# Combinatorics

What can be said about the $q$-analogue of the Sheffer identity?

In \cite{Roman} page 25 we read that,  a sequence $s_n(x)$ is Sheffer for $(g(t), f(t))$, for some invertible $g(t)$, if and only if

$$s_n(x+y)=\sum\limits_{k=0}^{\infty}\binom nk p_k(y) s_{n-k}(x),$$

for all $y$ in complex numbers,  where $p_n(x)$ is associated to $f(t)$.

Noting to the fact that $e_q(x+y) \neq e_q(x)e_q(y)$, leads to conclude that $s_{n,q}(x+y) \neq e_q(yt)s_{n,q}(x)$, and ,therefore, we do not have the $q$-analogue of the identity above directly. Is it possible to express the $q$-analogue of the above mentioned identity in any other way, or should we neglect such an identity for $q$-Sheffer sequences at all?

Any contribution is appreciated in advance.

\bibitem{Roman}Roman S., Rota G. The umbral calculus. Advances Math. 1978;27:95–188.

Marzieh Eini Keleshteri · Eastern Mediterranean University

Dear Dr. Waldemar W Koczkodaj

Thanks for your response. I faced with this question, for the first time, while I was studying the sequence of $q$-Appell polynomials. Since that time, this question has been in my mind for a long time and so far I have not been able to make myself convinced by a perfect answer to it. The reason to ask this question here, actually, is to consult with the experts of this field and read about their different ideas which are originated, clearly, from different points of view. Although I am enthusiastic for learning more and more and go forward through the science, in case that I found a good answer which helps me to publish my studies, I make you sure that I will definitely obey the publication rules and humanity.

What is the number of subsets of a finite set of non-negative integers which are neither the sumsets or the summands of other subsets of $X$?

Suppose that X is a finite set of positive integers. The sumset of two subsets A and B of X is defined as A+B={a+b:a\in A, b\in B}. Then, what is the number of subsets of X which are neither the non-trivial sumsets of any two other subsets of X nor the non-trivial summands of any other subsets of X?  Also, please suggest some useful references in this area.

Sudev

Joe Mccollum · US Forest Service

I'm still trying to understand the question.

Suppose X = {1,2,3,5}.

Then X has 2^4 possible subsets.  But {2,3} is out because it is the sumset of {1,2} and {1}.  Then are {1,2} and {1] out because they are summands of {2,3}?

If so, we're left with just two subsets - {1,3,5} and {1, 2, 3, 5}.

Just where do we draw the distinctive line between traditional data analysis and present day (Big) data analytics?

Some authorities like Davenport have already explained that traditional (Small?) data relates to corporate operations while Big data relates to corporate products and services.

Henning Barthel · Fraunhofer Institute for Experimental Software Engineering IESE

High Dele,

you might also draw the line by taking over the perspective of available big data solutions and how they operate.

With really large data sets you soon get into problems if your system architecture is based on vertical scaling only, the approach taken the last decades. The idea behind big data solutions (big data =big data sets + data analysis) is to split the data set into pieces and distribute it across a cluster of commodity hardware (see e.g. HDFS or NoSQL-Solutions like Cassandra, MongoDB, etc.). The analysis part then requires distributed computing algorithms (e.g. MapReduce, Mahout, Storm, Spark) which can deal with the partitioned and distributed data set in order to get a high analysis performance even for big data sets.

But not every "traditional" analysis/mining algorithm can be parallelized, or it might be hard work to do so. So, for big data, there is a great need to have analysis algorithms working in a massive parallel fashion.
From a technical perspective, that's where you can draw the line too.

Cheers,
Henning.

Can we use discrete mathematics in modeling of control and dynamical systems?

We usually use differential equations, ordinary and partial, difference and delayed. But, could dynamics be captured using discrete mathematics structures, or combinatorics?

George Stoica · University of New Brunswick

Dear Sonam,

It works in stochastic set-up as well, using appropriate modifications of the discrete approximations. The convergence schemes of the stochastic approximations are very well established; the rates of convergence therein are however slightly inferior to the deterministic ones.

