%I
%S 1,1,3,1,16,10,1,55,165,35,1,156,1386,1456,126,1,399,8456,25368,11970,
%T 462,1,960,42876,289920,393030,95040,1716,1,2223,193185,2577135,
%U 7731405,5525091,741741,6435,1,5020,803440,19411480,111675850,176644468
%N Triangle read by rows: T(n,k)=A(n,k)*binomial(n+k1,n), where A(n,k) are the Eulerian numbers (A008292).
%C Andrews, Theory of Partitions, (1976), discussion of multisets.
%C 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]
%D R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd edition, AddisonWesley, Reading, Mass., 1994, p. 269 (Worpitzky's identity).
%D Miklos Bona, Combinatorics of Permutations,Chapman and Hall,2004,page 6. [From _Geoffrey Critzer_, May 19 2010]
%e 1;
%e 1,3;
%e 1,16,10;
%e 1,55,165,35;
%e 1,156,1386,1456,126;
%e ...
%e 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]
%p 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);
%t Table[Table[Eulerian[n, k] Binomial[n + k, n], {k, 0, n  1}], {n, 1,10}] (* _Geoffrey Critzer_, Jun 13 2013 *)
%Y Cf. A001700, A014449, A000312.
%Y Row sums yield A000312 (Worpitzky's identity).
%Y Cf. A008292.
%K nonn,tabl
%O 1,3
%A _Alford Arnold_
%E More terms from _Emeric Deutsch_, May 08 2004
