A263327 -id:A263327 - cp's OEIS frontend

A263328 Permutation of {0, ..., 1023} corresponding to lexicographic ordering of numbers with decreasing digits (A009995). Inverse of A263327.

0, 1, 2, 11, 3, 12, 13, 56, 4, 14, 15, 57, 16, 58, 59, 176, 5, 17, 18, 60, 19, 61, 62, 177, 20, 63, 64, 178, 65, 179, 180, 386, 6, 21, 22, 66, 23, 67, 68, 181, 24, 69, 70, 182, 71, 183, 184, 387, 25, 72, 73, 185, 74, 186, 187, 388, 75, 188, 189, 389, 190

Offset: 0

Reinhard Zumkeller, Oct 15 2015

For n = 1..1023: A009995(a(n)) = A262557(n).

The fixed points and cycles of this permutation (up to reversal the same as for its inverse A263327) are listed in A263355. - M. F. Hasler, Dec 11 2019

Reinhard Zumkeller, Table of n, a(n) for n = 0..1023

Cf. A009995, A262557, A263327 (inverse), A263329 (fixed points).

a263328 0 = 0
a263328 n = head [x | x <- [1..1023], a009995' x == a262557 n]

A263328=vecsort(A263327,,1) \\ Does not include a(0)=0. - M. F. Hasler, Dec 11 2019

A263329 Fixed points of permutations A263327 and A263328.

Original entry on oeis.org

0, 1, 2, 17, 18, 84, 939, 1005, 1006, 1021, 1022, 1023

Offset: 1

Reinhard Zumkeller, Oct 15 2015

a(k) = A263355(k,1) for k such that A263383(k) = 1.

Cf. A263327, A263328, A263355, A263383.

a263329 n = a263329_list !! (n-1)
a263329_list = [x | x <- [0..1023], a263327 x == x]

A263355 Table read by rows: cycles of the permutation A263327, sorted in increasing order of their largest element. The elements in each cycle are listed in decreasing numerical order.

Original entry on oeis.org

0, 1, 2, 16, 12, 5, 17, 18, 84, 192, 75, 68, 65, 64, 56, 38, 28, 26, 7, 939, 978, 908, 881, 853, 852, 840, 809, 798, 782, 777, 776, 772, 760, 758, 756, 746, 736, 717, 711, 708, 703, 698, 690, 669, 666, 662, 647, 622, 610, 595, 585, 564, 555, 553, 547, 531

Offset: 1

Reinhard Zumkeller, Oct 16 2015

A263383(n) gives the number of terms in row n.
Fixed points: T(k,m) in A263329 <=> A263383(k) = 1 = m. [Corrected by M. F. Hasler, Dec 11 2019]
The permutations A263327 and its inverse A263328 have 18 cycles, of which 12 are fixed points (listed in A263329), two are 3-cycles (rows 4 and 14 of this table), two are 10-cycles (rows 8 & 13), one is a 74-cycle (row 10) and one is a 912-cycle. - M. F. Hasler, Dec 11 2019
Normally one would list the elements in each cycle in the order in which they appear when the permutation is applied, but that is not the order used here. - N. J. A. Sloane, Dec 11 2019

			   n | Cycles: A263355(n, k=1..A263383(n))                    | A263383(n)
  ---+--------------------------------------------------------+-----------
   1 | (0)                                                    |       1
   2 | (1)                                                    |       1
   3 | (2)                                                    |       1
   4 | (16, 12, 5)                                            |       3
   5 | (17)                                                   |       1
   6 | (18)                                                   |       1
   7 | (84)                                                   |       1
   8 | (192, 75, 68, 65, 64, 56, 38, 28, 26, 7)               |      10
   9 | (939)                                                  |       1
  10 | (978, 908, 881, 853, 852, 840, ..., 142, 115, 45)      |      74
  11 | (1005)                                                 |       1
  12 | (1006)                                                 |       1
  13 | (1016, 997, 995, 985, 967, 959, 958, 955, 948, 831)    |      10
  14 | (1018, 1011, 1007)                                     |       3
  15 | (1020, 1019, 1017, 1015, 1014, ..., 10, 9, 8, 6, 4, 3) |     912
  16 | (1021)                                                 |       1
  17 | (1022)                                                 |       1
  18 | (1023)                                                 |       1
A263327(5) = 16, A263327(16) = 12, A263327(12) = 5, so (5 16 12) = (16 12 5) is a 3-cycle. For all other cycles of length > 1, the order in which the terms occur under the map (e.g. 1018 -> 1007 -> 1011 -> 1018 for row 14) is different from the decreasing order given above. - _M. F. Hasler_, Dec 11 2019

