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 sz = 1 To maxsize

   For i = 1 To sz

    For j = 1 To sz

      If i = j Or i = sz + 1 - j Then

        tri(i, j) = 2018

      Else

        tri(i, j) = 1

      End If

    Next

   Next

   Text1.Text = Text1.Text & mform(sz, "##0") & " " & det(sz, tri()) & Str(Sqr(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

finds

 n det  square root
  1 2018 44.9221548904324
  2 0 0
  3 0 0
  4 0 0
  5 0 0
  6 0 0
  7 0 0
  8 0 0
  9 201964039878876 14211405.2745981
 10 0 0

I attribute the anomalous value at n=9 to rounding errors due to the large sizes that the numbers get to at such large matrices.

For n>1, the determinant seems always to be zero, which is a perfect square. So the sought smallest value of n is 2, where the determinant of the matrix is easily seen to be zero as the matrix consists solely of 2018's.