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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.
1
vote
Prove that when converge, the following expansions are equal
The sum in the denominator of $f_3(x)$ evaluates to
$$\frac{(-1)^n}{(x-n)\binom{x}{n}n!}.$$
Hence, the equality $f_2(x) = f_3(x)$ is straightforward.
- 29.5k
9
votes
Accepted
Enumerating subsets with no triple appearing together more than once
In fact, Wikipedia article on Steiner systems that you linked already provides an answer to your question:
An $S(3,4,n)$ is called a Steiner quadruple system. A necessary and sufficient condition for …
- 29.5k
7
votes
Sufficient conditions for the coefficients of a generating function to dominate those of its...
Take any series $g$ with zero free term and all other coefficients being nonnegative, and solve $f-f^2=g$ with respect to $f$, i.e.,
$$f=\frac{1-\sqrt{1-4g}}2.$$
This is the general form of all such $ …
- 29.5k
4
votes
Accepted
Collapsed partitions and generating functions
It is easy to see that
$$cp_k(n) = \sum_{i=1}^n p(n-i)\cdot i^k,$$
where $p(n-i)$ stands for the number of (collapsed) partitions of $n$ that contain $i$ as a part.
Since $i^k$ is the coefficient of …
- 29.5k
3
votes
Accepted
Coefficients $U_m(n,k)$ in the identity $n^{2m+1}=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(...
First off, as I explained in the comments, the identity (1.3) should contain $U_m(T,k)$ rather than $U_m(n,k)$ (the latter does not make any sense), and so the correct identity (1.3) (for polynomials …
- 29.5k
4
votes
Accepted
aproximate sum involving binomial coefficients
Just a rough idea. Let $\alpha, \beta$ be the zeros of $1+Ax+Bx^2$, then for $j\geq 1$
$$c_j = \left.\left(\frac{\partial}{\partial s}\right)^j \log( B(\alpha - e^s)(\beta-e^s) )\right|_{s=0} = \left. …
- 29.5k
10
votes
Accepted
Maximum permuted row/column sum of a matrix
This is the maximum weight matching problem in the weighted complete bipartite graph $K_{n,n}$, also known as the assignment problem. It does have a few polynomial-time algorithmic solutions.
- 29.5k
6
votes
Accepted
Divisibility of (finite) power sum of integers
Notice that
$$2S_a(b) \equiv 1^{2b} + 2^{2b} + \dots + (6a-4)^{2b} \pmod{6a-3}.$$
From Faulhaber's formula, we have
$$1^{2b} + 2^{2b} + \dots + (6a-4)^{2b} \equiv B_{2b} (6a-3) \pmod{6a-3}.$$
It follo …
- 29.5k
17
votes
Accepted
2-adic valuation of a certain binomial sum
First we notice that
\begin{split}
a_n & = n \int_0^1 x^n (1+x)^{n-1}{\rm d}x \\
& = n \int_0^1 (1-x)^n (2-x)^{n-1}{\rm d}x \\
& = n\sum_{k=0}^{n-1} \binom{n-1}{k}2^k (-1)^{n-1-k} \int_0^1 (1-x)^n x^ …
- 29.5k
7
votes
Accepted
A variant of numeric Vandermonde which failed symbolically?
Matrix $A_n$ can be generalized to $\big[(j+f(i))^{i-1}\big]_{i,j=1}^n$ for any function $f(i)$ not just $f(i)=i-1$.
Since
$$(j+f(i))^{i-1} = \sum_{k=0}^{i-1} c_{i,k}\cdot j^k,$$
where $c_{i,k} := \bi …
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4
votes
Accepted
Is there an efficient generalized algorithm to find at least one binary word with the maximu...
The function $f(x)$ is closely related to the notion of autocorrelation, which for a binary sequence $x$ of length $|x|=N$ and shift $w$ can be expressed as
$$\textbf{AC}_x(w) := N - 2H(x\oplus R(x,w) …
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9
votes
Accepted
Prove positivity of rational functions
Notice that
$$F_r(z) = \frac{1}{(1-z)^{r-1}} - \sum_{k=0}^{r-1} \left(\frac{z}{1-z}\right)^k$$
and therefore for $r\geq 4$ and $n\geq 1$, we have
\begin{split}
[x^n]\ F_r(z) &= \binom{n+r-2}{r-2} - \s …
- 29.5k
1
vote
Accepted
System of equations - Proof that a solution exists
We can recover $a$ as soon as $\det(X_{i,k})=1$ over the field $\mathbb{F}_2:=\{0,1\}$ for all pairs $i<k$ from $\{1,2,3,4,5\}$, where $X_{i,k}$ is the $6\times 6$ matrix formed by rows $x_{i,j}$ and …
- 29.5k
2
votes
Accepted
Knapsack problem with value range constraint
It is easy to enforce this constraint for a generic knapsack problem.