George

What are the graphs whose total graphs are complete graphs?

The total graph T(G) of a given graph G is a graph such that the vertex set of T corresponds to the vertices and edges of G and two vertices are adjacent in T if their corresponding elements are either adjacent or incident in G. Can we have non-trivial graphs whose total graphs are complete?

Natig Aliyev · Auburn University

I checked  response of Prof. Singaram Dharmalingam now, quick proof :)

Thank you for the question though.

Good Luck!

How can we relate set theory with networks theory?

Would you please help me to identify some applications of set theory, in particular sumset theory, in networks such as  in social and biological networks? Please suggest some useful reading too/

Qefsere Doko Gjonbalaj · University of Prishtina

I will suggest you this link

Connected Dominating Set: Theory and Applications - Springer
http://www.springer.com/mathematics/applications/book/978-1-4614-5241-6

What are the applications of additive combinatorics in the field of engineering and architecture?

I would like to know some practical applications of additive combinatorics in the field of engineering and architecture. Please suggest some useful reference.

Thank you very much Respected Professors, for your valuable remarks and inputs..

Can anybody suggest any references for combinatorial studies on perturbation theory?

Are there any combinatorial studies on perturbation theory? Can we relate graph theory to that area? If so, please suggest some useful references for a beginner like me.

Wiwat Wanicharpichat · Naresuan University

Thank you Prof. Sudev.

Is it right to take the sumset A + ∅ = ∅?

Is the concept A + ∅ = ∅ correct? A is any set of non-negative integers. I
think that if it is so, it contradicts the condition on the cardinality of sum sets
that |A| + |B| − 1 ≤ |A + B| ≤ |A| |B|. Please give your expert opinions.

Tapas Chatterjee · Indian Institute of Technology Ropar

I guess, Kemperman's theorem is for non-empty sets.

Is there any possibility for the sumset of two sets of integers, which are not arithmetic progressions, to be an arithmetic progression?

The notion of sumset of two sets is defined as A+B={a+b:a in A, b in B}. If the elements of A and B are not in AP,  then can the elements of A+B be an arithmetic progression? If so, what are the conditions required for that? Please can you recommend me some good references.

Attila Por · Western Kentucky University

Don't know if that was already pointed out, but the example by Alex Ravsky is as "bad" as that neither A nor B contain a an arithmetic progression of length 3.

Why does the four colorability of planar graphs not ensure the non-biplanarity of K_9?

My earlier expectation regarding the sufficiency of 4CT to adress this issue, is not correct.

Sanjib Kuila · Panskura Banamali College

My latest publication in http://www.ijeijournal.com/pages/v4i6.html may help to give some insight on this issue.

Compute the number of path from a node to a different one in a weighted undirected connected 3-regular graph.
The graph is a weighted undirected connected 3-regular graph. The number of nodes is N.
For each node there is one loop with weight $= \frac{1}{N}, and two other edges which goes from the node to its two nearest neighburs with weight$= \frac{\epsilon}{N}, where \^epsilon is a small parameter.
Therefore, we would like to know for a given value of $N$ how many different paths of lenght $2T$ are possible to go from one node to another passing through $k$ edges with weight $= \frac{\epsilon}{N}$ ?

I agree with Prof. Patrick Solé.  Maple is a very useful software for any kind of mathematical operations.

The number of paths of length n on the pair of given vertices (vi, vj) is the (i, j)th entry of An where A is the adjacency matrix of the given graph G.

You may refer refer the following link.

B Roberts, D P Kroese, Estimating the Number of s-t Paths in a Graph, http://jgaa.info/accepted/2007/RobertsKroese2007.11.1.pdf

Also visit the following page for a recursive algorithm to find all paths between two given vertices of a given graph.  http://www.technical-recipes.com/2011/a-recursive-algorithm-to-find-all-paths-between-two-given-nodes/

Are there any specific studies on the maximal bipartite sub-graphs of different products of two regular graphs?

Could you suggest me some of the articles on the maximal bipartite sub-graphs of different products of graphs, especially two regular graphs?

Dear Prof. Vitaly Voloshin, Thank you for the links...

