

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.


5



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
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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(n5*r+k1, n1)


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+k1, n1], {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