Indeed, it is enough just to multiply all $w_h$ and $B$ by $c:=\max_h \frac{v_h}{w_h}$. Then for any $h$,
$$v_h = \frac{v_h}{w_h} …
- 29.5k
7
votes
Accepted
Conjecture on bernoulli numbers and binomial coefficients
UPDATE. My earlier answer was incorrect due to miscalculation. Below I give a proof of the conjecture.
First, we notice that $k^{R(i)} = (k/2+i-1)_i\cdot 2^i = \binom{k/2+i-1}i\cdot i!\cdot 2^i$. Simi …
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3
votes
Magic circle (instead of magic square)
UPDATE. Solution exists for all $n\geq 2$. I have verified it computationally for $n\leq 12$ and below I give a sketch of existence proof for larger $n$.
It is easy to see that when $\varphi$ exists, …
- 29.5k
1
vote
Accepted
The combinatorics of $(f \partial)^n$ in the noncommutative setting?
In fact, Comtet's formula works almost directly in noncommutative case under an appropriate ordering of the products. Here is just a bit deeper look under the hood.
In umbral form $(R_f \partial)^n f$ …
- 29.5k
11
votes
Сlosed formula for $(g\partial)^n$
In OEIS A124796 I considered a similar problem of computing the coefficients of $(\partial_z\circ M_g)^n$, where $M_g$ is the operator of multiplying by $g(z)$.
It turns out that the coefficients rep …
- 29.5k
2
votes
Bell polynomial with variables 1 and 0
From the generating function:
$$\sum_{n\geq k \geq 0} B_{n,k}(x_1,\ldots,x_{n-k+1}) \frac{t^n}{n!} u^k = \exp\left( u \sum_{j=1}^\infty x_j \frac{t^j}{j!} \right)$$
it follows that
$$\sum_{n\geq k \ge …
- 29.5k
6
votes
Alternating binomial-harmonic sum: evaluation request
Denote the sum in question by $f(b,n)$, then
$$\sum_{b,n\geq 0} f(b,n) y^b z^n = \frac{\log(1+y)\log(1-\frac{yz}{1-z})}{1-z-yz}.$$
- 29.5k
1
vote
Accepted
Integral zeros of a multivariate polynomial
First, I will consider the case of $k\geq 4$.
Let $mk=aq^2$, where $q$ is an odd prime.
Since $f(x_1,\dots,x_m)=0$ implies that $mk$ divides $(\sum x_i)^2$, we will look for a zero with $\sum x_i = a …
- 29.5k
3
votes
Accepted
A special class of weighted Motzkin paths
Let $F_k(x)$ be a generating function for the total weight of Motzkin paths of length $n$ starting and ending at height $k$ and not going below that height. Let $G_k(x)$ be a similar generating functi …
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8
votes
Accepted
Vandermonde $V_n$ mod $n$
Per comments above, for a counterexample we have with necessity $\pi_n=n$ and prime $n$. The case $n=2$ is trivial, so I assume that $n$ is an odd prime.
The elements $U:=\{ 1,2,\dots,n-1\}$ form the …
- 29.5k
1
vote
Im looking for an algorithm that can solve or approximate the solution to a problem
This problem is known as the Minimum k-Union problem. Let $D_j$ be the set of doors requiring the key $j$, then we need to pick a set of keys $T$ with $|T|=M-k$ minimizing the size of $\bigcup_{j\in T …
- 29.5k
1
vote
Accepted
Partity of partitions with distinct parts of parts $>1$
We have
$$\prod_{k\geq 2} (1+x^k) = \frac{\prod_{k\geq 1} (1+x^k)}{1+x} \equiv \sum_{j\geq 0} \frac{x^{j(3j+1)/2} + x^{(j+1)(3j+2)/2}}{1+x}$$
$$\equiv \sum_{j\geq 0} x^{j(3j+1)/2}\frac{1 - x^{2j+1}}{1 …
- 29.5k
2
votes
Accepted
Distribution of non-overlapping words in randomly generated text
I'm not sure about the general case, but your particular one is relatively easy.
