{t_k+j_{k-1}} 1$$
…
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4
votes
The number of permutations with specified number of cycles and fixed points
The answer is $$\binom nh d_{h,N},$$ where $d_{h,N}$ is the number of derangements of size $h$ with $N$ cycles. By inclusion-exclusion we have:
$$d_{h,N} = \sum_{i=0}^N (-1)^i\binom hi c(h-i,N-i),$$
…
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3
votes
Accepted
On the existence of symmetric matrices with prescribed number of 1's on each row
Replace each $-1$ with $0$. Then the matrix you look for is the adjacency matrix of a graph with a given degree sequence $r_1, r_2, \dots$. Reconstruction of such a graph (and its adjacency matrix) is …
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4
votes
Minimal possible cardinality of a $(a_1, ..., a_k)$-distributable multiset
Here is a Mixed Integer Linear Problem (MILP) formulation that may be solved in practice for some instances with MILP-solvers like CPLEX.
For every integer vector $(i_1,\dots,i_k)\in [1,a_1]\times\cd …
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1
vote
Constant term of a power modulo a polynomial
Let $C$ be the companion matrix of $q(x)$ over $F_p$, and $I$ be the identity matrix of the same size. Then the constant term of $(x+k)^m\bmod q(x)$ can be explicitly expressed as
$$u (C+kI)^m u^T,$$
…
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8
votes
Accepted
A moment sequence and Motzkin numbers. Modular coincidence?
Let's start with $b_n$. Since Catalan number $C_k$ is odd iff $k=2^m-1$, from Lucas theorem it follows that
$$b_n=\sum_{k=0}^n \binom{n}{2k}C_k \equiv\sum_{m\geq 0}\binom{n}{2(2^m-1)}\equiv 1+\nu_2(\l …
- 29.5k
2
votes
Accepted
Enumerating possible number of satisfied linear equations
First off, there are quite a few errors in the question presentation as mentioned in the comments.
An integer $N$ can represent the number of satisfied equations iff it can be written as
$$N = \binom{ …
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13
votes
Accepted
When is the number of areas obtained by cutting a circle with $n$ chords a power of $2$?
Denoting $k:=2^{\lfloor m/2\rfloor}$, we get two cases to consider:
$k^2 = f(n)$ and $2k^2 = f(n)$, or making the coefficients integer:
$$(12k)^2 = 144f(n)\qquad\text{and}\qquad (12k)^2 = 72f(n).$$
Th …
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1
vote
Accepted
What is this numerically-generated function?
First, we notice that the first character must be 1. Let's take it off and focus on the remaining $n-1$ characters, where the number of 1's in each prefix must be at least as many as the number of the …
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14
votes
What is Lagrange Inversion good for?
In combinatorics, applications are more general than just counting trees. In general context, Lagrange inversion is used to obtain a generating function $\sum c_n t^n$ for the numbers $c_n$ of the for …
2
votes
Accepted
Some determinants which are closely related to recurrences
The question concerns the determinant of a Hankel matrix, or a fixed element of a Hankel matrix transform of a shifted sequence $a(n,k)$ for a fixed $k$, although I do not see how this fact alone can …
- 29.5k
0
votes
Number of Permutations?
Let me copy here an answer from Russian forum dxdy.ru that I obtained using the approach outlined in my paper.
Two given rows of a $3\times N$ matrix define a permutation of order $N$. Let $c_i$ ($i= …
- 29.5k
3
votes
Accepted
$n$-distant permutations more than not
There is an injective mapping from $B$ to $A$, which can be constructed as follows:
for a permutation $\pi=(p_1,p_2,\dots,p_{2n})\in B$, find the element $p_k$ that pairs with $p_1$ (i.e., $|p_k-p_1|= …
- 29.5k
2
votes
Accepted
How to solve this conditional recurrence relation?(two variable and conditions)
Under the assumption that $F(2i,n) = 0$ when $i<4$ or $2i>n$, the first and second cases in the recurrence become partial cases of the third one. Hence, the recurrence reduces to
$$F(2i,n) =
\begin{ca …
- 29.5k
8
votes
An explicit representation for polynomials generated by a power of $x/\sin(x)$
Faà di Bruno's formula implies that
$$d_k(n) =
\sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot (n)_{m_1+\cd …
- 29.5k
3
votes
Accepted
Expanding into monomials
Let $\mathrm{CC}_n$ denotes the set of Catalan codes of length $n$, i.e.
$$\mathrm{CC}_n = \left\{ (c_0, \dots, c_{n-1})\in \mathbb{Z}_{\geq 0}^n\ :\ c_0+\dots+c_i\leq i\ \text{for all}\ i=0,1,\dots,n …
- 29.5k
2
votes
Is this model of converting integers to Gray code correct?
Yes, the proposed scheme always produces a Gray code. This can be proved by induction on $k$ as follows.
Proof.
The base cases of $k=1,2$ are trivial. Assume that a Gray code is produced for integers …
- 29.5k
17
votes
Accepted
what is the status of this problem? an equivalent formulation?
In a recent paper A set of 12 numbers is not determined by its set of 4-sums (in Russian), Isomurodov and Kokhas identify an error in the argument of Ewell (1968) and present two distinct sets of 12 i …
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3
votes
Accepted
Ratios of polynomials and derivatives under a certain functional
Here a SageMath code that provides a function V(m) computing $V_m(p)$ in terms of elementary symmetric functions of $x_1,\dots,x_n$ (i.e. coefficients of $p$).
For example, if $p(x) = x^n - e_1 x^{n-1 …
- 29.5k
3
votes
Accepted
Maximum number of subsets in which people co-exist with their friends
The proof of the maximum is rather straight forward.