Where can I find useful literature on graph theoretical applications to biological networks?

Please provide me information regarding the recent developments in the mathematical, especially graph theoretical, studies on biological networks. Please give some good reference too..

Sudev

Michael Patriksson · Chalmers University of Technology

Perhaps this: https://www.researchgate.net/publication/239328836_A_Graph_Theory_Approach_to_Demographic_Loop_Analysis

And then there is the theory of branching processes to investigate growth and extinction of species. Here is a classic book:

And this, from Nature:

http://www.nature.com/nrg/journal/v4/n7/glossary/nrg1112.html

And I am sure there is more! :-)

Where can I find some real life or scientific applications of the sumsets of two sets of real numbers?

Please provide me some practical/real life applications for the sumsets of the sets of integers or real numbers. Please suggest me some useful references too..

Mihai Prunescu · Institute of Mathematics of the Romanian Academy

The sum is sometimes a direct sum, like in coordinate axes. R2 = (1,0) R + (0,1) R and this fact has a lot of applications (!) Another interesting fact is that R = Z + [0,1] and the decomposition is again unique. This leads to the functions integer part and fractional part. Also interesting, the sum Z + sqrt(2) Z is dense in R. These are the most easy (trivial) examples, but there are a lot of nice sumsets of reals!

Is it possible to find an infinite arithmetic progression of value of k such that we can get primes of the form 6k+1 successively?

for k=1, 2, 3 we get 7, 13, 19 as primes.

for k=5,  6, 7 we get  31, 37,  43,  as primes.

Patrick Solé · Institut Mines-Télécom

Even if true this is likely beyond current technology. It would imply Green Tao theorem:

http://en.wikipedia.org/wiki/Primes_in_arithmetic_progression

How to decrease the bounded gaps between primes?

Last year, Dr. Yitang Zhang has published a paper for the upper bound of twin primes, which is 7*10^7. Does anyone has idea to achieve lower bounded gap?

Christopher Stover · Florida State University

Terence Tao and his collaborators did tons of work on this in Polymath 8 (mentioned ambiguously in the above comment). You can read part of what was done here:

http://terrytao.wordpress.com/2014/09/30/the-bounded-gaps-between-primes-polymath-project-a-retrospective/

Tao's complete corpus of work on that project is:

http://terrytao.wordpress.com/tag/polymath8/

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

Thank you very much. I also like the job done on mutiple issues.

How can we express the recurrence formula about the labeled tree in Maple16?

In Maple 16,  how can we with the software combstruct,  to give the sentence about the recurrence formula,

A(x)=1+x[A(x)3+3A(x)A(x2)+2A(x3)]/6

Patrick Solé · Institut Mines-Télécom

Very strange, first the gf then the object to count!

Are there any statisticians to collaborate on a combinatorics challenge for a paper?

I and another three colleagues have an ongoing paper about the application of sociometry to multiple human resource allocation to multiple projects. The problem is mathematically quite challenging but we are on a dead end concerning one of the outcomes we are dealing with.

Particularly, we have been applying meta-heuristics and evolutionary algorithms to solve the problem of allocating groups of people so as to maximize cohesion among them. The research is quite interesting and we think it is going to open multiple and very interesting research and industry application in the near future.

However, we are stuck in one part. We are trying to calculate the number of viable combinations of people who can work either full-time, part-time or not work at all in several simultaneous projects and we need a person with advanced knowledge in combinatorics to give us a hand. We are willing to pay or to put his/her name as co-author on the paper.

Getting straight to the point, the problem statement is as follows:

There are N people (i=1...N) who can be selected to work in P simultaneous projects (j=1...P). Each person can have a dedication of work full-time (1), half-time (0.5) or not work (0), that is, three possible allocations (0, 0.5, 1).

Now, we know that each project j requires Rj people. How many different and viable combinations are there?

