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 1-3 of 3 results.

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

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Author

Clark Kimberling, Dec 17 2004

Keywords

Comments

Row sums = n^n. T(n,1) = n, T(n,n) = n!.

Examples

			First rows:
1;
2,   2;
3,  18,   6;
4,  84, 144,  24;
		

References

  • H. Picquet, Note #124, L'Intermédiaire des Mathématiciens, 1 (1894), pp. 125-127. - N. J. A. Sloane, Feb 28 2022

Crossrefs

Programs

  • Mathematica
    Table[Table[StirlingS2[n, k] Binomial[n, k] k!, {k, 1, n}], {n, 1, 8}] // Grid

Formula

T(n, h) = C(n, h)*U(n, h), where U(n, h) is the array in A019538. Thus T(n, h) = C(n, h)*h!*S(n, h), where S(n, h) is a Stirling number of the second kind (given by A048993 with zeros removed).
T(2n,n) = A288312(n). - Alois P. Heinz, Jun 07 2017

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

Views

Author

Alford Arnold, Oct 20 2010

Keywords

Examples

			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;
		

Crossrefs

Cf. A000169 (row sums), A000081 (unlabeled rooted trees) A179438 (a similar refinement), A054589, A135278, A019538, A101817, A101818

Formula

Sum_{k=1.. A000041(n)} a(n,k) = A000169(n). (Row sums)
a(n,k) = A098546(n,k) *A049019(n,k) /n. - Compare with the formula in A101818.

Extensions

Edited by R. J. Mathar, May 17 2016

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

Views

Author

Geoffrey Critzer, Jun 24 2023

Keywords

Comments

Let H_f denote the H-class in the semigroup of partial transformations containing f. Then H_f contains an idempotent iff the image of f is a transversal for the kernel of f.
Let H_f ~ H_g iff the image of f is contained in the image of g and the kernel of f is more coarse than the kernel of g. Then ~ is a partial order on the H-classes, hence a preorder (quasi-order) on the semigroup. The poset is isomorphic to the Segre product of the Boolean lattice of rank n and the partition lattice of [n+1].

Examples

			Triangle begins:
 1;
 1,   1;
 1,   6,   1;
 1,  21,  18,   1;
 1,  60, 150,  40,  1;
 1, 155, 900, 650, 75, 1;
 ...
		

References

  • O. Ganyushkin and V. Mazorchuk, Classical Finite Transformation Semigroups, 2009, Chapter 4.4 - 4.6.

Crossrefs

Columns k=0-1 give: A000012, A066524.
Row sums give A134055(n+1).
T(n,n-1) gives A002411.

Programs

  • Maple
    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
  • Mathematica
    Table[Table[Binomial[n, k] StirlingS2[n + 1, k + 1], {k, 0, n}], {n,0, 5}] // Grid

Formula

T(n,k) = A007318(n,k)*A008277(n+1,k+1).
Sum_{k=0..n} T(n,k)*k! = (n+1)^n = A000169(n+1).
T(n,1) = A101818(n,1) = A066524(n) = n*(2^n - 1). (Every partial function of rank 1 is idempotent.)
Showing 1-3 of 3 results.