

A038675


Triangle read by rows: T(n,k)=A(n,k)*binomial(n+k1,n), where A(n,k) are the Eulerian numbers (A008292).


2



1, 1, 3, 1, 16, 10, 1, 55, 165, 35, 1, 156, 1386, 1456, 126, 1, 399, 8456, 25368, 11970, 462, 1, 960, 42876, 289920, 393030, 95040, 1716, 1, 2223, 193185, 2577135, 7731405, 5525091, 741741, 6435, 1, 5020, 803440, 19411480, 111675850, 176644468
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OFFSET

1,3


COMMENTS

Andrews, Theory of Partitions, (1976), discussion of multisets.
Let a = a_1,a_2,...,a_n be a sequence on the alphabet {1,2,...,n}. Scan a from left to right and create an npermutation by noting the POSITION of the elements as you come to them in order from least to greatest. See example. T(n,k) is the number of sequences that correspond to such a permutation having exactly nk descents. [From Geoffrey Critzer, May 19 2010]


REFERENCES

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd edition, AddisonWesley, Reading, Mass., 1994, p. 269 (Worpitzky's identity).
Miklos Bona, Combinatorics of Permutations,Chapman and Hall,2004,page 6. [From Geoffrey Critzer, May 19 2010]


LINKS



EXAMPLE

1;
1,3;
1,16,10;
1,55,165,35;
1,156,1386,1456,126;
...
If a = 3,1,1,2,4,3 the corresponding 6permutation is 2,3,4,1,6,5 because the first 1 is in the 2nd position, the second 1 is in the 3rd position,the 2 is in the 4th position, the first 3 is in the first position, the next 3 is in the 6th position and the 4 is in the 5th position of the sequence a. [From Geoffrey Critzer, May 19 2010]


MAPLE

A:=(n, k)>sum((1)^j*(kj)^n*binomial(n+1, j), j=0..k): T:=(n, k)>A(n, k)*binomial(n+k1, n): seq(seq(T(n, k), k=1..n), n=1..10);


MATHEMATICA

Table[Table[Eulerian[n, k] Binomial[n + k, n], {k, 0, n  1}], {n, 1, 10}] (* Geoffrey Critzer, Jun 13 2013 *)


CROSSREFS

Row sums yield A000312 (Worpitzky's identity).


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AUTHOR



EXTENSIONS



STATUS

approved



