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A general recursion triangle with third part a power triangle:m=2; Power triangle: f(n,k,m)=If[n*k*(n - k) == 0, 1, n^m - (k^m + (n - k)^m)]; Recursion: A(n,k,m)=(m*(n - k) + 1)*A(n - 1, k - 1, m) + (m*k + 1)*A(n - 1, k, m) + m*f(n, k, m)*A(n - 2, k - 1, m).
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%I #2 Mar 30 2012 17:34:34

%S 1,1,1,1,10,1,1,43,43,1,1,148,590,148,1,1,469,5018,5018,469,1,1,1438,

%T 34047,91492,34047,1438,1,1,4351,204813,1187731,1187731,204813,4351,1,

%U 1,13096,1149652,12609880,27234646,12609880,1149652,13096,1,1,39337

%N A general recursion triangle with third part a power triangle:m=2; Power triangle: f(n,k,m)=If[n*k*(n - k) == 0, 1, n^m - (k^m + (n - k)^m)]; Recursion: A(n,k,m)=(m*(n - k) + 1)*A(n - 1, k - 1, m) + (m*k + 1)*A(n - 1, k, m) + m*f(n, k, m)*A(n - 2, k - 1, m).

%C Row sums are:

%C {1, 2, 12, 88, 888, 10976, 162464, 2793792, 54779904, 1206055680, 29460493056,...}.

%F m=0:Pascal:m=1Eulerian numbers;

%F m=2;

%F Power triangle:

%F f(n,k,m)=If[n*k*(n - k) == 0, 1, n^m - (k^m + (n - k)^m)];

%F Recursion:

%F A(n,k,m)=(m*(n - k) + 1)*A(n - 1, k - 1, m) +

%F (m*k + 1)*A(n - 1, k, m) +

%F m*f(n, k, m)*A(n - 2, k - 1, m).

%e {1},

%e {1, 1},

%e {1, 10, 1},

%e {1, 43, 43, 1},

%e {1, 148, 590, 148, 1},

%e {1, 469, 5018, 5018, 469, 1},

%e {1, 1438, 34047, 91492, 34047, 1438, 1},

%e {1, 4351, 204813, 1187731, 1187731, 204813, 4351, 1},

%e {1, 13096, 1149652, 12609880, 27234646, 12609880, 1149652, 13096, 1},

%e {1, 39337, 6188356, 117961172, 478838974, 478838974, 117961172, 6188356, 39337, 1},

%e {1, 118066, 32448653, 1015124312, 7053594482, 13257922028, 7053594482, 1015124312, 32448653, 118066, 1}

%t A[n_, 0, m_] := 1; A[n_, n_, m_] := 1;

%t A[n_, k_, m_] := (m*(n - k) + 1)*A[n - 1, k - 1, m] + (m*k + 1)*A[n - 1, k, m] + m*f[n, k, m]*A[n - 2, k - 1, m];

%t Table[A[n, k, m], {m, 0, 10}, {n, 0, 10}, {k, 0, n}];

%t Table[Flatten[Table[Table[A[n, k, m], {k, 0, n}], {n, 0, 10}]], {m, 0, 10}]

%t Table[Table[Sum[A[n, k, m], {k, 0, n}], {n, 0, 10}], {m, 0, 10}];

%Y A142459, A154336, A157277

%K nonn,tabl,uned

%O 0,5

%A _Roger L. Bagula_, Mar 03 2009