%I #18 Dec 11 2019 21:58:43
%S 0,1,2,16,12,5,17,18,84,192,75,68,65,64,56,38,28,26,7,939,978,908,881,
%T 853,852,840,809,798,782,777,776,772,760,758,756,746,736,717,711,708,
%U 703,698,690,669,666,662,647,622,610,595,585,564,555,553,547,531
%N 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.
%C A263383(n) gives the number of terms in row n.
%C Fixed points: T(k,m) in A263329 <=> A263383(k) = 1 = m. [Corrected by _M. F. Hasler_, Dec 11 2019]
%C 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
%C 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
%H Reinhard Zumkeller, <a href="/A263355/b263355.txt">Rows n = 1..18 of triangle, flattened</a>, 1024 terms
%e n | Cycles: A263355(n, k=1..A263383(n)) | A263383(n)
%e ---+--------------------------------------------------------+-----------
%e 1 | (0) | 1
%e 2 | (1) | 1
%e 3 | (2) | 1
%e 4 | (16, 12, 5) | 3
%e 5 | (17) | 1
%e 6 | (18) | 1
%e 7 | (84) | 1
%e 8 | (192, 75, 68, 65, 64, 56, 38, 28, 26, 7) | 10
%e 9 | (939) | 1
%e 10 | (978, 908, 881, 853, 852, 840, ..., 142, 115, 45) | 74
%e 11 | (1005) | 1
%e 12 | (1006) | 1
%e 13 | (1016, 997, 995, 985, 967, 959, 958, 955, 948, 831) | 10
%e 14 | (1018, 1011, 1007) | 3
%e 15 | (1020, 1019, 1017, 1015, 1014, ..., 10, 9, 8, 6, 4, 3) | 912
%e 16 | (1021) | 1
%e 17 | (1022) | 1
%e 18 | (1023) | 1
%e 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
%o (Haskell)
%o import Data.List ((\\), sort)
%o a263355 n k = a263355_tabf !! (n-1) !! (k-1)
%o a263355_row n = a263355_tabf !! (n-1)
%o a263355_tabf = sort $ cc a263327_list where
%o cc [] = []
%o cc (x:xs) = (reverse $ sort ys) : cc (xs \\ ys)
%o where ys = x : c x
%o c z = if y /= x then y : c y else []
%o where y = a263327 z
%o (PARI) {M=0; (C(x,L=[x])=until(x==L[1], M+=1<<x; x&&L=concat(L,x=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
%Y Cf. A263327, A263383 (row lengths), A263329.
%K nonn,fini,full,tabf
%O 1,3
%A _Reinhard Zumkeller_, Oct 16 2015
%E Edited by _M. F. Hasler_, Dec 11 2019