Numerical Example:

People available  Project j=1     Project j=2      Project j=3=P

i=1                       0 or 0.5 or 1   0 or 0.5 or 1    0 or 0.5 or 1     Row sum =<1

i=2                       0 or 0.5 or 1   0 or 0.5 or 1    0 or 0.5 or 1     Row sum =<1

i=3                       0 or 0.5 or 1   0 or 0.5 or 1    0 or 0.5 or 1     Row sum =<1

i=4                       0 or 0.5 or 1   0 or 0.5 or 1    0 or 0.5 or 1     Row sum =<1

i=5                       0 or 0.5 or 1   0 or 0.5 or 1    0 or 0.5 or 1     Row sum =<1

Requirements      R_1=2            R_2=1             R_3=1              Row sum=4

,that is, we need 2+1+1 people working in these 3 projects and 1 out of the five available people is not used. We can use each person totally (1), half-time (0.5) or not use him/her, but the combinations must be feasible, i.e., each person cannot be assigned over 1 (full-time) and the requirements have to be fulfilled.

This problem is easier when the people can only work full-time or not work, but with half-times is far more complicated.

If anyone think that he/she is able to solve it, or just want further details, please  contact me.

Thanks

Peter T Breuer · Birmingham City University

Yes, notice that I added a few words in my answer saying effectively "well, I think you meant to be generous, and allow work to be shared among too many workers by letting many do only part-work, but ended up accidentally saying that you wanted parsimonious allocations, where as many workers as possible are fully employed; still never mind because the one reduces to the other". The reduction is

[generous]  fg(p;n1,n2)=fp(p;n1,n2,p-n1-n2) [parsimonious]

There's an extra dummy project to absorb the efforts of under-employed workers.

So one can stick to the easier (parsimonious) calculation, which in the simplest picture is basically a 3-way recursion from p;n1,n2 to the sum of three smaller values in n1,n2 at p-1. The three ways are how to to distribute two half-person efforts among two projects. Those are 1+0, 0.5+0.5, 0+1. That leads to a recursion from p;n1,n2 to p-1;n1-1,n2-0 and p-1;n1-0.5;n2-0.5 and p-1;n1-0;n2-1. One sums those contributions.

If you are going to consider 1/3 allocations, and 1/4 allocations, etc, the same principle applies. For each person you need to consider how many ways there are of distributing his m/m of available effort among exactly n projects.

That's the number of partitions of a set of m objects into up to n labeled partitions (some of the partitions may be empty). Each partition represents the allocation of effort among the n projects by that person. That is the sum of Stirling numbers of the second kind  mS0+mS1+mS2+...+mSn. Unfortunately you don't so much as want that number as actually the partitions themselves. Each partition x1+...+xn=m is a point in the parameter space (p-1;n1-x1/m,...nn-xn/m) to recurse to and the sum of the Stirling numbers that I mentioned is merely a count of the number of terms in the recursion!

I should say that when n=m, i.e. you allow 1/nths of effort to be distributed among exactlu n projects, the count of terms above that I expressed as a sum of Stirling numbers is a well-known number, the Bell number Bn. It's big, as you might imagine. There are horrible recursions for it.

All that indicates that you don't actually want to calculate the numbers you say you want to calculate, because they're just nasty and meaningless. They're also unscaled, in that the number goes up if you allow each person to divide their efforts more finely. That should not happen! You should probably be scaling by 1/mn, where n is the number of  projects, and 1/m is the minimum unit of effort. Then your numbers would maybe stay on-scale.

I know that you want numbers, but I'm saying they aren't meaningful.

-----------

I'd rather look at approximations. In particular imagine that each person can give away a bit dp of their effort to any of n projects:

f(p;n1,n2,..)=f(p-dp;n1-dp,n2,..)+f(p-dp;n1,n2-dp,...)+...                  (*)

[This is an overestimate. In reality not every assignment of a dp to a project comprises a different overall assignment in the end, since if I assign 2dp to project 1 that is one choice, not two].That's not going to make sense unless I scale by n, the number of projects:

f(p;n1,n2,..)=(f(p-dp;n1-dp,n2,..)+f(p-dp;n1,n2-dp,...)+...)/n

in which case one gets

f(p;n1,n2,...)=f(p;n1,n2,...) -df/dp- 1/n df/dn1- 1/n df/dn2-...

which says

df/dp = -1/n(df/dn1+...+df/dnn)

This is a partial differential equation. My PDEs are rusty, but I believe that is saying that if I were to move by k dp and k ndn1, or k dp and k ndn2, etc, the nett change in f would be zero. That means that this f is an arbitrary function of p-(n1+n2+...)/n . I.e. f(p;n1,n2,...)=F(p-(n1+n2+...)/n) for some F.

