By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. How many triangles can be created from a grid of certain dimensions? The lines of negative slope contribute the same amount as those of positive slope, so it suffices to compute the contribution from the latter.

We get:. We're in luck! I note that the formulas given there have the form of summations although with slightly lower complexity than the ones I've written downso it's unlikely you'll get a closed form.

a generating function for the diagonal t2n,n in triangles

Let's attack this problem from a different angle. Instead of looking for lines, let us note that if we wish to find all the lines we can simply look for all possible rectangles such that the lines which contain two opposite corners pass through at least one point on the interior.

That's one direction. If I'm right, then that answers your question in the negative. Please let me know if anything is unclear or if you spot any mistakes. Thanks, I enjoyed that question! Sign up to join this community.

The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Asked 4 years, 11 months ago. Active 4 years, 11 months ago. Viewed times. This question may have been asked before, but the closest question I could find was How many triangles can be created from a grid of certain dimensions? Active Oldest Votes. Tad Tad 5, 1 1 gold badge 10 10 silver badges 25 25 bronze badges. Archaick Archaick 2, 7 7 silver badges 25 25 bronze badges.

Doesn't this count lines and thus triplets multiple times? Do you actually hit all lines? Will edit now. I've done it slightly differently and the answers don't seem to agree yet.

Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password. Post as a guest Name. Email Required, but never shown.A tree is an ordered tree in which every vertex has at most two children. Ties are counted even when candidates A and B lose the election. NoeMar 16 Row sums of triangle A Weisstein, Mar 14 It has apparently not been proved that no [other] prime central trinomials exist. See the papers by Ekhad-Zeilberger and Zeilberger. A and B are disjoint and ii.

A and B contain the same number of elements. Barcucci, R. Pinzani, R. Sprugnoli, The Motzkin family, P. A, Vol. Comtet, Advanced Combinatorics, Reidel,pp. Teubner, Leipzig,Series 1Vol. Graham, D. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA,p. Henrici, Applied and Computational Complex Analysis. Wiley, NY, 3 vols.

Riordan, Combinatorial Identities, Wiley,p. Stanley, Enumerative Combinatorics, Cambridge, Vol. Lin Yang and S. Yang, The parametric Pascal rhombus. See p. Noe G. Andrews, Euler's 'exemplum memorabile inductionis fallacis' and q-trinomial coefficientsJ. Armen G. Barry, Continued fractions and transformations of integer sequencesJIS 12 A rooted binary tree is a type of graph that is particularly of interest in some areas of computer science.

a generating function for the diagonal t2n,n in triangles

A typical rooted binary tree is shown in figure 3. The root is the topmost vertex. The vertices below a vertex and connected to it by an edge are the children of the vertex. It is a binary tree because all vertices have 0, 1, or 2 children.

Notice that any rooted binary tree on at least one vertex can be viewed as two possibly empty binary trees joined into a new tree by introducing a new root vertex and making the children of this root the two roots of the original trees; see figure 3.

To make the empty tree a child of the new vertex, simply do nothing, that is, omit the corresponding child.

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Once we know the trees on 0, 1, and 2 vertices, we can combine them in all possible ways to list the trees on 3 vertices, as shown in figure 3. Note that the first two trees have no left child, since the only tree on 0 vertices is empty, and likewise the last two have no right child.

Now by Newton's Binomial Theorem 3. In exercise 7 in section 1. There are many counting problems whose answers turns out to be the Catalan numbers. Enumerative Combinatorics: Volume 2by Richard Stanley, contains a large number of examples. One such path is indicated by arrows. In how many different ways can this be done? Recall from section 1. Hence, this exercise gives us a lower bound on the total number of partitions. In how many ways can we arrange for each person to shake hands with another person at the table such that no two handshakes cross?

