cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 61-70 of 70 results.

A211363 Permutation corresponding to the inversion sets interpreted as binary numbers (A211362) ordered by value.

Original entry on oeis.org

0, 1, 3, 2, 4, 5, 9, 11, 8, 10, 16, 17, 6, 7, 13, 15, 12, 14, 18, 19, 21, 20, 22, 23, 33, 35, 41, 39, 45, 47, 32, 34, 40, 38, 44, 46, 64, 65, 70, 71, 30, 31, 37, 36, 42, 43, 61, 63, 67, 69, 60, 62, 66, 68, 90, 91, 93, 92, 94, 95, 24, 25, 27
Offset: 0

Views

Author

Tilman Piesk, Jun 03 2012

Keywords

Comments

A211362 lists the binary interpretations of inversion sets ordered by the reverse colexicographic order of permutations (A055089). This permutation orders them by value. Its inverse begins: 0, 1, 3, 2, 4, 5, 12, 13, 8, 6, 9, 7, 16, 14, 17, 15, 10, 11, 18, 19, 21, 20, 22, 23, ...

Examples

			These are the first 24 finite permutations. The inversion sets interpreted as binary numbers on the right form the sequence A211362, which is not monotonic:
No.  permutation   inversion set  A211362
00     1 2 3 4     0  0 0  0 0 0     0
01     2 1 3 4     1  0 0  0 0 0     1
02     1 3 2 4     0  0 1  0 0 0     4
03     3 1 2 4     1  1 0  0 0 0     3
04     2 3 1 4     0  1 1  0 0 0     6
05     3 2 1 4     1  1 1  0 0 0     7
06     1 2 4 3     0  0 0  0 0 1    32
07     2 1 4 3     1  0 0  0 0 1    33
08     1 4 2 3     0  0 1  0 1 0    20
09     4 1 2 3     1  1 0  1 0 0    11
10     2 4 1 3     0  1 1  0 1 0    22
11     4 2 1 3     1  1 1  1 0 0    15
12     1 3 4 2     0  0 0  0 1 1    48
13     3 1 4 2     1  0 0  1 0 1    41
14     1 4 3 2     0  0 1  0 1 1    52
15     4 1 3 2     1  1 0  1 0 1    43
16     3 4 1 2     0  1 1  1 1 0    30
17     4 3 1 2     1  1 1  1 1 0    31
18     2 3 4 1     0  0 0  1 1 1    56
19     3 2 4 1     1  0 0  1 1 1    57
20     2 4 3 1     0  0 1  1 1 1    60
21     4 2 3 1     1  1 0  1 1 1    59
22     3 4 2 1     0  1 1  1 1 1    62
23     4 3 2 1     1  1 1  1 1 1    63
This is the same list ordered by the inversion sets, so the right column is monotonic now. The left column is the beginning of the permutation p, i.e., this sequence:
No.  permutation   inversion set  A211362*p
00     1 2 3 4     0  0 0  0 0 0     0
01     2 1 3 4     1  0 0  0 0 0     1
03     3 1 2 4     1  1 0  0 0 0     3
02     1 3 2 4     0  0 1  0 0 0     4
04     2 3 1 4     0  1 1  0 0 0     6
05     3 2 1 4     1  1 1  0 0 0     7
09     4 1 2 3     1  1 0  1 0 0    11
11     4 2 1 3     1  1 1  1 0 0    15
08     1 4 2 3     0  0 1  0 1 0    20
10     2 4 1 3     0  1 1  0 1 0    22
16     3 4 1 2     0  1 1  1 1 0    30
17     4 3 1 2     1  1 1  1 1 0    31
06     1 2 4 3     0  0 0  0 0 1    32
07     2 1 4 3     1  0 0  0 0 1    33
13     3 1 4 2     1  0 0  1 0 1    41
15     4 1 3 2     1  1 0  1 0 1    43
12     1 3 4 2     0  0 0  0 1 1    48
14     1 4 3 2     0  0 1  0 1 1    52
18     2 3 4 1     0  0 0  1 1 1    56
19     3 2 4 1     1  0 0  1 1 1    57
21     4 2 3 1     1  1 0  1 1 1    59
20     2 4 3 1     0  0 1  1 1 1    60
22     3 4 2 1     0  1 1  1 1 1    62
23     4 3 2 1     1  1 1  1 1 1    63
		

Crossrefs

Cf. A211362.

A211364 Inversion sets of finite permutations that have only 0's and 1's in their inversion vectors.

