A306428 Decimal representation of permutations of lengths 1, 2, 3, ...

1, 21, 312, 132, 231, 321, 4123, 1423, 2413, 4213, 1243, 2143, 3412, 4312, 1342, 3142, 4132, 1432, 2341, 3241, 4231, 2431, 3421, 4321, 51234, 15234, 25134, 52134, 12534, 21534, 35124, 53124, 13524, 31524, 51324, 15324, 23514, 32514, 52314, 25314, 35214, 53214, 12354, 21354, 31254

Offset

0,2

Comments

One way to generate the permutations is by using the factorial base (not to be confused with the Lehmer code).

Here is a detailed example showing how to compute a(2982).

We have i = 2982 = (4, 0, 4, 1, 0, 0, 0) in the factorial base.

So the initial vector "0" is (1, 2, 3, 4, 5, 6, 7), using seven active digits.

The factorial base vector is reversed, giving (0, 0, 0, 1, 4, 0, 4).

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:

Start on the right; on the vector "0", a rotation of 4 units is made

(0, 0, 0, 1, 4, 0, [4])

(1, 2, 3, 4, 5, 6, 7)

The result is:

(4, 5, 6, 7, 1, 2, 3)

The 3 is retained, one column is advanced.

Next a rotation of 0 units is made (the null rotation)

(0, 0, 0, 1, 4, [0], 4)

(4, 5, 6, 7, 1, 2, 3)

The result is:

(4, 5, 6, 7, 1, 2, 3)

The 2 is retained, one column is advanced.

Now a rotation of 4 units is made

(0, 0, 0, 1, [4], 0, 4)

(4, 5, 6, 7, 1, 2, 3)

The result is:

(5, 6, 7, 1, 4, 2, 3)

The 4 is retained, one column is advanced.

Now a rotation of 1 units is made

(0, 0, 0, [1], 4, 0, 4)

(5, 6, 7, 1, 4, 2, 3)

The result is:

(1, 5, 6, 7, 4, 2, 3)

The 7 is retained, one column is advanced.

Now 3 null rotations are made.

All remaining values are retained: 6, 5, and 1

Thus 2982 represents the permutation: (1, 5, 6, 7, 4, 2, 3)

Or a(2982) = 1567423.

Links

Table of n, a(n) for n=0..44.

Example

The sequence may be regarded as a triangle, where each row consists of permutations of N terms; i.e., we have
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;...
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:
1/2,3,4,5,6,7,8,9,...
2,1/3,4,5,6,7,8,9,...
3,1,2/4,5,6,7,8,9,...
1,3,2/4,5,6,7,8,9,...
2,3,1/4,5,6,7,8,9,...
3,2,1/4,5,6,7,8,9,...
4,1,2,3/5,6,7,8,9,...
1,4,2,3/5,6,7,8,9,...
2,4,1,3/5,6,7,8,9,...
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.

Prog

' The following program is developed in Visual Basic, and works for N = 0 to N = 9.
' This restriction is imposed by the number of lines in Excel spreadsheets.
' In this example, N = 5.
' To modify N, you should only change the definition of Dim A(5) to Dim A(N), and
change the value in N = 5.
Option Explicit
Dim Fila As Double, N As Double
Dim A(5) As Double
Sub Factorial()
    Dim J As Double, M As Double, O As Double, Y As Double, I As Double
    Dim Z As Double, R As Double, V As Double, Aux As Double
    Fila = 0
    M = 1
    N = 5
    For J = 1 To N
        A(J) = J
        M = M * A(J)
    Next J
    For Fila = 0 To M - 1
        V = M
        Z = N
        Do While Z > 1
            V = V / Z
            If Fila Mod V = 0 Then
                R = Z
                Z = 1
            Else
                Z = Z - 1
            End If
        Loop
        Y = R \ 2
        If Y > 0 Then
            For Z = 1 To Y
                Aux = A(Z)
                A(Z) = A(R - Z + 1)
                A(R - Z + 1) = Aux
            Next Z
        Call PrintData
        End If
    Next Fila
End Sub
Sub PrintData()
    Dim I As Integer
    For I = 1 To N
    Range("A1:G5040").Cells(Fila + 1, I) = N + 1 - A(I)
    Next I
End Sub

Crossrefs

Sequence in context: A018054 A021484 A198376 * A317201 A281254 A157088

Adjacent sequences: A306425 A306426 A306427 * A306429 A306430 A306431

Keyword

nonn

Author

Raúl Mario Torres Silva, Feb 14 2019

Status

approved