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%I #15 May 29 2024 13:48:19
%S 1,21,312,132,231,321,4123,1423,2413,4213,1243,2143,3412,4312,1342,
%T 3142,4132,1432,2341,3241,4231,2431,3421,4321,51234,15234,25134,52134,
%U 12534,21534,35124,53124,13524,31524,51324,15324,23514,32514,52314,25314,35214,53214,12354,21354,31254
%N Decimal representation of permutations of lengths 1, 2, 3, ...
%C One way to generate the permutations is by using the factorial base (not to be confused with the Lehmer code).
%C Here is a detailed example showing how to compute a(2982).
%C We have i = 2982 = (4, 0, 4, 1, 0, 0, 0) in the factorial base.
%C So the initial vector "0" is (1, 2, 3, 4, 5, 6, 7), using seven active digits.
%C The factorial base vector is reversed, giving (0, 0, 0, 1, 4, 0, 4).
%C The instructions are to read from the factorial base vector, producing rotations to the right by as many steps as the column says, in the following order:
%C Start on the right; on the vector "0", a rotation of 4 units is made
%C (0, 0, 0, 1, 4, 0, [4])
%C (1, 2, 3, 4, 5, 6, 7)
%C The result is:
%C (4, 5, 6, 7, 1, 2, 3)
%C The 3 is retained, one column is advanced.
%C Next a rotation of 0 units is made (the null rotation)
%C (0, 0, 0, 1, 4, [0], 4)
%C (4, 5, 6, 7, 1, 2, 3)
%C The result is:
%C (4, 5, 6, 7, 1, 2, 3)
%C The 2 is retained, one column is advanced.
%C Now a rotation of 4 units is made
%C (0, 0, 0, 1, [4], 0, 4)
%C (4, 5, 6, 7, 1, 2, 3)
%C The result is:
%C (5, 6, 7, 1, 4, 2, 3)
%C The 4 is retained, one column is advanced.
%C Now a rotation of 1 units is made
%C (0, 0, 0, [1], 4, 0, 4)
%C (5, 6, 7, 1, 4, 2, 3)
%C The result is:
%C (1, 5, 6, 7, 4, 2, 3)
%C The 7 is retained, one column is advanced.
%C Now 3 null rotations are made.
%C All remaining values are retained: 6, 5, and 1
%C Thus 2982 represents the permutation: (1, 5, 6, 7, 4, 2, 3)
%C Or a(2982) = 1567423.
%e The sequence may be regarded as a triangle, where each row consists of permutations of N terms; i.e., we have
%e 1/,2,1/,3,1,2;1,3,2;2,3,1;3,2,1/4,1,2,3;1,4,2,3;2,4,1,3;...
%e Append to each an infinite number of fixed terms and we get a list of rearrangements of the natural numbers, but with only a finite number of terms permuted:
%e 1/2,3,4,5,6,7,8,9,...
%e 2,1/3,4,5,6,7,8,9,...
%e 3,1,2/4,5,6,7,8,9,...
%e 1,3,2/4,5,6,7,8,9,...
%e 2,3,1/4,5,6,7,8,9,...
%e 3,2,1/4,5,6,7,8,9,...
%e 4,1,2,3/5,6,7,8,9,...
%e 1,4,2,3/5,6,7,8,9,...
%e 2,4,1,3/5,6,7,8,9,...
%e Alternatively, if we take only the first n terms of each such infinite row, then the first n! rows give all permutations of the elements 0,1,2,...,n-1.
%o (Visual Basic)
%o ' The following program is developed in Visual Basic, and works for N = 0 to N = 9.
%o ' This restriction is imposed by the number of lines in Excel spreadsheets.
%o ' In this example, N = 5.
%o ' To modify N, you should only change the definition of Dim A(5) to Dim A(N), and
%o change the value in N = 5.
%o Option Explicit
%o Dim Fila As Double, N As Double
%o Dim A(5) As Double
%o Sub Factorial()
%o Dim J As Double, M As Double, O As Double, Y As Double, I As Double
%o Dim Z As Double, R As Double, V As Double, Aux As Double
%o Fila = 0
%o M = 1
%o N = 5
%o For J = 1 To N
%o A(J) = J
%o M = M * A(J)
%o Next J
%o For Fila = 0 To M - 1
%o V = M
%o Z = N
%o Do While Z > 1
%o V = V / Z
%o If Fila Mod V = 0 Then
%o R = Z
%o Z = 1
%o Else
%o Z = Z - 1
%o End If
%o Loop
%o Y = R \ 2
%o If Y > 0 Then
%o For Z = 1 To Y
%o Aux = A(Z)
%o A(Z) = A(R - Z + 1)
%o A(R - Z + 1) = Aux
%o Next Z
%o Call PrintData
%o End If
%o Next Fila
%o End Sub
%o Sub PrintData()
%o Dim I As Integer
%o For I = 1 To N
%o Range("A1:G5040").Cells(Fila + 1, I) = N + 1 - A(I)
%o Next I
%o End Sub
%K nonn
%O 0,2
%A _Raúl Mario Torres Silva_, Feb 14 2019
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