Collapse menu 1 Fundamentals 1. Examples 2. Combinations and permutations 3. Binomial coefficients 4. Bell numbers 5. Choice with repetition 6. The Pigeonhole Principle 7. Sperner's Theorem 8. Stirling numbers 2 Inclusion-Exclusion 1. The Inclusion-Exclusion Formula 2. Forbidden Position Permutations 3 Generating Functions 1.

Newton's Binomial Theorem 2. Exponential Generating Functions 3. Partitions of Integers 4. Recurrence Relations 5. Catalan Numbers 4 Systems of Distinct Representatives 1.In mathematics, a transformation of a sequence's generating function provides a method of converting the generating function for one sequence into a generating function enumerating another.

These transformations typically involve integral formulas applied to a sequence generating function see integral transformations or weighted sums over the higher-order derivatives of these functions see derivative transformations. The main article gives examples of generating functions for many sequences. Other examples of generating function variants include Dirichlet generating functions DGFsLambert seriesand Newton series.

In this article we focus on transformations of generating functions in mathematics and keep a running list of useful transformations and transformation formulas. Then we have the formula [1]. The next formulas for powers, logarithms, and compositions of formal power series are expanded by these polynomials with variables in the coefficients of the original generating functions.

a generating function for the diagonal t2n,n in triangles

One particular formula results in the case of the double factorial function example given immediately below in this section. The single factorial function2 n! Fractional integrals and fractional derivatives see the main article form another generalized class of integration and differentiation operations that can be applied to the OGF of a sequence to form the corresponding OGF of a transformed sequence.

This class of polylogarithm-related integral transformations is related to the derivative-based zeta series transformations defined in the next sections. This result, which is proved in the reference, follows from a variant of the double factorial function transformation integral for the Stirling numbers of the second kind given as an example above. In particular, since.

In general, the Hadamard product of two rational generating functions is itself rational. For example, the Hadamard product of the two generating functions. Ordinary generating functions for generalized factorial functions formed as special cases of the generalized rising factorial product functionsor Pochhammer k-symboldefined by. Moreover, since the single factorial function is given by both n!

This observation suggests an approach to approximating the exact formal Laplace—Borel transform usually given in terms of the integral representation from the previous section by a Hadamard product, or diagonal-coefficient, generating function. Examples of sequences enumerated through these diagonal coefficient generating functions arising from the sequence factorial function multiplier provided by the rational convergent functions include.

Other examples of sequences enumerated through applications of these rational Hadamard product generating functions given in the article include the Barnes G-functioncombinatorial sums involving the double factorial function, sums of powers sequences, and sequences of binomials. These weighted-harmonic-number expansions are almost identical to the known formulas for the Stirling numbers of the first kind up to the leading sign on the weighted harmonic number terms in the expansions.

The next series related to the polylogarithm functions the dilogarithm and trilogarithm functions, respectivelythe alternating zeta function and the Riemann zeta function are formulated from the previous negative-order series results found in the references.

It is known that the first-order harmonic numbers have a closed-form exponential generating function expanded in terms of the natural logarithmthe incomplete gamma functionand the exponential integral given by. For example, the second-order harmonic numbers have a corresponding exponential generating function expanded by the series. A further generalization of the negative-order series transformations defined above is related to more Hurwitz-zeta-likeor Lerch-transcendent-likegenerating functions.

Specifically, if we define the even more general parametrized Stirling numbers of the second kind by. Several other series for the zeta-function-related cases of the Legendre chi functionthe polygamma functionand the Riemann zeta function include.

Additionally, we can give another new explicit series representation of the inverse tangent function through its relation to the Fibonacci numbers [19] expanded as in the references by.

The remainder of the results and examples given in this section sketch some of the more well-known generating function transformations provided by sequences related by inversion formulas the binomial transform and the Stirling transformand provides several tables of known inversion relations of various types cited in Riordan's Combinatorial Identities book.