Original entry on oeis.org

0, 1, 4, 3, 32, 33, 20, 11, 512, 513, 516, 515, 288, 289, 148, 75, 16384, 16385, 16388, 16387, 16416, 16417, 16404, 16395, 8704, 8705, 8708, 8707, 4384, 4385, 2196, 1099, 1048576, 1048577, 1048580, 1048579, 1048608, 1048609, 1048596
Offset: 0

Views

Author

Tilman Piesk, Jun 03 2012

Keywords

Comments

The finite permutations whose position in reverse colexicographic order is A059590(n) (compare A055089, A195663) have the special feature that their inversion vectors (compare A007623) have only zeros and ones, and give 2*n when interpreted as binary numbers. As the inversion vectors are special, one may also take a look at the inversion sets. This sequence shows them, interpreted as binary numbers (compare A211362).

Examples

			These are the 8 permutations of 4 elements that have only 0's and 1's in their inversion vectors. The left column shows their numbers (compare A055089, A195663), i.e., the beginning of A059590. The right column shows the inversion sets interpreted as binary numbers, i.e., the beginning of this sequence.
No.  permutation  inv. vector  inversion set     a
00     1 2 3 4     0 0 0 0     0  0 0  0 0 0     0
01     2 1 3 4     0 1 0 0     1  0 0  0 0 0     1
02     1 3 2 4     0 0 1 0     0  0 1  0 0 0     4
03     3 1 2 4     0 1 1 0     1  1 0  0 0 0     3
06     1 2 4 3     0 0 0 1     0  0 0  0 0 1    32
07     2 1 4 3     0 1 0 1     1  0 0  0 0 1    33
08     1 4 2 3     0 0 1 1     0  0 1  0 1 0    20
09     4 1 2 3     0 1 1 1     1  1 0  1 0 0    11
		

Crossrefs

Formula

a(n) = A211362(A059590(n)).

A211369 Array read by antidiagonals: T(m,n) = m*(m+n-1)! + Sum( n <= i <= m+n-2 ) i!

Original entry on oeis.org

1, 2, 5, 6, 14, 21, 24, 54, 80, 105, 120, 264, 390, 512, 633, 720, 1560, 2304, 3030, 3752, 4473, 5040, 10800, 15960, 21024, 26070, 31112, 36153, 40320, 85680, 126720, 167160, 207504, 247830, 288152, 328473, 362880, 766080, 1134000
Offset: 1

Views

Author

Tilman Piesk, Jul 07 2012

Keywords

Comments

Index numbers (compare A055089) of transpositions.

Examples

			T(3,2) = 3*4! + Sum( 2 <= i <= 3 ) i!
= 3*4! + 2! + 3!
= 3*24 + 2 + 6 = 80.
The array starts:
     1,     2,     6,    24,   120,...
     5,    14,    54,   264,  1560,...
    21,    80,   390,  2304, 15960,...
   105,   512,  3030, 21024,167160,...
   633,  3752, 26070,207504,1860600,...
		

Crossrefs

Cf. A055089, A000142 (row 1), A052649 (row 2)

Programs

  • Maple
    A211369 := proc(m,n)
        m*(m+n-1)!+add(i!,i=n..m+n-2) ;
    end proc: # R. J. Mathar, May 10 2013

Formula

T(m,1) = A001563(m) + A007489(m-1). - R. J. Mathar, May 11 2013

A141826 Bisect sequence A057113.

Original entry on oeis.org

0, 4, 3, 12, 16, 19, 20, 7, 8, 15, 11, 23, 77, 73, 74, 85, 86, 78, 82, 90, 94, 89, 93, 81, 63, 60, 64, 71, 67, 51, 48, 56, 59, 68, 55, 52, 108, 112, 111, 104, 107, 119, 115, 99, 96, 103, 100, 116, 46, 45, 42, 29, 25, 34, 33, 41, 37, 26, 38, 30
Offset: 0

Views

Author

Alford Arnold, Aug 07 2008

Keywords

Comments

Since A057113 is generated based on A057112(n) and successive transpositions, we can be assured that A141826(n) will always map (using A055089)to permutation with an even number of transpositions (applicable, e.g. to the Icosahedron and truncated icosahedron studied in the Geometry Center at http://www.scienceu.com/geometry/facts/solids/handson.html ).

Examples

			A057113(n)is 0 1 4 5 3 2 12 13 16 22 19 18 20 10 7 6 8 14 15 9 11 21 23 17 77... so the sequence begins 0 4 3 12 16 19 20 7 8 15 11 23 77 ...
		

Crossrefs

Extensions

More terms from Alford Arnold, Aug 18 2008

A195665 Consecutive bit-permutations of nonnegative integers.