Reinhard Zumkeller, Rows n = 1..18 of triangle, flattened, 1024 terms

Cf. A263327, A263383 (row lengths), A263329.

import Data.List ((\\), sort)
a263355 n k = a263355_tabf !! (n-1) !! (k-1)
a263355_row n = a263355_tabf !! (n-1)
a263355_tabf = sort $ cc a263327_list where
   cc [] = []
   cc (x:xs) = (reverse $ sort ys) : cc (xs \\ ys)
      where ys = x : c x
            c z = if y /= x then y : c y else []
                  where y = a263327 z

{M=0; (C(x,L=[x])=until(x==L[1], M+=1<A263327[x]));L); vecsort(vector(18,i,vecsort(C(valuation(M+1,2)),,12)))} \\ append [^15] to remove the long row 15. - M. F. Hasler, Dec 11 2019

Edited by M. F. Hasler, Dec 11 2019

A263383 Lengths of cycles of permutation A263327.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 10, 1, 74, 1, 1, 10, 3, 912, 1, 1, 1

Offset: 1

Reinhard Zumkeller, Oct 16 2015

Cycle type of permutation A263327 = (1^12, 3^2, 10^2, 74, 912);
A263355(k,1) = A263329(k) for k such that a(k) = 1.
row lengths of A263355;
number of terms = 18; sum of terms = 1024.

			See A263355.

Cf. A263327, A263329, A263355.

Haskell
```
a263383 = length . a263355_row
```

A009995 Numbers with digits in strictly decreasing order. From the Macaulay expansion of n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97, 98, 210, 310, 320, 321, 410, 420, 421, 430, 431, 432, 510, 520, 521, 530

Offset: 1

N. J. A. Sloane

There are precisely 1023 terms (corresponding to every nonempty subset of {0..9}).
A178788(a(n)) = 1. - Reinhard Zumkeller, Jun 30 2010
A193581(a(n)) > 0 for n > 9. - Reinhard Zumkeller, Aug 10 2011
A227362(a(n)) = a(n). - Reinhard Zumkeller, Jul 09 2013
For a fixed natural number r, any natural number n has a unique "Macaulay expansion" n = C(a_r,r)+C(a_{r-1},r-1)+...+C(a_1,1) with a_r > a_{r-1} > ... > a_1 >= 0. If r=10, concatenating the digits a_r, ..., a_1 gives the present sequence. The representation is valid for all n, but the concatenation only makes sense if all the a_i are < 10. - N. J. A. Sloane, Apr 05 2014
a(n) = A262557(A263327(n)); a(A263328(n)) = A262557(n). - Reinhard Zumkeller, Oct 15 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..1023
B. Sury, Macaulay Expansion, Amer. Math. Monthly 121 (2014), no. 4, 359--360. MR3183022. [See p. 359. - _N. J. A. Sloane_, Apr 05 2014]
Eric Weisstein's World of Mathematics, Digit.

Cf. A009993.

Cf. A262557 (sorted lexicographically), A263327, A263328.

import Data.Set (fromList, minView, insert)
a009995 n = a009995_list !! n
a009995_list = 0 : f (fromList [1..9]) where
   f s = case minView s of
         Nothing     -> []
         Just (m,s') -> m : f (foldl (flip insert) s' $
                              map (10*m +) [0..m `mod` 10 - 1])
-- Reinhard Zumkeller, Aug 10 2011

Sort@ Flatten@ Table[FromDigits /@ Subsets[ Range[9, 0, -1], {n}], {n, 10}] (* Zak Seidov, May 10 2006 *)

is(n)=fromdigits(vecsort(digits(n),,12))==n \\ Charles R Greathouse IV, Apr 16 2015

A294648 Irregular triangle read by rows, representing a family of sequences L(n), for n=1, 2, 3, ... The sequence L(n) (i.e., the n-th row) is the ordinance of vectors of the n-dimensional Boolean cube (hypercube) {0,1}^n in accordance with their (Hamming) weights, where the lexicographic order is chosen as a second criterion for an ordinance the vectors of equal weights.