Let $r$ be a word composed of $n$ instances of $u:=a$ and $v:=ab$ chosen randomly.
We notice that each non-overlappi …
- 29.5k
7
votes
3
answers
372
views
A hypergeometric identity related to Bessel functions
The identity in my recent answer can be stated in a particularly neat form:
$${}_2F_0\left({-n, n+1\atop{}};\frac{x}{2}\right) ~\cdot~ {}_2F_0\left({-n, n+1\atop{}};-\frac{x}{2}\right) ~=~ {}_3F_0\lef …
- 29.5k
25
votes
3
answers
1k
views
Removal of non-isomorphic edges results in the same graph
There exists a (simple unlabeled) graph on 6 nodes with a pair of non-isomorphic edges (i.e., there is no graph automorphism that sends one edge into the other) such that removal of either of them res …
- 29.5k
3
votes
Accepted
Why does this "factorial sequence" appear in the OEIS?
We have
$$\frac{1}{x}f(\frac1x) = \sum_{n\geq 1} (F_{2n-3}-1)x^n,$$
where $F_{2n+1}$ are Fibonacci numbers.
Per answers to your previous question, it follows that
$$c_{n+1} = \sum_{i=0}^n s(n,i) (F_{2 …
- 29.5k
12
votes
Looking for a combinatorial proof for a Catalan identity
Not sure if this is what you look for, but still:
$$\sum_{k=1}^n\left[\frac{k}n\binom{2n}{n-k}\right]^2= \sum_{k=1}^n \big[1-\frac{(n+k)(n-k)}{n^2}\big]\binom{2n}{n+k}\binom{2n}{n-k}$$
$$=\sum_{k=1}^n …
- 29.5k
7
votes
Dividing a cake between $n-1$, $n$, or $n+1$ guests
$f(6) = 13$ with a cake of size $210$ and piece sizes:
$$\{3, 5, 8, 10, 12, 13, 17, 18, 20, 22, 25, 27, 30\},$$
where
$$17+25 = 5+10+27 = 20 + 22 = 3+8+13+18 = 12+30,$$
$$10+25 = 5+30 = 3+12+20 = 17+1 …
- 29.5k
3
votes
Chebyshev polynomials and ballot numbers
It can be seen that the left-hand side for $j<n$ equals the coefficient of $x^j$ in
$$\frac12(1+\sqrt{1+x})^n=\frac12\left(\frac2{C(\tfrac{-x}4)}\right)^n,$$
where $C(x):=\frac{1-\sqrt{1-4x}}{2x}$ is …
- 29.5k
4
votes
Two conjectural series for $\pi$ involving the central trinomial coefficients
Not an answer, but a reduction to a definite integral.
First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$
a …
- 29.5k
3
votes
Concepts in topology successfully transferred to graph theory and combinatorics with non-tri...
Last year I co-authored a paper whose title speaks for itself:
Comparative Genomics Meets Topology: a Novel View on Genome Median and Halving Problems
The introduced topological framework allowed to …
2
votes
Finding the set of all $0$-$1$ vectors in an affine subspace
You may also like to pose this problem as a Closest Vector Problem and use LLL algorithm to solve it.
Namely, if matrix $A$ has size $m\times n$, multiply it by sufficiently large constant $c$ and e …
- 29.5k
2
votes
A formula combining Euler $\phi$ and $\gcd$
Below I characterize set of values achievable by $\psi(a_1,\dots,a_n;N)$.
Let $d_i=\gcd(a_1,\dots,a_i+1,\dots,a_n,N)$. First notice that $d_1,\dots,d_n$ are divisors of $N$ and they are pairwise co-p …
- 29.5k
16
votes
Accepted
In search of a combinatorial reasoning for a vanishing sum
For fixed $s,j>0$, $\sum_{\mathcal{A}_{j,s}}\binom{s}{n_1,\dots,n_j}$ enumerates the compositions of $j$ into $s$ positive parts by ordering the corresponding partitions. That is, we have
$$\sum_{\mat …
- 29.5k
4
votes
A conjecture harmonic numbers
The sequence OEIS A260695 can be seen as the interweaving of nonzero Hultman numbers $\mathcal{H}(n,k)$ for $k=1$ and $k=2$. Namely,
$$a(n) = \mathcal{H}(n,1+(n\bmod 2)).$$
It is known that $\mathcal{ …
- 29.5k
5
votes
Accepted
Solving a Diophantine equation related to Algebraic Geometry, Steiner systems and $q$-binomi...
