• Ghislain R. Franssens added an answer:
    What is the power series expansion at zero of the secant to the power of three?

    It is well known that the secant $\sec z$ may be expanded at $z=0$ into the power series
    \sec z=\sum_{n=0}^\infty(-1)^nE_{2n}\frac{z^{2n}}{(2n)!}
    for $|z|<\frac\pi2$, where $E_n$ for $n\ge0$ stand for the Euler numbers which are integers and may be defined by
    \frac{2e^z}{e^{2z}+1}=\sum_{n=0}^\infty\frac{E_n}{n!}z^n =\sum_{n=0}^\infty E_{2n}\frac{z^{2n}}{(2n)!}, \quad |z|<\pi.
    What is the power series expansion at $0$ of the secant to the power of $3$? In other words, what are coefficients in the following power series?
    \sec^3z=\sum_{n=0}^\infty A_{2n}\frac{z^{2n}}{(2n)!}, \quad |z|<\frac\pi2.
    It is clear that the secant to the third power $\sec^3z$ is even on the interval $\bigl(-\frac\pi2,\frac\pi2\bigr)$.

    Ghislain R. Franssens · Belgian Institute for Space Aeronomy

    Dear Dr. Feng,

    Take a look at:

    Ghislain R. Franssens, On a Number Pyramid Related to the Binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.1

    in particular section: 7.1 Multinomial Euler numbers

    The results in section 7.1 are for powers of sech, so you'll need to substitute t -> it.

    The above paper is freely available from the Journal's website, and also from my ResearchGate pages.

    Best wishes,

    Ghislain Franssens

  • Sergey Perepechko added an answer:
    What is the formula to find the number of simple cycles in a graph? Is this problem NP Complete?
    I would like to list out all simple cycles in a connected graph.
    Sergey Perepechko · Petrozavodsk State University

    You need to give additional details about your problem. If you know adjacency matrix of your graph and want to count number of simple cycles of fixed length look at slides and papers in my profile.

    Sergey Perepechko

  • Peter T Breuer added an answer:
    Can anyone help me with the inequality for the ratio of two Bernoulli numbers?

    Dear Colleagues

    I need an inequality for the ratio of two Bernoulli numbers, see attached picture. Could you please help me to find it? Thank you very much.

    Best regards

    Feng Qi (F. Qi)

    Peter T Breuer · University of Birmingham

    Yes, but you haven't understood one of the problems - the English you have used is meaningless or wrong. You asked for something you did not want and did not ask for what you did want.  Please _alter the question_ as advised so that it asks for what you WANT.

    (now it seems that you wanted a PROOF of the inequality, not a NAME for it, as I surmised - but I was guessing as to what you meant).

    PS .. yes, I see from your attachment that you are following the line of reasoning that I pointed at, about the ratio being given by the ratio of the Riemann Zeta function values at consecutive points, times an explicit factor.  Your proof is correct (and nice). The words "in n" are missing after "is decreasing" at the end.

  • Geetha Thangavelu added an answer:
    Which Software Package is the best for computations in Codes over Rings?
    What is the best software package available for carrying out computations related to algebraic coding theory, especially codes over rings.
    Geetha Thangavelu · The Institute of Mathematical Sciences

    You could try Macaulay 2 which is freely available.

  • Samuli Piipponen added an answer:
    Let A & B be two square matrices such that A^2 is not equal to B^2, A is not equal to B but A^3=B^3 & A^2B=AB^2. What is determinant of A^2-B^2?
    If A and B Are non-singular then we have determinant of (A^2-B^2)=-determinant of AB , Am I correct? If so, then the problem is If A and B are singular, how can we prove?
    Samuli Piipponen · University of Eastern Finland

    Dear Prasanth

    Let's start with A2B=AB2.  By rules of matrix computations we 
    get : Det(A)Det(A)Det(B)=Det(A)Det(B)Det(B), moreover
    det(A)^3=det(B)^3 so either both are of full rank or neither is.

    Then consider equation B(A- B2)A=BA-A2B=0 (BY ASSUMPTION BA^2=A^2B) 

    Again by basic rules of matrix comp. 

    Det(A)Det(A2 - B2)Det(B)=Det(A)2Det(A2 - B2)=0 (Since Det(A)=Det(B)), 

    O.k so after some thought it is quite clear  from assumtions that A and B can't be of full rank so that det(A)=det(B)=0.

