A213743 Triangle T(n,k), read by rows, of numbers T(n,k)=C^(4)(n,k) of combinations with repetitions from n different elements over k for each of them not more than four appearances allowed.

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 121, 1, 6, 21, 56, 126, 246, 426, 1, 7, 28, 84, 210, 455, 875, 1520, 1, 8, 36, 120, 330, 784, 1652, 3144, 5475, 1, 9, 45, 165, 495, 1278, 2922, 6030, 11385, 19855, 1, 10, 55, 220, 715, 1992, 4905, 10890, 22110, 41470, 72403

Offset: 0

Author

Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

Keywords

• nonn
• tabl

Comments

The left side of triangle consists of 1's, while the right side is formed by A187925. Further, T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A000332(n) for n>=7, T(n,5)=A027659(n) for n>=3, T(n,6)=A064056(n) for n>=4, T(n,7)=A064057(n) for n>=5, T(n,8)=A064058(n) for n>=6, T(n,9)=A000575(n) for n>=6.

Examples

			Triangle begins
  n/k.|..0.....1.....2.....3.....4.....5.....6.....7
  ==================================================
  .0..|..1
  .1..|..1.....1
  .2..|..1.....2.....3
  .3..|..1.....3.....6....10
  .4..|..1.....4....10....20....35
  .5..|..1.....5....15....35....70....121
  .6..|..1.....6....21....56...126....246...426
  .7..|..1.....7....28....84...210....455...875....1520
T(4,2)=C^(4)(4,2): From 4 elements {1,2,3,4}, we have the following 10 allowed combinations of 2 elements: {1,1}, {1,2}, {1,3}, {1,4}, {2,2}, {2,3}, {2,4}, {3,3}, {3,4}, {4,4}.
		

Links

• Peter J. C. Moses, Rows n = 0..50 of triangle, flattened

Crossrefs

Cf. A007318, A005725, A111808, A187925, A213742, A000217, A000292, A000332, A027659, A064056, A064057, A064058, A000575.

Programs

Mathematica

Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[5] (* Peter J. C. Moses, Apr 16 2013 *)

Formula

C^(4)(n,k) = Sum_{r=0...floor(k/5)} (-1)^r*C(n,r)*C(n-5*r+k-1, n-1).