Original entry on oeis.org

0, 1, 0, 1, 2, 3, 0, 1, 2, 4, 3, 5, 6, 7, 0, 1, 2, 4, 8, 3, 5, 6, 9, 10, 12, 7, 11, 13, 14, 15, 0, 1, 2, 4, 8, 16, 3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 15, 23, 27, 29, 30, 31, 0, 1, 2, 4, 8, 16, 32, 3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 33, 34, 36, 40, 48, 7, 11, 13, 14, 19, 21

Offset: 1

Valentin Bakoev, Nov 06 2017

The sequences represent the ordinance of vectors of the n-dimensional Boolean cube (hypercube) {0,1}^n in accordance with their (Hamming) weights. The lexicographic order is chosen as a second criterion for the ordinance of the vectors of equal weights. We refer to this order as a Weight-Lexicographic Order (WLO). The WLO is represented by the (serial) numbers of the vectors, instead of the vectors itself. It is well known that if the vectors of {0,1}^n are in lexicographic (standard) order, their numbers form the sequence of natural numbers 0, 1, 2, ..., 2^n-1. So, the WLO means a permutation of the numbers 0, 1, 2, ..., 2^n-1, such that the corresponding vectors are in WLO. This sequence (permutation) is denoted by L(n). It consists of (n+1) subsequences, corresponding to the layers of the Boolean cube.

			The lexicographic order of {0,1}^3 is: (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1), and the corresponding sequence of (serial) numbers is: 0, 1, 2, ..., 7. The WLO of these vectors is given by the sequence L(3)= 0, 1, 2, 4, 3, 5, 6, 7.
The triangle starts:
n=1: 0, 1;
n=2: 0, 1, 2, 3;
n=3: 0, 1, 2, 4, 3, 5, 6, 7;
n=4: 0, 1, 2, 4, 8, 3, 5, 6, 9, 10, 12, 7, 11, 13, 14, 15;
n=5: 0, 1, 2, 4, 8, 16, 3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 15, 23, 27, 29, 30, 31;
n=6: 0, 1, 2, 4, 8, 16, 32, 3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 33, 34, 36, 40, 48, 7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 35, 37, 38, 41, 42, 44, 49, 50, 52, 56, 15, 23, 27, 29, 30, 39, 43, 45, 46, 51, 53, 54, 57, 58, 60, 31, 47, 55, 59, 61, 62, 63;

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Valentin Bakoev, Some problems and algorithms related to the weight order relation on the n-dimensional Boolean cube, Discrete Mathematics, Algorithms and Applications, Vol. 13 No 3, 2150021 (2021); arXiv preprint, arXiv:1811.04421 [cs.DM], 2018-2020.
Valentin Bakoev, Fast Computing the Algebraic Degree of Boolean Functions, arXiv:1905.08649 [cs.DM], 2019.

A051459 gives the orders' number of the vectors of {0,1}^n in accordance with their weights.
Cf. A007318, A047869 (8th row), A091444 (as bits), A187769, A263327 (10th row).
Cf. A382467.

with(ListTools): conc := (a,b,c) -> Flatten([op(a),[seq(op(j)+c, j in b)]], 1):
rec := proc(n,k) option remember; `if`(k=0, [0], `if`(k=n, [2^n-1], conc(rec(n-1,k), rec(n-1,k-1), 2^(n-1)))) end:
L := n -> `if`(n=1, [0,1], Flatten([seq(rec(n,k),k=0..n)], 1)):
Flatten([seq(L(n), n = 1..6)], 1); # Peter Luschny, Nov 06 2017

conc[a_List, b_List, c_] := Join[a, b + c];
rec[n_, k_] := rec[n, k] = If[k == 0, {0}, If[k == n, {2^n - 1}, conc[rec[n - 1, k], rec[n - 1, k - 1], 2^(n - 1)]]];
L[n_] := If[n == 1, {0, 1}, Flatten[Table[rec[n, k], {k, 0, n}]]];
Array[L, 6] // Flatten (* Jean-François Alcover, Jul 26 2018, after Peter Luschny *)
Table[SortBy[Range[0, 2^n - 1], DigitSum[#, 2] &], {n, 6}] (* Paolo Xausa, Jul 28 2025 *)

cmph(x, y) = my(d=hammingweight(x)-hammingweight(y)); if (d, d, x-y);
row(n) = my(v=[0..2^n-1]); vecsort(v, cmph); \\ Michel Marcus, Sep 16 2023

from itertools import product
def sortby(x): return (len(x), x.count('1'), x)
def agen(maxbindigs):
    for i in range(1, maxbindigs+1):
        for t in sorted([p for p in product("01", repeat=i)], key=sortby):
            yield int("".join(t), 2)
print([an for an in agen(6)]) # Michael S. Branicky, Aug 13 2021