At least my scaling does "remove the infinities", but what am I counting? I scaled by 1/n when allocating a dp of a person, so if I had been allocating 1/mths of a person I would have scaled by 1/nm when allocating all of one person. Therefore my number is 1/nmp times too small for what you intend to count.

If you start with p=n1+...+nn exactly (supply equals demand), I conclude that your numbers grow as F(p-p/n) nmp  in the limit when you can allocate small amounts 1/m of each of p persons and distribute them among n projects.

The work ni needed by each particular project does not seem to matter.

I don't know what the constant of proportionality F is, but the prediction is that it depends closely only on the number p of persons available (p-p/n, exactly speaking). I presume it is monotonic increasing, so you would be ok in taking it as a bound not an asymptote and writing F(p) instead of F(p-p/n).  If one thinks about the one-project case (that needs p workers), the implication is that it is exactly 1.

[Note: this is an overestimate, deriving from the overestimate at (*) at the beginning of this approximation].

-----------

There's a more accurate way of expressing the intention of yours that I expressed as (*) above, namely:

f(p;n1,n2,..)= Integral f(p-1;n1-x1,n2-x2,..) dA

where the integral is the sum that appeared in (*) extended to take account of the whole possible set of assignments of a unit of work from p to the projects 1 to n, so the integral is over the sums x1+x2+...=1 for xi in the range [0,1]. That's the "diagonal" plane of the unit hypercube, so I suppose the area is something like 2n/2, but you can tell me!

This would make sense as a stochastic expectation if one were to divide by that 2n/2. I suggest the xi are uniformly distributed over the range [0,1] taking into account the x1+x2+...=1 stipulation, which is to say that each equally sized subunit of the area of that hyperplane through the unit cube is equally probable. That's why dividing by the total area A=2n/2 make sense. It makes the distribution stochastic (total probability = 1).

The expectation is a kind of average count of the weighted number of ways of dividing up the work, counting a tiny fraction weight for each possible way.

If one extends this integral to an assignment of all of p, then

f(p;n1,n2,..)= Integral 1 dA

where the integral is over all xi in the range [0,ni] such that x1+x2+...=p. I believe one should weight all these possibiities equally again to get an expectation, which is to say that each subunit of equal area of the hyperplane x1+x2+...=p=n1+n2+... in the scaled cube [0,n1]x[0,n2]x... gets equal weight [but I'm willing to take suggestions!]. I imagine the total area of that plane segment is 2n/2 n1n2... or something very like that, so I have divided by that in getting an expectation, which will be E[1]=1 and the count represented by the unvarnished integral above is that multiplied by what I divided by, which is 2n/2 n1 n2 ....

However, the same scaling arguments as I applied for my earlier estimation applies again. Whereas I have been applying an integral in which each "way" of dividing up the effort counts for a tiny increment in the integral above, you have been counting it as "1" for each different way.... so I need to scale up by something more again to get to the kind of count you were doing. I'll work it out later, but I guess the extra scaling number is mn-1. So the estimate here is something like 2n/2 n1 n2 ...nn mn-1 for your numbers. This replaces the mnp over-estimate I had before [or will when I check my working and perfect the working .. watch this space].

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)
James F Peters · University of Manitoba

This is a good question with lots of interesting paths to consider in looking for answers.    This question leads to a consideration of metric spaces that are Ptolemaic and Buseman convex spaces, stemming from

H. Busemann, The Geometry of Geodesics, 1955

A good place to start is

T. Foertsch, A. Lytchak, V. Schroeder, Non-positive curvature and the Ptolemy inequality:

http://carma.newcastle.edu.au/jon/Preprints/Papers/CAT(0)/Papers/fls-ptolemy.pdf

These authors prove that a metric space is CAT(0) if and only it is Ptolemy and Busemann convex, p. 11.  Ultrarays assocated with geodesics and their enclosed angles are considered, starting on page 8.

