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

Offset

0,5

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.

Links

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

Formula

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

Example

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}.

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 *)

Crossrefs

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

Sequence in context: A247046 A081422 A213742 * A213744 A213745 A213808

Adjacent sequences: A213740 A213741 A213742 * A213744 A213745 A213746

Keyword

nonn,tabl

Author

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

Status

approved