For n=1, 2, 3, ..., L(n) is defined by the recurrence:
if n=1, L(1)= 0, 1;
else L(n)= l(n, 0), l(n, 1), ..., l(n, k), ..., l(n, n), where the subsequences are defined as follows:
   l(n, k)= 0, if k=0, else
   l(n, k)= 2^n - 1, if k=n, else
   l(n, k)= l(n-1, k), l(n-1, k-1) + 2^{n-1}, for 0 < k < n.
Comments:
1) l(n, k)= l(n-1, k), l(n-1, k-1) + 2^{n-1} means that l(n, k) is a concatenation of two subsequences: l(n-1, k) and l(n-1, k-1) + 2^{n-1}. The second one is obtained after addition of the number 2^{n-1} to each member of l(n-1,k-1).
2) The computing of the members of L(n) resembles the computing (and filling in) the binomial coefficients in Pascal's triangle. The binomial coefficients determine the lengths of the subsequences l(n, k), 0 <= k <= n, in L(n). Thus the beginning of each subsequence can be computed easily.
3) The inductive formula, corresponding to the recurrence, is much more useful for implementations.

A262557 Numbers with digits in strictly decreasing order, sorted lexicographically.

Original entry on oeis.org

0, 1, 10, 2, 20, 21, 210, 3, 30, 31, 310, 32, 320, 321, 3210, 4, 40, 41, 410, 42, 420, 421, 4210, 43, 430, 431, 4310, 432, 4320, 4321, 43210, 5, 50, 51, 510, 52, 520, 521, 5210, 53, 530, 531, 5310, 532, 5320, 5321, 53210, 54, 540, 541, 5410, 542, 5420, 5421

Offset: 1

N. J. A. Sloane, Oct 14 2015

Original name: "Countdown sequences, allowing gaps."
Only digits 0 through 9 are used. The last term is 9876543210.
Equals A009995, sorted lexicographically. - Reinhard Zumkeller, Oct 14 2015
There are 2^k terms starting with digit k >= 0, they start at index 2^k. The countdown sequences, i.e., digits of the n-th term, are given in rows of A272011. - M. F. Hasler, Dec 11 2019

Donald S. McDonald, Email message to N. J. A. Sloane, Oct 14 2015.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1023

Cf. A009995, A263327, A263328.

a262557 n = a262557_list !! (n-1)
a262557_list = 0 : f [[0]] where
   f xss = if x < 9 then (map (read . concatMap show) zss) ++ f zss else []
           where zss = (map (z :) $ map tail xss) ++ (map (z :) xss)
                 z = x + 1; x = head $ head xss
-- Reinhard Zumkeller, Oct 14 2015

A262557[n_] := FromDigits[BitLength[n] - Flatten[Position[IntegerDigits[n, 2], 1]]]; Array[A262557, 100] (* or *)
A262557full = Rest[Map[FromDigits, LexicographicSort[Subsets[Range[9, 0, -1]]]]] (*  Paolo Xausa, Feb 13 2024 *)

is_A262557 = is_A009995
apply( A262557(n)=fromdigits(Vecrev(vecextract([0..exponent(n+!n)],n))), [1..99])
# A262557=concat(apply(x(i)=concat(vector(i%10+1,j,if(j>1,x(i*10+j-2),i))),[0..9])) \\ M. F. Hasler, Dec 11 2019

from itertools import combinations
afull = list(map(int, sorted("".join(c) for i in range(1, 11) for c in combinations("9876543210", i)))) # Michael S. Branicky, Feb 13 2024

a(n) = A009995(A263328(n)); a(A263327(n)) = A009995(n). - Reinhard Zumkeller, Oct 15 2015

New name from M. F. Hasler, Dec 11 2019

cp's OEIS Frontend

A263328 Permutation of {0, ..., 1023} corresponding to lexicographic ordering of numbers with decreasing digits (A009995). Inverse of A263327.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

PARI

A263329 Fixed points of permutations A263327 and A263328.

Views

Author

Keywords

Comments

Crossrefs

Programs

Haskell

A263355 Table read by rows: cycles of the permutation A263327, sorted in increasing order of their largest element. The elements in each cycle are listed in decreasing numerical order.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

PARI

Extensions

A263383 Lengths of cycles of permutation A263327.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Haskell

A009995 Numbers with digits in strictly decreasing order. From the Macaulay expansion of n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

A262557 Numbers with digits in strictly decreasing order, sorted lexicographically.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Python

Formula

Extensions