    This is because the fact that since if det(A)=det(B) \neq 0 then this would imply the existence of the inverse of A and B so that the equation

    A2 B=AB2 would result after multiplication by A-1 from the right and after multiplication by B-1 from the left to equation A=B, which would be a contradiction to the original claim A does not equal B. So the only possibility is det(A)=det(B)=0 ???

    So looked like I only complicated things...

    So  in fact I think the last commentor  Nitin Bisth is correct in his claim.
    Nothing can be said about det(A^2-B^2).

    Prasanth are you sure you have the facts dorrect ?  Could you double chek them ? Of course it might be that my mind is again playing tricks on me :-).

    best regards !


  • Panagiotis Stefanides added an answer:
    What is your favourite combinatorial object, or mathematical object you find the most fascinating?
    I always am curious of what types of combinatorial or mathematical objects (in general) interest other researchers. For myself I have a pretty keen interest in types of restricted weak integer compositions (weak implies 0 is applied in integer sequence whereas the lack of this description means you can't use 0 in integer sequences, and restricted means finding subsets of these objects). What combinatorial or mathematical object (preferably more elusive ones) fascinate you the most, or are your favourite
    Panagiotis Stefanides · University of London

    My calculations indicate that pi [π] needs reconsideration by the involvement of the implications of the qualities of the "quadrature triangle" [ref: Panagiotis Stefanides "Plato's Timaeus Most beautiful Triangle [similar to the Kepler/Magirus triangle, but not the same and not as beautiful].
    By this Pi is related to the square root of the golden ratio and so the other constants should follow.

    Please refer to my extensive link relationships, for possible examination by those interested in this work :

