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I am trying to define an $n \times n$ matrix for even $n$ using

d1 = 
 Table[
   Which[
     i < j, 0, i == j, i!/(2^i 0! i!), 
     EvenQ[i - j] && j == 0, (i)!/(2^i (i/2)! j!), 
     EvenQ[i - j] && OddQ[i], (i)!/(2^i ((i + 1)/2 - 1)! j!), 
     EvenQ[i - j] && EvenQ[i], (i)!/(2^i (i/2 - 1)! j!), 
     True, 0], 
   {i, 0, n}, {j, 0, n}]

Is my code correct? Will it be correct when $n$ is odd?

enter image description here

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  • $\begingroup$ Probably you don't need a separate case for j==0. Also, is there are additional j dependencies in the other two cases (I don't know, because I don't know exactly what you are trying to do). What happens when n is odd? $\endgroup$ – mjw Mar 1 at 20:45
  • $\begingroup$ Dear mjw, I added both cases even and odd...thanks $\endgroup$ – user62716 Mar 1 at 20:54
  • $\begingroup$ Okay, so i is the row, and j is the column, right? The term's dependence on j (other than whether it zero or not) is the j! in the denominator, yes? $\endgroup$ – mjw Mar 1 at 21:01
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    $\begingroup$ How should we interpret the image you show? Is it the output are now getting or the output you want to get? $\endgroup$ – m_goldberg Mar 2 at 0:34
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    $\begingroup$ I suggest using SparseArray and Band. $\endgroup$ – Αλέξανδρος Ζεγγ Mar 3 at 3:02
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This works for both even and odd $N$=n. Notice that the generated matrix is actually $(N+1)\times(N+1)$ as the indices run from 0 to $N$. You can see this in your images, which show an even-sized matrix for odd $N$ and vice-versa.

M[n_Integer /; n >= 0] :=
  SparseArray[{{i_,j_} /; EvenQ[i-j] && i>=j -> (i-1)!/(2^(i-1)((i-j)/2)!(j-1)!)},
    {n + 1, n + 1}]

If you need a non-sparse matrix, use Normal.

I'm not sure why the bottom-right matrix element in your odd-$N$ image is zero. Shouldn't it be $2^{-N}$ as it lies on the diagonal?

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  • $\begingroup$ Or use // MatrixForm to view the matrix. $\endgroup$ – mjw Mar 3 at 3:54
  • $\begingroup$ Dear all, many thanks for all comments, I will check and back...best regards $\endgroup$ – user62716 Mar 3 at 20:08
  • $\begingroup$ Many thanks Roman, you solved my problem, thanks mjw $\endgroup$ – user62716 Mar 3 at 20:30
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d1[n_] := Table[
            Which[i < j, 0,EvenQ[i - j], (i)!/(2^i ((i-j)/2)! j!), True, 0], 
                  {i, 0, n}, {j, 0, n}
               ] 

The output seems to be what you want. Here are a couple of examples:

d1[3] // MatrixForm:

enter image description here

d1[4] // MatrixForm:

enter image description here

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  • $\begingroup$ This is not the matrix from the images. In yours, element (3,1) is $3/8$ whereas in the original it is $3/4$. $\endgroup$ – Roman Mar 3 at 6:19
  • $\begingroup$ Defining a matrix with MatrixForm is not a good idea as it interferes with subsequent calculations and confuses beginners. $\endgroup$ – Roman Mar 3 at 6:19
  • $\begingroup$ You are right about the 3/4 vs 3/8. I"ll look at it again when I have some time. $\endgroup$ – mjw Mar 3 at 13:45
  • $\begingroup$ Defining a matrix with MatrixForm... I did not know about the interference, please elaborate. I am also a beginner! $\endgroup$ – mjw Mar 3 at 13:48
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    $\begingroup$ MatrixForm is only a display wrapper, like InputForm and TeXForm etc. If you define A={{1,1},{1,1}}//MatrixForm then a call like Eigenvalues[A] will fail because it is interpreted as Eigenvalues[MatrixForm[{{1,1},{1,1}}]] instead of Eigenvalues[{{1,1},{1,1}}]. Use MatrixForm only to display a matrix, not to define it. See mathematica.stackexchange.com/questions/3098/… $\endgroup$ – Roman Mar 3 at 14:32

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