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.
Lie Zhu · Suzhou University

Design Theory, Zhe-Xian Wan, World Scientific Publishing Co Pte Ltd, 2009

The maximum number of cycles of the product of a given permutation and a n-cycle?

Given a permutation A on the set [n], is there a way to determine the maximum number of disjoint cycles of AC where C ranges over all n-cycles on [n]? For which class of permutations A, this problem has been studied before?  Thanks

Ricky X. F. Chen · University of Southern Denmark

The titles of the papers I mentioned are something like:

Zagier, on the distribution of cycles....

Stanley, two enumerative results on product of cycles...

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.
Lie Zhu · Suzhou University

The reference provided 10 minutes ago may not be appropriate.

But, one may try some direct product construction as follows.

Start from skew starter S_1 in Z_{23} and skew starter S_2 in Z_{29}.

Find a permutation P=(p_0,p_1,...,p_{28}) such that
P-I = (p_0-0, p_1-1, ..., p_{28}-28)
And
P+I = (p_0+0, p_1+1, ..., p_{28}+28)
Are both permutations in Z_{29}. For example, take p_s=2s.

For each pair {x, y} in S_1, construct 29 pairs {(x,p_s), (y,s)}, s=0,1,…,28.
This generates 11x29 pairs.
Add 14 more pairs {(0,u),(0v)} such that {u,v} is from S_2.

It may be easy to verify if the 11x29+14 pairs form a skew starter in Z_{23}xZ_{29}.
If it still does not work, please let me know.

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
\label{secant-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

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?

It is clear that the secant to the third power $\sec^3z$ is even on the interval $\bigl(-\frac\pi2,\frac\pi2\bigr)$.

Feng Qi · Tianjin Polytechnic University

Dear All, How are you going? I have a good news to tell you: Several days ago I found a method to give a closed form for coefficients of MacLaurin series of the function $(sec z)a$, where $a$ may be any given number. When I complete themanuscript, I would announce here.

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

Not sure if the following qualifies but this phi spiral?

Which operations can be constructed using only xy and x+y+z?

What method can I use to charcterize which operations repeated multiplication xy and 3-input addition x+y+z are able to construct mod 4, or mod 8, or .. or mod 232?

The reason I ask is because one can prove that it is impossible to construct a known constant from unknown inputs using these operators (no formula f(x,y,..)=K for all x,y,..), and one can also characterise those other operators that one can add into the mix with them that preserve the property.

But I don't know what the totality of operations one can construct with xy and x+y+z is. Indeed, I haven't figured it out for plain old xy and x+y ! (yes, I can describe those: "all polynomials"; but how many are there and what operations cannot be formed as polynomials?)

Any clever ways to go about this? I am trying symbolic computation to generate the orbits of  "x" and "y" under substitution in the binary operators xy and x+y+z mod 4.

Peter T Breuer · Birmingham City University

Strangely, I see 1436 operator tables constructed by xy and x+y mod 4. That's multiplication and the ordinary addition, not the more rarified three-input addition.

Shouldn't these plain old operations construct all polynomials in x,y with 0 constant term? Going by the coefficients, there should be 415 of them (15 coefficients taking the values 0,1,2,3). But I suppose there are polynomials with different coefficients that are functionally equal, since there are divisors of zero mod 4? The difference might be zero mod 4, but have nonzero coefficients?

Can anyone count the number of functionally distinct multinomials in x,y mod 4?

How many polynomials are there? 64, according to the computer:

[0,1]x3+[0,1]x2+[0,1,2,3]x+[0,1,2,3]

That's 16 that are zero at zero.

That was due to reductions from 2x3=2x2=2x, and x4=x2. I'll try counting the multinomials. There are 4 possible coefficients of x, y and xy. There are 2 possible coefficients of x2, y2, x2y,xy2,x2y2. There are 2 possible coefficients of x3,y3,x3y,xy3,x3y3, x3y2,x2y3. That is all. That's 43 212, or 218. It's no good .. I'll have to count them computationally.

