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.

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

Author

Reinhard Zumkeller, Oct 16 2015

Keywords

• nonn
• fini
• full
• tabf

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
		

Links

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

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