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Triangle read by rows: let f(n,m)=Product[k + m + 1, {k, 0, n}]; t0(n,m)=f(n, m)/(f(n - m, m)); then t(n,m)=t0(n,m)+t0(n,n-m).
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%I #3 Mar 30 2012 17:34:33

%S 2,4,4,21,8,21,211,35,35,211,3025,342,84,342,3025,55441,5047,560,560,

%T 5047,55441,1235521,95048,7992,1440,7992,95048,1235521,32432401,

%U 2162169,154530,12870,12870,154530,2162169,32432401,980179201,57657610,3603710

%N Triangle read by rows: let f(n,m)=Product[k + m + 1, {k, 0, n}]; t0(n,m)=f(n, m)/(f(n - m, m)); then t(n,m)=t0(n,m)+t0(n,n-m).

%C Row sums are: {2, 8, 50, 492, 6818, 122096, 2678562, 69523940, 2083398482, 70781242248, 2688204949586,...}.

%C The sequence has an interesting inverted center structure and is based on a generalized permutation product form.

%F f(n,m)=Product[k + m + 1, {k, 0, n}];

%F t0(n,m)=f(n, m)/(f(n - m, m));

%F t(n,m)=t0(n,m)+t0(n,n-m).

%e {2},

%e {4, 4},

%e {21, 8, 21},

%e {211, 35, 35, 211},

%e {3025, 342, 84, 342, 3025},

%e {55441, 5047, 560, 560, 5047, 55441},

%e {1235521, 95048, 7992, 1440, 7992, 95048, 1235521},

%e {32432401, 2162169, 154530, 12870, 12870, 154530, 2162169, 32432401},

%e {980179201, 57657610, 3603710, 241560, 34320, 241560, 3603710, 57657610, 980179201},

%e {33522128641, 1764322571, 98018052, 5767476, 384384, 384384, 5767476, 98018052, 1764322571, 33522128641},

%e {1279935820801, 60949324812, 3047466396, 160395144, 8943480, 1048320, 8943480, 160395144, 3047466396, 60949324812, 1279935820801}

%t Clear[f, t, n, m, k];

%t f[n_, m_] := Product[k + m + 1, {k, 0, n}];

%t t[n_, m_] := f[n, m]/(f[n - m, m]);

%t Table[Table[t[n, m] + t[n, n - m], {m, 0, n}], {n, 0, 10}];

%t Flatten[%]

%K nonn,tabl

%O 0,1

%A _Roger L. Bagula_, Jan 25 2009

%E Edited by _N. J. A. Sloane_, Jan 31 2009