    Regards from Athens,

    Panagiotis Stefanides

  • Albert Manfredi added an answer:
    Can anyone help me find amortized splay tree operation derivation?
    Please help me with a derivation of "amortized cost of a splay tree operation" . I am waiting for that please consider it fast and let me know.
    Albert Manfredi · The Boeing Company
    Both of these articles include proofs. Not sure whether this is what you're asking. Hope they help.
  • Cheng Tianren asked a question:
    How do I use the Ptolemy inequality to study the geodesic angle?
    here, we try to use the ptolemy inequality to study the geodesic angle. and we introduce 2 preliminary results : 1. the ptolemy theorem in euclid spherical geometry by studying the matrix of 4 quadruple points on the euclid spherical, we find the fact that, since there are 6 lines which connect these 4 points, so if 5 of the 6 lines are equal. then we will get that the sixth line's length is double as the other equally 5 lines;this result imply that the angle between the lines is included in $\frac{\sqrt{a}}{2r}$$\in(2k\pi+\pi/5,2k\pi+5\pi/6)$ ; then we substitute this result into the discriminant, and we get an inequality about the radius in $n+1$ dimension: $op_{n+1}^2$$\ge$$\frac{r^2}{t^2}(1+\frac{r^2}{n})-\frac{r^2}{n}$ 2. ptolemy inequality in minkowski geometry here ,we use the centroid method to study the n-polygon problem in minkowski geometry. firstly, we introduce a well known problem such that: if every angle of a polygon is equal, and the sidelines are : $1^2,2^2,......,N^2$, then $\sum{n_{s}}e^{isa}=0$ by this theorem, we can factorize the mass on the vertex into pairs and the number of pairs is primes $N/2=\Pi{p_{i}^{a_{i}}}$ ,the weight of each pair is $\sum{4k-1}$,then we can divide these pairs into groups and every group has prime points too(page 21 in[1]); consequently we can divide the sidelines into 2 parts: $1^2,3^2,......$ and $2^2,4^2,......$; the next step is to construct regular n-polygon and use the ptolemy inequality to make a regulation for the average of the sum of the mass in different group, and we can rearrange these groups of mass to ensure the first part $1^2,3^2,......$ is larger than the average and the second part $2^2,4^2,......$ is less than it. therefore we can apply this average of sum to the distance formula in minkowski geometry in polar coordinate (page 24 in [1]). here we also use combinatorics method (result we get in step 2) to study the natural logorithm in distance formula of minkowski geometry(page 25). our goal is to represent the polar angle in minkowski geometry as the product of the mass lie on different vertex (page 26 in [1]). so our question is how to apply the 2 results above to the geodesic angle? by the inequality for the radius in $n+1$ dimension euclid spherical, we can ensure $v_{n+1}\ge0$; consequently we can substitute the representation of the polar angle in step 2 to spherical equation , which imply that we can also restrict the range of $cos^2{\varphi}$ that is $[\sqrt{33}-4l,3]$ lastly,we apply the property of ptolemy space to get our estimate for geodesic angle,the bounde is : $1-4e^2d/3+\frac{2}{3-4/3e^{2d}}$ is this method feasible? for more detail, you can refer to : application of the ptolemy theorem (3) (page 15-27) the analysis techniques for convexity: CAT-spaces (3)
  • Alina Abraham added an answer:
    In integral theory, how can you integrate all four perspectives in human development into a model?
    According to Wikipedia,, if you have 3 objects [a,b,c] combinatorics there are 6 possible permutations. But positioning in space and time are not considered. What if a 'c' is placed 0,5xx behind [a,b] or 'a' and 'b' separately either horizontal/vertical, or 'c' and 'b' are isolated back to back and 'a' spins around and around for 30 seconds more until reaching a stable position? I am looking at Integral theory and trying to integrate all four perspectives in human development (internal-individual; internal-collective; external-individual; external-collective) into a model. I am not satisfied with those similar to the above ''max 6 possible permutation of the three elements". I believe that for 3 objects there are more than 1+2+3 permutations; and for 4 objects (my case) more than 10.
    Alina Abraham · ICL Business School
    Adapting Ken Wilber/Terri O'Fallon model, using a spiraling trajectory
  • Christopher Landauer added an answer:
    In terms of combinatorics, could (6677,333,166) be cyclical on the torus?
    Would like to know if (6677,333,166) could be cyclical on the torus.
    Christopher Landauer · The Aerospace Corporation
    The notation is ambiguous - what combinatorial object are these the parameters for? Is it some kind of design? If so, what kind, and what parameters are indicated? The problem is that there are hundreds of different parameterized combinatorial designs that use the same kind of notation
  • Mohan Shrikhande added an answer:
    Is there a database available on the net of symmetric designs?
    Symmetric designs=combinatorial 2-designs with as many points as blocks.
    Mohan Shrikhande · Central Michigan University
    Nice to hear from you, Patrick!
  • Daniel Crespin added an answer:
    Spin - could you provide some clarification?
    I always thought that the sum of angular momentum should be zero. I am wondering if this applies on a sub-atomic scale as well as a macroscopic scale.
    Daniel Crespin · Central University of Venezuela