Original entry on oeis.org

0, 1, 0, 2, 1, 3, 0, 1, 4, 5, 2, 3, 6, 7, 0, 2, 4, 6, 1, 3, 5, 7, 0, 4, 1, 5, 2, 6, 3, 7, 0, 4, 2, 6, 1, 5, 3, 7, 0, 1, 2, 3, 8, 9, 10, 11, 4, 5, 6, 7, 12, 13, 14, 15, 0, 2, 1, 3, 8, 10, 9, 11, 4, 6, 5, 7, 12, 14, 13, 15, 0, 1, 4, 5, 8, 9, 12
Offset: 0

Views

Author

Tilman Piesk, Sep 23 2011

Keywords

Comments

All rows of this array are infinite permutations of the nonnegative integers. Row m (counted from 0) is always generated by modifying the sequence of nonnegative integers in the following way: The sequence of integers is written in reverse binary. Than the finite permutation p_m (row m of array A055089) is applied on the digits of all entries.
The rows of the top left n! X 2^n submatrix describe the rotations and reflections of the n-hypercube that preserve the binary digit sums of the vertex numbers. With permutation composition these permutations form the symmetric group S_n.
Applying such a permutation on the binary string of a Boolean function gives the string of a function in the same big equivalence class (compare A227723).
Triangle row m has length 2^n for m in the interval [(n-1)!,n![. The rest of the array row repeats the same pattern. The first digit of the rest is the digit before plus one.

Examples

			Top left corner of array:
0 1 2 3 4 5 6 7
0 2 1 3 4 6 5 7
0 1 4 5 2 3 6 7
0 2 4 6 1 3 5 7
0 4 1 5 2 6 3 7
0 4 2 6 1 5 3 7
The entry in row 2, column 5 (both counted from 0) is 3: 5 in reverse binary is 101, permutation p_2 applied on 101 gives 110, 110 from reverse binary to decimal is 3.
Corresponding rows of the triangle:
0 1
0 2 1 3
0 1 4 5 2 3 6 7
0 2 4 6 1 3 5 7
0 4 1 5 2 6 3 7
0 4 2 6 1 5 3 7
		

Crossrefs

The finite permutations in A055089 are applied on the reverse binary digits.
Row 0: A001477.
Row 1: A080412.
Row n!-1 of the triangle is the n-bit bit-reversal permutation. Compare A030109.

Extensions

Huge edit by Tilman Piesk, Aug 01 2013

A211365 Array read by antidiagonals: T(m,n) = Sum(1<=i<=m) i * (2m+n-1-i)!

Original entry on oeis.org

1, 2, 10, 6, 36, 186, 24, 168, 1032, 6936, 120, 960, 6840, 53040, 462120, 720, 6480, 52560, 461520, 4499280, 48453840, 5040, 50400, 458640, 4495680, 48449520, 571404960, 7321381200, 40320, 443520, 4475520, 48424320, 571374720, 7321345920
Offset: 1

Views

Author

Tilman Piesk, Jun 22 2012

Keywords

Comments

Index numbers (compare A055089) of permutations like (2,4,6,...,1,3,5...).

Examples

			T(3,2) = Sum( 1 <= i <= 3 ) [ i * (7-i)! ]
= 1*(7-1)! + 2*(7-2)! + 3*(7-3)!
= 1*720 + 2*120 + 3*24
= 1032
		

Crossrefs

Cf. A055089.

A211366 Array read by antidiagonals: T(m,n) = Sum(1<=i<=m) i * ( n + 2(i-1) )!

Original entry on oeis.org

1, 2, 13, 6, 50, 373, 24, 246, 2210, 20533, 120, 1464, 15366, 163490, 1834933, 720, 10200, 122424, 1466886, 18307490, 241335733, 5040, 81360, 1098840, 14637624, 201050886, 2892317090, 43830481333, 40320, 730800, 10967760, 160766040
Offset: 1

Views

Author

Tilman Piesk, Jul 07 2012

Keywords

Comments

Index numbers (compare A055089) of permutations like (n,1, n+1,2, n+2,3, ...).

Examples

			T(3,2) = Sum( 1 <= i <= 3 ) i * ( 2 + 2(i-1) )!
= 1*(2+0)! + 2*(2+2)! + 3*(2+4)!
= 1*2 + 2*24 + 3*720
= 2210
		

Crossrefs

Cf. A055089.

Programs

  • MATLAB
    % See Links section (the Wikiversity page).

A211367 Array read by antidiagonals: T(m,n) = m * Sum(1<=i<=m) (m+n-2+i)!

