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.

Showing 1-5 of 5 results.

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

Views

Author

Reinhard Zumkeller, Oct 15 2015

Keywords

Comments

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

Crossrefs

Programs

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

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

Views

Author

Keywords

Comments

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

Crossrefs

Cf. A009993.
Cf. A262557 (sorted lexicographically), A263327, A263328.

Programs

  • Haskell
    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
    
  • Mathematica
    Sort@ Flatten@ Table[FromDigits /@ Subsets[ Range[9, 0, -1], {n}], {n, 10}] (* Zak Seidov, May 10 2006 *)
  • PARI
    is(n)=fromdigits(vecsort(digits(n),,12))==n \\ Charles R Greathouse IV, Apr 16 2015

A263327 A permutation of {0, 1, ..., 1023} corresponding to lexicographical ordering A262557 of numbers with decreasing digits A009995.

Original entry on oeis.org

0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 33, 34, 36, 40, 48, 65, 66, 68, 72, 80, 96, 129, 130, 132, 136, 144, 160, 192, 257, 258, 260, 264, 272, 288, 320, 384, 513, 514, 516, 520, 528, 544, 576, 640, 768, 7, 11, 13, 14
Offset: 0

Views

Author

Reinhard Zumkeller, Oct 15 2015

Keywords

Comments

For n = 1..1023, A262557(a(n)) = A009995(n).
Cycle type = (1^12, 3^2, 10^2, 74, 912), i.e., this permutation has 12 fixed points, two 3-cycles and two 10-cycles, and two more cycles of length 74 and 912. See A263355 for the list of these cycles, A263383 for the length of the n-th cycle (ordered by increasing largest element).

Crossrefs

Cf. A009995, A262557, A263328 (inverse), A263329 (fixed points), A263383, A263355 (cycles).
Row 10 of A294648.

Programs

  • Haskell
    a263327 0 = 0
    a263327 n = head [x | x <- [1..1023], a262557 x == a009995' n]
    
  • Mathematica
    SortBy[Range[0, 1023], DigitCount[#, 2, 1] &] (* Paolo Xausa, Mar 31 2025 *)
  • PARI
    A263327=vecsort(A262557,,1) \\ Does not include a(0)=0. - M. F. Hasler, Dec 11 2019

Extensions

Edited by M. F. Hasler, Dec 11 2019

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

Views

Author

N. J. A. Sloane, Oct 14 2015

Keywords

Comments

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

References

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

Crossrefs

Programs

  • Haskell
    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
    
  • Mathematica
    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 *)
  • PARI
    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
    
  • Python
    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

Formula

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

Extensions

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

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

Views

Author

Reinhard Zumkeller, Oct 16 2015

Keywords

Comments

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

Examples

			   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
		

Crossrefs

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

Programs

  • Haskell
    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
    
  • PARI
    {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

Extensions

Edited by M. F. Hasler, Dec 11 2019
Showing 1-5 of 5 results.