The cardinality
\begin{split}
&\left|\left\{S\in\binom{P}{r}: |A\cap S| \geq r' \text{ and } (A^c\cap S)\subseteq \bigcap\limits_{i\in A\cap S}F_i\ …
- 29.5k
4
votes
Accepted
Sign Enumeration
The g.f. equals
$$\frac1{1-y}\prod_{i=1}^n \left(xy^i + (xy^i)^{-1}\right).$$
That is, the number of solutions is given by the coefficient of $x^cy^b$.
- 29.5k
4
votes
Reference book for primality testing
C. Pomerance, R. Crandall. Prime Numbers: A Computational Perspective
4
votes
Lower bound for sum of binomial coefficients?
Here is a relevant paper:
T. Worsch. "Lower and upper bounds for (sums of) binomial coefficients".
http://digbib.ubka.uni-karlsruhe.de/volltexte/181894
- 29.5k
3
votes
Accepted
Avoiding multiples of $p$
As I mentioned in the comments above, the number of permutations of elements $1,2,\dots,p$ (i.e., including $p$) is just by factor of $p-2$ larger than the amount in question (in fact, this is true fo …
- 29.5k
1
vote
Weighted counting of circular codes
The formula expressing $c_n$ in terms of $p_d$ follows from the Redfield-Polya enumeration theorem, which also has a weighted mutivariate version. See
https://en.wikipedia.org/wiki/Pólya_enumeration_t …
- 29.5k
0
votes
Prove a family of series having integer coefficients
As a follow-up to js21's answer, the following explicit formula for $F_r(x^2)$ holds:
$$F_r(x^2) = \int_0^{\infty} \cos(xt)^r e^{-t} dt.$$
- 29.5k
1
vote
Accepted
Is this recurrent sequence decreasing?
Since $x_0=0$, it will be convenient to do summation starting from $t=0$.
Denoting $r:=1-p-q$, we have
\begin{split}
S_n &= \frac1n\sum_{t=0}^n \left(pr^{2t} + (q-p)r^{t} - q\right) \\
&=\frac{p}{1-r^ …
- 29.5k
1
vote
sum over all integer partitions, of the product of the factorials of the terms
Is equals the coefficient of $x^ny^k$ in
$$\left(\prod_{i=1}^{\infty} (1-i!x^iy)\right)^{-1}.$$
- 29.5k
8
votes
Accepted
How to calculate: $\sum\limits_{k=0}^{n-m} \frac{1}{n-k} {n-m \choose k}$
Notice that $\frac{1}{n-k} = \int_0^1 x^{n-k-1} dx$. Hence,
$$\sum_{k=0}^{n-m} \frac{1}{n-k}\binom{n-m}k = \int_0^1 dx \sum_{k=0}^{n-m} x^{n-k-1}\binom{n-m}k = \int_0^1 x^{m-1}(1+x)^{n-m}dx = (-1)^m B …
- 29.5k
3
votes
Accepted
Relation between $\sum_{k\ge 0}\binom {n+k}{k}a_k $ and $\sum_{k\ge 0}\binom {n+k}{k}\frac{a...
Since the second sum cannot start at $k=0$, I assume that both sums start at $k=1$.
Consider a particular example: $a_k = k\alpha^k$ with $|\alpha|<1$. Then
$$\sum_{k\geq 1} \binom{n+k}k \frac{a_k}k = …
- 29.5k
6
votes
Accepted
Conjecture on sum over permutations of products of Catalan numbers
The problem naturally fits in the framework of breakpoint graphs (per Peter Taylor's observation), which makes it possible to obtain a differential equation for the generating function
$$H(x,u,s_1,s_2 …
- 29.5k
4
votes
Accepted
Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}, \ m=1,2,....
EDIT 2018-04-16: Formulae are corrected.
I'm not sure about connection with $\beta_{mv}$, but we can obtain a recurrence formula for $A_{j,m}$ as follows.
First let us fix the unused values $A_{j,m} …
- 29.5k
5
votes
Decorated permutations and subset permutations
Here is a direct proof for the formula via generating functions.
Using the formula for derrangement numbers, we have $$\begin{split}
F(n,k)=&\binom nk\cdot !(n-k)\\
=&\frac{n!}{k!}\cdot [x^{n-k}]\ \f …
- 29.5k
2
votes
Accepted
One question about nega-cyclic Hadamard matrices
Such matrices do not exist as from the parity consideration already first two rows cannot be orthogonal.
- 29.5k
4
votes
The number of ways to merge a permutation with itself
Not sure how it's useful but here is an explicit formula for $N_{2k-1}^{\sigma}$.
For a given permutation $\sigma=(\sigma_1,\dots,\sigma_k)$, we have
$$N_{2k-1}^{\sigma} = \sum_{i=1}^k \sum_{j=1}^k \b …
- 29.5k
2
votes
Accepted
Sum with Stirling numbers of the second kind
The same idea of grouping terms by the number of unit bits, as well as grouping by the value of $\mathrm{wt}(j)$ (representing $j$ via individual bits) works here:
\begin{split}
s(n,m) &= \sum_{\ell=0 …
- 29.5k
3
votes
Another generalization of parity of Catalan numbers
Notice that
$$\frac{1}{(j-1)n+1}\binom{j n}{n} = \frac1{jn+1}\binom{jn+1}{n}\equiv \binom{jn+1}{n}\pmod{j}.$$
Suppose that $j=p^k$ for prime $p$. Then from $\binom{jn+1}{n}\equiv 1\pmod{j}$ we can con …
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