    Hello Donald, A reasoning similar to yours implies that the total linear momentum of the Universe is zero. And furthermore, there should be a motionless center of mass of the Universe. But how many technical details regarding these arguments can be worked out is not so clear. It may be worth a trial. The term "spin" in Physics usually refers to angular momentum of atomic scale objects. Technically, for n >= 3 the special orthogonal group SO(n) is connected and has first homotopy group (also called fundamental group) isomorphic to Z_2, the integers modulo 2. Therefore its universal covering space, denoted Spin(n), is a well defined connected topological group (and a Lie group as well) with fibers consisting of two points, equivalently, it is a connected double cover. If you consider instead the orthogonal group O(n), this is non-connected, has two connected components and the connected component of the identity is SO(n). The universal cover of O(n) is a topological space known as Pin(n), necessarily a non-connected double cover of O(n). But the group structure of Pin(n) is not unique. This subtlety is mentioned in See references there. In Classical Physics, the natural formalism for rotations in ordinary three dimensional space is based on SO(3), its tangent bundle, and the Lie algebra so(3) = tangent space to SO(3) at the identity. A good reference is the book Classical Mechanics by V. Arnold. Then comes Quantum Mechanics and descendants. Schrödinger time dependent equation involves complex numbers. This forces the introduction of complex valued wave functions \psi. To go beyond the mathematical formalism of QM, the mathematical object \psi requires physical interpretation. The wave "amplitude" |\psi|^2 is interpreted "physically" as a probability distribution. This makes the "phase" disappear physically, because |\psi|^2 and |\exp(-i n t) \psi|^2 are one and the same physical state. But the phase is the natural way to consider rotations. Thus, rotations disappear in QM. The quantistic way to recover something resembling rotation is to use Pauli matrices, equivalently, to use spin(3). I have been unable to make sense of these as rotations. Here the quantum dictum "Shut up and calculate" is acutely present. A nearby neon sign says "Abandon hope all ye who enter here". In my opinion the unfortunate and mistaken choice of a unitary evolution equation ---that is, of Schrödinger evolution equation--- for the hydrogen atom made it impossible to understand microscopic rotational phenomena. Even worse, transitions themselves are contradicted by this equation. On the other hand, Schrödinger eigenvalue equation is one of the most impressive wonders of science. Most cordially and with best regards, Daniel Crespin
  • Feng Qi added an answer:
    Can anyone help me with a combinatorial interpretation?
    I am asking for a combinatorial interpretation of a formula for Bell numbers in terms of Kummer confluent hypergeometric functions and Stirling numbers of the second kind. See the formula (8) and Theorem 1 in the attached PDF file or you please help me? Thank a lot.
    Feng Qi · Tianjin Polytechnic University
    Yes, th formula (8) is not the only such formula. Some mathematician asked me to provide a combinatorial interpretation, but I do not know the combinatorial meanings of the formula (8). I think that "the formula (8) just represents one more expression for those numbers, using Kummer confluent hypergeometric function" may be not a combinatorial interpretation of the formula (8).
  • Marcel Van de Vel added an answer:
    Does the discrete n-circle (n even) admit a partition into n/2 pairs, all with a distinct diameter?
    A (discrete) n-circle is the set of complex n-th roots of unity, or: the vertices of a regular n-gon. The above question arose as part of a (nearly finished) research project on a method to produce unpredictable number sequences. Although my partial answers are no longer needed for the project, the simple-looking and still unsettled problem keeps intriguing me. I proved that if a partition exists into pairs of distinct diameters, then n must be of type 8k or 8k+2 (k>0 integer). Computer generated examples confirm that for n <= 112, these types are *exactly* the sizes that work. The computer was stopped after running for two days on the case n=114 (having inspected nearly 0.000...001% (about 300 zeros) of the total search space). The only hope on further information must come from construction methods other than brute-force search with back-tracking and from proofs. Specifically, the problem becomes this: Design an algorithm that is guaranteed to produce a partition (as desired) whenever there exists one and reports failure otherwise. Unlike the current backtracking brute-force search, the algorithm should provide answers in a reasonable time. [Added 09-12-2013: solved] The problem is certainly NP (Nondeterministic Polynomial), but chances are that it is NP-complete. [Added 09-12-2013: not NP complete] A weaker problem is to find a number b <= n/2 such that *any* b vertex pairs with different diameters can be rotated apart in the n-circle for *any* (even) n. It might be "(n/2)-1", I haven't checked on this yet. Ultimately, one should be able to determine the best b for each individual n (including the odd case). [Added 09-12-2013: this is still wide open. Exhaustive computer search is getting quite demanding, even for fairly low n]
    Marcel Van de Vel · VU University Amsterdam
    I have finished my paper containing the questions that I collected in my original posts. There are several more questions (and partial answers) in it, which all arose with the development of one major result on a theoretical method to produce unpredictable numbers. As this paper is intended for publication in a regular journal, I cannot place it it here in public. If anyone is interested in receiving a copy (25 pp), please let me know. I'll send a pdf file by e-mail. The paper classifies mainly as combinatorial mathematics.
