A101817
Triangle read by rows: T(n,h) = number of functions f:{1,2,...,n}->{1,2,...,n} such that |Image(f)|=h; h=1,2,...,n, n=1,2,3,... . Essentially A090657, but without zeros.
Original entry on oeis.org
1, 2, 2, 3, 18, 6, 4, 84, 144, 24, 5, 300, 1500, 1200, 120, 6, 930, 10800, 23400, 10800, 720, 7, 2646, 63210, 294000, 352800, 105840, 5040, 8, 7112, 324576, 2857680, 7056000, 5362560, 1128960, 40320, 9, 18360, 1524600, 23496480, 105099120
Offset: 1
First rows:
1;
2, 2;
3, 18, 6;
4, 84, 144, 24;
- H. Picquet, Note #124, L'Intermédiaire des Mathématiciens, 1 (1894), pp. 125-127. - N. J. A. Sloane, Feb 28 2022
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Table[Table[StirlingS2[n, k] Binomial[n, k] k!, {k, 1, n}], {n, 1, 8}] // Grid
A181415
Irregular triangle a(n,k) = A049009(n,k)/n, read by rows 1<=k<=A000041(n).
Original entry on oeis.org
1, 1, 1, 1, 6, 2, 1, 12, 9, 36, 6, 1, 20, 40, 120, 180, 240, 24, 1, 30, 75, 50, 300, 1200, 300, 1200, 2700, 1800, 120, 1, 42, 126, 210, 630, 3150, 2100, 3150, 4200, 25200, 12600, 12600, 37800, 15120, 720, 1, 56, 196, 392, 245, 1176, 7056, 11760, 8820, 11760, 11760, 88200
Offset: 1
Row three is calculated as follows:
( 3 18 6) divided by (3 3 3) yielding (1 6 2)
1;
1,1;
1,6,2;
1,12,9,36,6;
1,20,40,120,180,240,24;
1,30,75,50,300,1200,300,1200,2700,1800,120;
1,42,126,210,630,3150,2100,3150,4200,25200,12600,12600,37800,15120,720;
A363849
Triangular array read by rows. T(n,k) is the number of Green's H-classes of rank k in the semigroup of partial transformations, n >= 0, 0 <= k <= n.
Original entry on oeis.org
1, 1, 1, 1, 6, 1, 1, 21, 18, 1, 1, 60, 150, 40, 1, 1, 155, 900, 650, 75, 1, 1, 378, 4515, 7000, 2100, 126, 1, 1, 889, 20286, 59535, 36750, 5586, 196, 1, 1, 2040, 84700, 435120, 486570, 148176, 12936, 288, 1, 1, 4599, 335880, 2864820, 5358150, 2876202, 493920, 27000, 405, 1
Offset: 0
Triangle begins:
1;
1, 1;
1, 6, 1;
1, 21, 18, 1;
1, 60, 150, 40, 1;
1, 155, 900, 650, 75, 1;
...
- O. Ganyushkin and V. Mazorchuk, Classical Finite Transformation Semigroups, 2009, Chapter 4.4 - 4.6.
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T:= (n, k)-> binomial(n, k)*Stirling2(n+1, k+1):
seq(seq(T(n, k), k=0..n), n=0..10); # Alois P. Heinz, Jun 24 2023
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Table[Table[Binomial[n, k] StirlingS2[n + 1, k + 1], {k, 0, n}], {n,0, 5}] // Grid
Showing 1-3 of 3 results.
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