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%I #11 Dec 14 2018 19:37:49
%S 1,1,1,1,1,4,4,1,4,1,1,11,11,11,36,11,1,11,11,1,1,26,26,66,196,66,26,
%T 196,196,26,1,26,66,26,1,1,57,57,302,848,302,302,1898,1898,302,57,848,
%U 1898,848,57,1,57,302,302,57,1,1,120,120,1191,3228,1191,2416,13644
%N Partitioned 2nd-order Eulerian numbers forming an "Eulerian pyramid" (tetrahedron).
%C For N+1 = i+j+k, let P(N+1;i,j,k) = (N+1-i)*P(N;i-1,j,k) + (N+1-j)*P(N;i,j-1,k) + (N+1-k)*P(N;i,j,k-1), with P(N;i,j,k) invariant upon permutation of the indices i,j,k, also P(N;N,0,0)=1 and P(N;i,j,k) = 0 if i or j or k is negative. The indexing of these values is shown explicitly in the examples.
%C The row sums are the second-order Eulerian numbers, A008517; precisely, Sum_{(j,k)|j+k=N-i} P(N;i,j,k) = <<N+1,i>> = T(N+1,i+1) of A008517. The row sum of row i=N of slice N is (N+1)!. The sum of all entries in slice N is (2*N+1)!!. The edges of the N-th triangular slice of the pyramid are row (N+1) of the first-order Eulerian triangle, A008292.
%e The first few slices of the tetrahedron (and row sums) are:
%e 1 (1); i=0, N=0, (j,k)=(0,0)
%e ------------------------
%e 1 (1); i=0, N=1, (j,k)=(0,0)
%e 1 1 (2); i=1, N=1, (j,k)=(1,0) (0,1)
%e ------------------------
%e 1 (1); i=0, N=2, (j,k)=(0,0)
%e 4 4 (8); i=1, N=2, (j,k)=(1,0) (0,1)
%e 1 4 1 (6); i=2, N=2, (j,k)=(2,0) (1,1) (0,2)
%e ------------------------
%e 1 (1); i=0, N=3, (j,k)=(0,0)
%e 11 11 (22); i=1, N=3, (j,k)=(1,0) (0,1)
%e 11 36 11 (58); i=2, N=3, (j,k)=(2,0) (1,1) (0,2)
%e 1 11 11 1 (24); i=3, N=3, (j,k)=(3,0) (2,1) (1,2) (0,3)
%e ------------------------
%e 1 (1); i=0, N=4, (j,k)=(0,0)
%e 26 26 (52); i=1, N=4, (j,k)=(1,0) (0,1)
%e 66 196 66 (328); i=2, N=4, (j,k)=(2,0) (1,1) (0,2)
%e 26 196 196 26 (444); i=3, N=4, (j,k)=(3,0) (2,1) (1,2) (0,3)
%e 1 26 66 26 1 (120); i=4, N=4, (j,k)=(4,0) (3,1) (2,2) (1,3) (0,4)
%K nonn,tabf
%O 0,6
%A _Gregory Gerard Wojnar_, Nov 13 2018