Let $t = 1+q+q^2+\dots+q^n $ then each of the equations (1) and (2) implies that $24t+1$ is a square (namely, $24t+1=(12k+1)^2$ and $24t+1=(12k+5)^2$, respectively). For $n=2$ that leads to a Pellian …
- 29.5k
0
votes
Recursions which define polynomials
Multiplying the recurrence relation $P_n(t)=a(t)P_{n-1}(t)+b_n(t)$ by $x^n$ and summing up over $n=1,2,\dots,\infty$, we get
$${\cal P}(x,t) - P_0(t) = a(t){\cal P}(x,t)x + {\cal B}(x,t)$$
where ${\ca …
- 29.5k
2
votes
How to label a tree with minimum cost?
This is small parsimony problem, which is solved by Fitch's algorithm -- e.g., see http://www.cs.tau.ac.il/~rshamir/algmb/98/scribe/html/lec09/node6.html
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11
votes
Accepted
Is there a recurrence for the coefficients of the Laurent series expansion of $\frac{1}{1-e^...
$e^{e^x-1}$ is the exponential generating function for Bell numbers ${\cal B}_n$:
$$e^{e^x-1} = \sum_{n\geq 0} {\cal B}_n \frac{x^n}{n!}.$$
Then
$$g(x) := \frac{e^{e^x-1}-1}{x} = \sum_{n\geq 0} {\cal …
- 29.5k
1
vote
0
answers
92
views
3D generalization of Gaussian q-binomial coefficient
It is known that the coefficient of $q^t$ in Gaussian binomial coefficient $\binom{m+n}m_q$ equals the number of permutations of the multiset $\{0^m, 1^n\}$ with $t$ inversions.
Is there a closed for …
- 29.5k
3
votes
Accepted
Special configurations on a circle from a homological algebra problem
There is a simple characterization of interesting configurations:
Lemma. A configuration $x_0=0< x_1 < x_2 < ... <x_r$ of Gorenstein dimension $g$ is interesting if and only if there exist indices $i, …
- 29.5k
16
votes
Accepted
The GCD-matrix: generalizing a result of Smith?
For Pell numbers, the answer appears to be
$$\det\left[\gcd(P_i,P_j)\right]_{i,j=1}^n = \prod_{k=1}^n \sum_{d|k} \mu(\frac{k}{d})\cdot P_d,$$
i.e. the product of first $n$ terms of the Moebius transfo …
- 29.5k
14
votes
Examples of integer sequences coincidences
Just another instance of the (second) Strong Law of Small Numbers:
We have A157656(n) = A059100(n-1) for all known terms (i.e., $n\leq 6$), but it's also known that A157656(29) > A059100(28). So, the …
8
votes
Accepted
Conjecture on palindromic numbers
The conjecture is true.
Let $b=a(n)=2^n+1$.
First, notice that the number in question is
$$N=(b^c-1)\frac{b^{m_1}+1}2\cdots \frac{b^{m_n}+1}2 = \frac{b^c-1}{b-1}(b^{m_1}+1)\cdots (b^{m_n}+1).$$
Sec …
- 29.5k
7
votes
Accepted
An identity for product of central binomials
We have
$$\prod_{j=1}^n \binom{2j}{j} = \frac{2!4!\cdots (2n)!}{(1!2!\cdots n!)^2}$$
and
$$\prod_{j=1}^n 2\binom{n+j}{2j} = 2^n\frac{(n+1)!(n+2)!\cdots (2n)!}{(2!4!\cdots (2n)!)\cdot (0!1!\cdots (n-1) …
- 29.5k
2
votes
Maximum partition of bipartite graph
The case $k=2$ is addressed in the paper https://arxiv.org/abs/cs/0108018 (Bipartite graph partitioning and data clustering by H. Zha, X. He, C. Ding, M. Gu, H. Simon)
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0
votes
Proof that $3^ns + \sum_{k=0}^{n-1} 3^{n-k-1}2^{a_k}=2^m.$
I found it interesting to consider the problem with the minus rather than plus sign in front of the sum:
$$(\star)\qquad 3^n s - \sum_{k=0}^{n-1} 3^{n-1-k} 2^{a_k} = 2^m.$$
Consider the function
$$f …
- 29.5k