  • Henning Meyerhenke added an answer:
    How close is spectral partitioning to the solution of the min-cut problem?
    There are many approaches using different matrices and eigenvectors to solve the min-cut problem. What is the best theoretical result providing a good approximation from a spectral cut to the solution?
  • Donald beverly Giles asked a question:
    Is there any interest in a very different spreadsheet algorithm for generating incidence matrices of projective planes?
    I have devised an algorithm for generating the incidence matrices of projective geometries not shown in traditional texts. Will include as an attachment. This design is self dual and relies on the 4 mols of order 5.Rather than use M*Mt can use M^2.
  • Christopher Landauer added an answer:
    Square Root of a Symmetric Matrices
    The square root of a 31 by 31 matrix with 6"s down the main diagonal and 1"s elsewhere is a symmetric binary matrix with six 1's in each row and column. If someone has an algorithm for this square root then perhaps they can apply it to a larger matix that Iam presently working on. This larger matrix is an 81 by 81 matrix with 16's down the main diagonal and 3's eveywhere else. The square root of this will be a binary symmetric martix with sixteen 1's in each row and column. For me this is not a simple problem. My knowledge of matrices does not extend to taking square roots of symmetric matrices and getting symmetric binary matrices as the answer.
    Christopher Landauer · The Aerospace Corporation
    since the given matrices are linear combinations of the identify and the matrix commonly denoted J (the all 1 matrix), the eigenvalues are simple to compute (the eigenvalues of the k x k matrix J are one k and the rest 0), but only one of them is a square (36 in the first 31 x 31 example, 256 in the 81 x 81 example), and the rest are all equal (5 in the first example and 13 in the second) - also, since J . J = kJ, a square root of J is J/sqrt(k) - therefore, it is easily shown that a square root of rI + sJ in k x k matrices can be found having the same form aI+bJ, with r = a^2, s = 2*a+b^2*k, which is easily solved for a and b (need r >= 0 for a to be real, and s-2*a >= 0 for b to be real) - in the two given examples, r=5, s=1 and r=13, s=3, then second condition does not hold and b will be imaginary there are, of course, other square roots, as other people pointed out
  • Issofa Moyouwou added an answer:
    Up to now, what is known on (the maximal) domains that guarantee the transitivity of the majority rule?
    It is well known that the majority rule may not be transitive for some configurations of individual preferences. Domain restrictions are possible ways out. But what is known about maximal such domains (i) with respect to the cardinality? (ii) via set inclusion?
    Issofa Moyouwou · University of Yaounde I
    That is very instructive. Are there some references for further readings?
  • Bahattin Yildiz added an answer:
    What do the three parameters represent i.e. (44,22,10)?
    Was thinking these were parameters of some type of combinatorial design
    Bahattin Yildiz · Fatih University
    Actually it is [44,22,10] and it describes the parameters of a binary linear code, of length 44, dimension 22 and minimum Hamming weight 10.
  • Daniel Page added an answer:
    Can anyone suggest some good reference papers for beginners in the field of Combinatorial Design?
    It can either be related to Key Distribution or any other application, just an overview needed.
    Daniel Page · University of Manitoba
    I can't recommend a book better than Combinatorial Designs by Douglas Stinson. It's how I learnt Design Theory. It is a little more expensive, but it is full of handy results. Another good book is the Handbook of Combinatorial Designs. My previous supervisor and my current one wrote a pretty interesting bit in there on Lotto Designs.
  • Donald beverly Giles asked a question:
    Can anyone recommend ways to find a skew starter for a room square of side 667?
    Skew room squares exist for all odd values greater than 5. If n is prime it is a simple matter to generate a skew starter. But 667 is not prime. 667=23*29 which means a computer search has to be done in order to generate one. I would be satisfied with the skew room square of side 667, even though we can show it exists we can't seem to construct it. Any suggestions on this particular problem are appreciated.
  • Donald beverly Giles added an answer:
    Why do mathematicians think that the four colour problem cannot be solved theoretically?
    I published a proof. If there is any error in my proof, kindly inform me.
    Donald beverly Giles · Board of Governors of the Federal Reserve System
    This is because Bill Tutte, the man who broke the German Tunny code on his own, spent a great deal of time trying to to prove the four colour theorem. He had a nice short proof for the five colour theorem which was not his proof. He also constructed the Tutte Fragment which ended Heawood's conjecture of trivalent graphs. Any graph can be triangualized.The dual of the triangualized graph is a trivalent graph. Now a Hamiltonian circuit with an even number of vertices can then be constructed and 2 coloured. The remaining third of the vertices can then be 3 coloured This implies that any map can be 4 coloured. Well the Tutte fragment shut this conjecture down flatlly. A strange proof he made was that the averaqe number of colours for all graphs was pi. This meant some were more than pi and some were fewer than pi. I was with Tutte as a student in 74 when he was pursuing this problem. It was a wonderful class on graph theory. Most of the students were professors. Only a few of us were there for a credit.
