pattern matching – Solving indexed ODE in Mathematica

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co.combinatorics – Sum involving determinants of binomial coefficients, indexed by partitions

Given an integer partition $lambdavdash n$, define a polynomial in $N$ whose roots are the negatives of the contents of the partition,
$$ (N)_lambda=prod_{square in lambda}(N+c(square)).$$ This polynomial is closely related to the value of a Schur function evaluated at the $Ntimes N$ identity matrix. On the other hand, given $nuvdash m$ and $rhovdash k$ contained in $nu$, Jacobi-Trudi applied to a skew-Schur function leads to a determinant of binomial coefficients
$$s_{nu/rho}(1_N)=det_{1le,i,jle m}left({N+nu_i-i-rho_j+j-1 choose nu_i-i-rho_j+j}right).$$ The final ingredient I need for my question is another determinant of binomials,
$$A_{lambdarho}=det_{1le,i,jle k}left({rho_i-i choose lambda_j-j}right).$$

Now, in the course of some physics calculation, I arrived at the quantity
$$ E_{lambdanu}(N)=sum_{lambdasubsetrhosubsetnu} A_{lambdarho}s_{nu/rho}(1_N).$$
I thought this was as far as I could push it, but experimentation convinced me that, as a function of $N$, this guy satisfies
$$ E_{lambdanu}(N)propto (N)_{nu/lambda}.$$
It is very surprising to me that this sum should factor like this.

For example, if $nu=(2,2,1)$ and $lambda=(1)$, the six terms in the sum are
$${frac{1}{24}N(N^2-1)(5N-6),-frac{1}{2}N^2(N-1) ,frac{1}{3}N(N^2-1) ,frac{1}{2}N(N-1),-N^2,N}.$$ When all these are added, the result is proportional to $N(N-2)(N^2-1)=(N)_{(2,2,1)/(1)}$.

Moreover, I think I know the proportionality constant when $nu$ and $lambda$ are both hooks:
$$E_{lambdanu}(N)= frac{1}{(m-n)!}{m-n choose m-n-ell(nu)+ell(lambda)}(N)_{nu/lambda}.$$

parallelization – Evaluating values from a indexed ParametricNDSolve solution in parallel

I have a ParametricNDsolve solution for a differential equation for a huge system given by s1. I have no problem in evaluating the value using the solution. It’s, in fact, speeding up the process using Parallel cores.

I have represented the variables by a, b, c and d, some of which are indexed by j, and the parameters are g, n1and n2.
My attempt is to create a table of all the required variables and replace s1 to them as in Tab1and finally evaluate t shown by Tab2

Tab1 = Table( Table(Table({a(j, t)(g, n1, n2), b(j, t)(g, n1, n2), 
   c(g, n1, n2)(t), d(g, n1, n2)(t)}, {j, 1, 2}), {g, 1, 3, 1}), {n1, 1, 4}, {n2, 1, 4}) /. s1; (*first step*)

Tab2=Tab1/.t->12 (say)

So,Tab1 is only a replacement of the solution s1 to the given variables a, b, c and d, which takes me more time to evaluate than Tab2. Is there any way I can parallel map or parallelEvaluate Tab1 using all cores available such that s1 is mapped to the table of variables simultaneously in parallel

Should tags always be indexed or not?

Hi, one question has always been on my mind. Unfortunately, there are many URLs on the webmaster’s account that are in the status section in the coverrage section:Excluded /this number : 10.1K

Most of these addresses are tags. When I check them in my webmaster account, there is no indexing at all. On the other hand, there are no settings to prevent indexing. Is this important? Should tags always be indexed or not?


replacement – Replacing an indexed term

Consider the expression:

s(i_, n_) := Sum((e(c(j)) + e(z(j)))*(l(i, j) + m(i, j)), {j, 1, n})

I’m trying to replace e(c(3)) by c(3) and have attempted

s(i, n) /. e(c(3)) -> c(3)
FullSimplify(s(i, n), c(3) == e(c(3)))
Eliminate({f == s(i, n), c(3) == e(c(3))}, e(c(3)))
Unevaluated(s(i, n)) /. e(c(3)) :>  c(3)
Unevaluated(s(i, n)) /. HoldPattern(e(c(3))) :> c(3)

But none produced the intended result. I think they all failed because the term e(c(3)) does not appear explicitly in the expression. How can we perform substitutions when that is the case?

Associations – Make an indexed list

I am trying to create an indexed list with Association[] But I only get a few keys when I access a particular key.
Here is the code:

In[] :=series = Association[
      "white" -> 1,
      "white" -> 0,
      "blue" -> 2,
      "green" -> 1,
      "green" -> 2,
      "yellow" -> 3
 Out[] :=<|"white" -> 0, "blue" -> 2, "green" -> 2, "yellow" -> 3|>

And when I access the white button, it only shows the number 0 instead of 1 and 0.

 In[]:= series["white"]
 Out:= 0