Original entry on oeis.org

1, 2, 16, 6, 60, 450, 24, 288, 2592, 23616, 120, 1680, 17640, 184800, 2045400, 720, 11520, 138240, 1635840, 20188800, 263727360, 5040, 90720, 1224720, 16148160, 219769200, 3137732640, 47249833680, 40320, 806400, 12096000, 175795200
Offset: 1

Views

Author

Tilman Piesk, Jul 07 2012

Keywords

Comments

Index numbers (compare A055089) of chains of transpositions.

Examples

			T(3,2) = 3 * Sum( 1 <= i <= 3 ) [ (3+i)! ]
= 3 * ( (3+1)! + (3+2)! + (3+3)! )
= 3 * ( 24 + 120 + 720 )
= 2592
		

Crossrefs

Cf. A055089.

A211368 Array read by antidiagonals: T(m,n) = Sum(1<=i<=m) ( n + 2(i-1) )!

Original entry on oeis.org

1, 2, 7, 6, 26, 127, 24, 126, 746, 5167, 120, 744, 5166, 41066, 368047, 720, 5160, 41064, 368046, 3669866, 40284847, 5040, 41040, 368040, 3669864, 40284846, 482671466, 6267305647, 40320, 367920, 3669840, 40284840, 482671464, 6267305646
Offset: 1

Views

Author

Tilman Piesk, Jul 07 2012

Keywords

Comments

Index numbers (compare A055089) of rows of adjacent transpositions.

Examples

			T(3,2) = Sum( 1 <= i <= 3 ) [ ( 2 + 2(i-1) )! ]
= (2+0)! + (2+2)! + (2+4)!
= 2 + 24 + 720
= 746
		

Crossrefs

Cf. A055089.

A362867 Irregular triangle read by rows; the n-th row is the n-th permutation of 0 to infinity, in reversed colexicographic ordering, terminating when the rest of the row equals k.

Original entry on oeis.org

0, 1, 0, 0, 2, 1, 2, 0, 1, 1, 2, 0, 2, 1, 0, 0, 1, 3, 2, 1, 0, 3, 2, 0, 3, 1, 2, 3, 0, 1, 2, 1, 3, 0, 2, 3, 1, 0, 2, 0, 2, 3, 1, 2, 0, 3, 1, 0, 3, 2, 1, 3, 0, 2, 1, 2, 3, 0, 1, 3, 2, 0, 1, 1, 2, 3, 0, 2, 1, 3, 0, 1, 3, 2, 0, 3, 1, 2, 0, 2, 3, 1, 0, 3, 2, 1, 0
Offset: 0

Views

Author

Davis Smith, May 06 2023

Keywords

Comments

Every row of this triangle represents a number which is normal in base b >= 2. Since there are infinitely many bases, every row represents a set of infinitely many numbers, each of which is normal in at least one base.
Treating A(n,k) as equal to k when k is greater than A084558(n) (A084558(n) is the length of the n-th row minus one), for any base b >= 2, the concatenation of the base-b expansion of the n-th row is normal in base b. In other words, for any n >= 0 and b >= 2, C(n,b) is normal in base b.
C(n,b) = Sum_{k >= 0} A(n,k)/(b^(Sum_{m=0..k} ceiling(log_b(A(n,m)+1))))
(The equation for Champernowne's constants using the n-th row of this triangle rather than 0,1,2,...)
C(0,b) is Champernowne's constant for base b (C_b).
Even though for any b and m >= 2, b != m, C(n,b) != C(n,m), it is possible for C(n,b) = C(m,p) where n != m and b != p. In such a case, C(n,b) is normal in two different bases. m will likely be significantly larger than n and p will likely be a power of b.

Examples

			The irregular triangle begins (the numbers enclosed in square brackets are the continuation of the row):
n/k  0, 1, 2, 3, 4, 5, ...
0  | 0 [1, 2, 3, 4, 5, ...]
1  | 1, 0 [2, 3, 4, 5, ...]
2  | 0, 2, 1 [3, 4, 5, ...]
3  | 2, 0, 1 [3, 4, 5, ...]
4  | 1, 2, 0 [3, 4, 5, ...]
5  | 2, 1, 0 [3, 4, 5, ...]
6  | 0, 1, 3, 2 [4, 5, ...]
7  | 1, 0, 3, 2 [4, 5, ...]
8  | 0, 3, 1, 2 [4, 5, ...]
9  | 3, 0, 1, 2 [4, 5, ...]
10 | 1, 3, 0, 2 [4, 5, ...]
		

Crossrefs

Programs

  • PARI
    A362867_row(n)=my(N=n,m=0);while(N\=m++, );Vecrev(vecextract(abs([-(m-1)..0]),numtoperm(m,n)))

Formula

A(n,k) = A055089(n,k) - 1.
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