What if a polynomial (zero at zero) is zero on all odd numbers? The coefficients satisfy a3+a2+a1=0=a3-a2+a1, so 2a2=0 and a2=0 or a2=2. In one case a3=-a1, and the other a3=2-a1. so one has -a1x3+a1x and (2-a1)x3+2x2+a1x. If the polynomial is equal at even numbers too, then 2a1=0 so a1=0 or a1=2. The polynomial is then 0, 2x3+2x, 2x3+2x2, 2x2+2x.

That gives four versions of each polynomial with the same functionality. This is the kernel of the embedding of polynomials into functions. It's a simple count .. there are 44/4=43=64 polynomials by functionality, 16 with zero constant term.

What if a multinomial x3p3(y)+x2p2(y)+xp1(y)+p0(y) is functionally zero?

p3(y)+p2(y)+p1(y)+p0(y)=0

-p3(y)+p2(y)-p1(y)+p0(y)=0

2p1(y)+p0(y)=0

p0(y)=0

So 2p3(y)=2p2(y)=2p1(y)=p0(y)=0.

So p3,p2,p1  take only even values (0 or 2) and p0(y)=0.

So p0 is one of the four functionally zero polynomials (that's 4 options).

That p takes only even values means it is b3x3+b2x2+b1x+[0,2], where one or three of b1,b2,b3 are even: [0,2]x3+[1,3]x2+[1,3]x+[0,2], [1,3]x3+[0,2]x2+[1,3]x+[0,2], [1,3]x3+[1,3]x2+[0,2]x+[0,2], [0,2]x3+[0,2]x2+[0,2]x+[0,2], for 244=26=64 options.

That is a total of 6434=220 functions.

That would bring the number of functionally distinct multinomials down to about 416/220 =212=4096! I don't seem to be able to get a stable estimate for this quantity :-(. My computer says the answer is 3686, 906 of which are zero at zero. That can't be right.

What are the values of the special Bell polynomials of the second kind?

Bell polynomials of the second kind Bn,k(x1,x2,...,xn-k+1) are also called the partial Bell polynomials, where n and k are positive integers. It is known that Bn,k(1,1,...,1) equals Stirling numbers of the second kind S(n,k).

What are the values of the special Bell polynomials of the second kind Bn,k(0,1,0,1,0,1,0,...) and Bn,k(1,0,1,0,1,0,...)? Where can I find answers to Bn,k(0,1,0,1,0,1,0,...) and Bn,k(1,0,1,0,1,0,...)? Do they exist somewhere?

Feng Qi · Tianjin Polytechnic University

Bai-Ni Guo and Feng Qi, Explicit formulas for special values of Bell polynomials of the second kind and Euler numbers, ResearchGate Technical Report, available online at http://dx.doi.org/10.13140/2.1.3794.8808.

Can anyone give me an interpretation or link about improper uniform prior as a prior distribution in Bayesian estimation?

I need a good reason why improper uniform prior could be use as a prior in Serial Numbered Population (SNP) problem. Or maybe someone can tell me about improper uniform prior itself. An pdf link might be very helpful for me.

I attached the journal.

Brian S. Blais · Bryant University

I'm not sure if Jochen's answer is entirely correct; "An improper prior can be used when the resulting posterior is proper."  I'd modify it to say that an improper prior can be used when the resulting posterior can be achieved with a *proper* prior and the limits of the distribution, after the posterior is calculated, is taken to infinity and the result is shown to be the same.  For example, you can use a proper, bounded uniform prior and then, after you calculate the posterior, take the limits as the bounds grow to infinity to show that if you used the improper prior in the first place it would give the same answer.

Seen this way, the improper prior is a short-cut to make the analysis cleaner.  For example, I personally find E T Jaynes' analyses very clear because he uses improper priors (when he can!), and warns that one has to be careful to do it properly if there is even a hint of trouble.  I find a paper like Bretthorst's "Difference of Means" paper to be a much harder read because he uses proper priors throughout, even in cases where I think the improper prior would work.