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Pascal Determinant (Posted on 2018-03-31) Difficulty: 3 of 5
Suppose that Pascal's triangle is written as follows:

1  1   1   1   1  .  .  .
1  2   3   4   5  .  .  .
1  3   6  10  15  .  .  .
1  4  10  20  35  .  .  .
1  5  15  35  70  .  .  .
.    .    .    .    .
.    .    .    .    .
.    .    .    .    .

The first row and column consist entirely of 1s, and every other number is the sum of the number to its left and the number above. For each positive number n, let D(n) denote the determinant of the matrix consisting of the first n rows and first n columns of this array. Compute D(n).

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution computer assisted solution | Comment 1 of 3
The program at bottom uses Cramer's rule to calculate the determinants up through D(10). All are 1. So now treat this as a sequences problem:

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...

As a sequence puzzle this indicates that D(n) = 1 for all n. As in all sequence puzzles, this is not a proof.

1  1   1   1   1  .  .  .
1  2   3   4   5  .  .  .
1  3   6  10  15  .  .  .
1  4  10  20  35  .  .  .
1  5  15  35  70  

We can manually verify the correctness of the Cramer calculation through n=3:

For 1, D(1) = 1

D(1 1)
 (1 2)   =  2 - 1 = 1
 
D(1 1 1
  1 2 3
  1 3 6) 
  
  Using diagonals:
  
  1 1 1 1 1
  1 2 3 1
  1 3 6 1 3
  
  = 1*2*6 + 1*3*1 + 1*1*3 - 1*2*1 - 3*3*1 - 6*1*1 
          = 12 + 3 + 3 - 2 - 9 - 6 = 1
          
Beyond this we need to trust Cramer's rule.          



DefDbl A-Z
Dim crlf$, tri(100, 100)


Private Sub Form_Load()
 Form1.Visible = True
 Text1.Text = ""
 crlf = Chr(13) + Chr(10)
 
 maxsize = 10
 
 For i = 1 To maxsize
   tri(1, i) = 1
   tri(i, 1) = 1
 Next i
 For row = 2 To maxsize
   For col = 1 To maxsize
     tri(row, col) = tri(row - 1, col) + tri(row, col - 1)
   Next col
 Next row
 
 For sz = 1 To maxsize
   Text1.Text = Text1.Text & mform(sz, "##0") & " " & det(sz, tri()) & crlf
   DoEvents
 Next
 
 
 
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

Function det(sz, triangle())
  If sz = 2 Then
    xx = xx
  End If
  If sz = 1 Then
    det = triangle(1, 1)
    Exit Function
  End If
  ReDim matrix(sz - 1, sz - 1)
  tot = 0
  For col = 1 To sz
    newrow = 0
    For r = 2 To sz
      newrow = newrow + 1
      newcol = 0
      For c = 1 To sz
        If c <> col Then
          newcol = newcol + 1
          matrix(newrow, newcol) = triangle(r, c)
        End If
      Next
    Next r
    term = triangle(1, col) * det(sz - 1, matrix())
    If col Mod 2 = 0 Then term = -term
    tot = tot + term
  Next col
  det = tot
End Function


Function mform$(x, t$)
  a$ = Format$(x, t$)
  If Len(a$) < Len(t$) Then a$ = Space$(Len(t$) - Len(a$)) & a$
  mform$ = a$
End Function



n=10 was the first value of n where it took any noticeable time at all to evaluate the determinant (a couple of seconds), but at n=11 it was taking longer than I wanted to wait.

  Posted by Charlie on 2018-03-31 14:28:37
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