A286416
Number T(n,k) of entries in the k-th last blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.
Original entry on oeis.org
1, 3, 1, 8, 6, 1, 24, 25, 10, 1, 83, 98, 63, 15, 1, 324, 399, 338, 135, 21, 1, 1400, 1746, 1727, 980, 257, 28, 1, 6609, 8271, 8874, 6426, 2455, 448, 36, 1, 33758, 42284, 47191, 40334, 20506, 5474, 730, 45, 1, 185136, 231939, 263458, 250839, 158827, 57239, 11128, 1128, 55, 1
Offset: 1
T(3,2) = 6 because the number of entries in the second last blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 0+2+2+1+1 = 6.
Triangle T(n,k) begins:
1;
3, 1;
8, 6, 1;
24, 25, 10, 1;
83, 98, 63, 15, 1;
324, 399, 338, 135, 21, 1;
1400, 1746, 1727, 980, 257, 28, 1;
6609, 8271, 8874, 6426, 2455, 448, 36, 1;
...
A285424
Sum of the entries in the last blocks of all set partitions of [n].
Original entry on oeis.org
1, 5, 19, 75, 323, 1512, 7630, 41245, 237573, 1451359, 9365361, 63604596, 453206838, 3378581609, 26285755211, 212953670251, 1792896572319, 15658150745252, 141619251656826, 1324477898999161, 12791059496663293, 127395689514237279, 1307010496324272157
Offset: 1
a(3) = 19 because the sum of the entries in the last blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 6+3+2+5+3 = 19.
-
a:= proc(h) option remember; local b; b:=
proc(n, m, s) option remember; `if`(n=0, s,
add(b(n-1, max(m, j), `if`(j
-
a[h_] := a[h] = Module[{b}, b[n_, m_, s_] := b[n, m, s] = If[n == 0, s, Sum[b[n-1, Max[m, j], If[j < m, s, h - n + 1 + If[j == m, s, 0]]], {j, 1, m + 1}]]; b[h, 0, 0]];
Array[a, 25] (* Jean-François Alcover, May 22 2018, translated from Maple *)
A286231
Sum T(n,k) of the entries in the k-th last cycles of all permutations of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.
Original entry on oeis.org
1, 5, 1, 25, 10, 1, 143, 79, 17, 1, 942, 634, 197, 26, 1, 7074, 5462, 2129, 417, 37, 1, 59832, 51214, 23381, 5856, 786, 50, 1, 563688, 523386, 269033, 80053, 13934, 1360, 65, 1, 5858640, 5813892, 3281206, 1111498, 232349, 29728, 2204, 82, 1
Offset: 1
T(3,2) = 10 because the sum of the entries in the second last cycles of all permutations of [3] ((123), (132), (12)(3), (13)(2), (1)(23), (1)(2)(3)) is 0+0+3+4+1+2 = 10.
Triangle T(n,k) begins:
1;
5, 1;
25, 10, 1;
143, 79, 17, 1;
942, 634, 197, 26, 1;
7074, 5462, 2129, 417, 37, 1;
59832, 51214, 23381, 5856, 786, 50, 1;
563688, 523386, 269033, 80053, 13934, 1360, 65, 1;
...
Showing 1-3 of 3 results.