In many cases, we omit the corresponding functional equations implied by the inversion relationships between two sequences this part of the article needs more work. The first inversion relation provided below implicit to the binomial transform is perhaps the simplest of all inversion relations we will consider in this section. These tables appear in chapters 2 and 3 in Riordan's book providing an introduction to inverse relations with many examples, though which does not stress functional equations between the generating functions of sequences related by these inversion relations.

The interested reader is encouraged to pick up a copy of the original book for more details.Documentation Help Center. Use diag to create a matrix with the elements of v on the main diagonal. The result is a 6-by-6 matrix. The result has one fewer element than the main diagonal. Calling diag twice returns a diagonal matrix composed of the diagonal elements of the original matrix.

Diagonal elements, specified as a vector. Data Types: single double int8 int16 int32 int64 uint8 uint16 uint32 uint64 logical char Complex Number Support: Yes.

Pascal’s Triangle

Input matrix. Diagonal number, specified as an integer. The trace of a matrix is equal to sum diag A. For variable-size inputs that are variable-length vectors 1-by-: or :-by-1diag :. For variable-size inputs that are not variable-length vectors, diag :. If the input is variable-size :m-by-:n and has shape 0-by-0 at run time, then the output is 0-by-1, not 0-by However, if the input is a constant size 0-by-0, then the output is [].

For variable-size inputs that are not variable-length vectors 1-by-: or :-by-1diag treats the input as a matrix from which to extract a diagonal vector. This behavior occurs even if the input array is a vector at run time. To force diag to build a matrix from variable-size inputs that are not 1-by-: or :-by-1, use:. If you supply kthen it must be a real and scalar integer value. This function fully supports GPU arrays. This function fully supports distributed arrays.

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Euler’s polygon triangulation problem

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Examples collapse all Create Diagonal Matrices. Open Live Script.

Binomial Theorem Expansion, Pascal's Triangle, Finding Terms \u0026 Coefficients, Combinations, Algebra 2

Get Diagonal Elements. Input Arguments collapse all v — Diagonal elements vector.Calculate in how many different ways the polygon can be divided into triangles using diagonals that do not intersect each other in the interior of P. This problem was proposed by Euler in to his friend Christian Goldbach. These decompositions are all distinct, therefore the total number of triangulations is given by the following formula:. This recurrence formula was introduced by Segner in For an exhaustive study about the properties and applications of Catalan numbers, see the text [1] available on Amazon.

Instead of using one side of the polygon to enumerate triangles, he used diagonals. The problem that arises is that not all decomposition triangles are counted if we start from a single diagonal. However, if we do this operation starting from every vertex of the polygon, then all triangles are certainly included, and we get this total number:.

However, in this way the triangles are counted several times. The box principle is attributed to the German mathematician Dirichlet It is also called the pigeonhole principle. As an example of application we describe the RSA algorithm for public key Read more…. The birth of probability theory is usually set in the mid-seventeenth century. At that time the two great mathematicians Blaise Pascal — and Pierre de Fermat — discussed together some gambling problems and defined Read more….

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Additional non-necessary cookies will be stored in your browser only if you use our services. We don't use third party cookies. Necessary Always Enabled.For some of these exercises, you may want to use the sage applet above, in example 3. How many different bunches of 10 balloons are there, if each bunch must have at least one balloon of each color and the number of white balls must be even? You may use Sage or a similar program.

From section 1. You can check your answer in Sage. Find the number of such partitions of It is easy to discover this formula directly; the point here is to see that the generating function approach gives the correct answer. Solve the recursion.

After each pair of rabbits was one month old, they produced another pair each month in perpetuity. Thus, after 1 month, he had the original pair, after two months 2 pairs, three months, 3 pairs, four months, 5 pairs, etc.

Suppose instead that each mature pair gives birth to two pairs of rabbits. Set up and solve a recurrence relation for the number of pairs of rabbits. One such path is indicated by arrows. In how many different ways can this be done? Recall from section 1. Hence, this exercise gives us a lower bound on the total number of partitions.

In how many ways can we arrange for each person to shake hands with another person at the table such that no two handshakes cross?


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