  • Sanjib Kuila added an answer:
    Why does the four colorability of planar graphs not ensure the non-biplanarity of K_9?
    I hope the proof of the four color theorem is sufficient to explain the answer.
    Sanjib Kuila · Panskura Banamali College
    I have already studied, that paper of Beineke. I also gone through the original works of J Battle et. al., and that of Tutte. In this topic, Harary wrote in his book, "No elegant or even reasonable proof is known."
  • Donald beverly Giles added an answer:
    Proof of the existence of balanced tournament designs?
    I'm writing my senior research paper on balanced tournament designs, and I am looking for the proof of the existence of them which is in this paper.
    Donald beverly Giles · Board of Governors of the Federal Reserve System
    Jennarose: Ron Mullin, Waterloo's first graduate student, proved the existence of room squares for all odd v greater than 5. The last one that needed proving was 257 which was constructed in the orient in the mid 70's. Further to this skew room squares exist for all odd v greater than 5. A skew room square is a room square where exactly one the cells (i,j) or (j,i) is occupied and the other cell is empty. Don Giles is converting skew room squares of order 4n+3 into symmetric block designs with parameters (4n+3,2n+1,n) which in turn are being converted into Hadamard matrices of order (4n+4). All of these combinatorial structures are interrelated. Paul Schellenberg developed the room square of order 25 as part of his PhD. He too was a student and then professor at Waterloo in the late 60's and early 70's Addition and multiplication theorems were used after a collection of smaller room squares had been established by a host of researchers. Hope this helps you with BALANCED TOURNAMENT DESIGNS> I can elaborate more on their usage if you wish.
  • Vitaly Voloshin added an answer:
    Open problem for anyone: can you color the points?
    Suppose you have n points and want to color them with the colors 1, 2, 3, ... But there are the following conditions. There are two families of subsets of these points. In each set of the first family, there must be 2 points of the same color; in each set of the second family, there must be 2 points of different colors. When can you color all n points?
    Vitaly Voloshin · Troy University
    Dear Collin, after I looked at your profile, I see that you work in Algebra. Here is an example of a hypergraph from Algebra and nothing is known about colorability. Regards! Vitaly
  • Matthew D Adams added an answer:
    Advice for a Heuristic Search?
    I am looking for a Heuristic Search to use in R for combinatorial optimization. I will need to choose a specified number of items from the set. For each combination I calculate a cost, which I want to minimize.
    Matthew D Adams · McMaster University
    At this current time I need to formalize some details before describing it in more detail. Thank you Stephen, Daniel and Gennady for taking the time to answer my question, however vague it was.
  • Vitaly Voloshin added an answer:
    Are there any graph theory models of cancer?
    Cancer cells grow in a similar way as graphs from some classes are reconstructed, step by step by adding vertices.
    Vitaly Voloshin · Troy University
    In P.S of P.S.: the paper is here: :
  • Mostafa Aminizahan asked a question:
    Can someone advise on minimal graph with two operators?
    We use signal flow graph for modeling of ODE systems in control. And we have z or s operator with summation and multiplication, when we speaking about the relation between input and output in control theory with Mason Gain we can speak about minimal graph, which can make the transfer function. How can the number of operators be minimized when my transfer function has two variable?
  • Philon Nguyen added an answer:
    What is the value of the Bell polynomial of the second kind? (See attached picture.)
    Where to find the answer?
    Philon Nguyen · Concordia University Montreal
    Check the attachment... It uses equations that can be found on the WIKIPEDIA page on Bell polynomials and the generalized chain rule formula (Faà di Bruno equation)... You could replace g(x) with your equation that uses s(n,k) and f(x) becomes f(g(x))=g(x)-m... But I am not sure how g(x) can be derivated at a glance... If it actually works out algebraically nicely, the equation would give you a correspondance between higher order derivatives and combinatorics.
  • Patrick Solé asked a question:
    Is there a website for the best q-ary constant weight codes (esp q=3,4)?
    There are--it seems-- such tables online but only for q=2

About Combinatorics

Combinatorics is a branch of mathematics concerning the study of finite or countable discrete structures. Aspects of combinatorics include counting the structures of a given kind and size (enumerative combinatorics), deciding when certain criteria can be met, and constructing and analyzing objects meeting the criteria (as in combinatorial designs and matroid theory), finding "largest", "smallest", or "optimal" objects (extremal combinatorics and combinatorial optimization), and studying combinatorial structures arising in an algebraic context, or applying algebraic techniques to combinatorial problems (algebraic combinatorics).

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