A176560 A symmetrical triangle recursion:q=5;t(n,m,0)=Binomial[n,m];t(n,m,1)=Narayana(n,m);t(n,m,2)=Eulerian(n+1,m);t(n,m,q)=t(n,m,g-2)+t(n,m,q-3).
1, 1, 1, 1, 6, 1, 1, 16, 16, 1, 1, 35, 85, 35, 1, 1, 71, 351, 351, 71, 1, 1, 140, 1295, 2590, 1295, 140, 1, 1, 274, 4488, 16108, 16108, 4488, 274, 1, 1, 537, 14943, 89409, 157953, 89409, 14943, 537, 1, 1, 1057, 48379, 457711, 1315645, 1315645, 457711, 48379
Offset: 0
Examples
{1}, {1, 1}, {1, 6, 1}, {1, 16, 16, 1}, {1, 35, 85, 35, 1}, {1, 71, 351, 351, 71, 1}, {1, 140, 1295, 2590, 1295, 140, 1}, {1, 274, 4488, 16108, 16108, 4488, 274, 1}, {1, 537, 14943, 89409, 157953, 89409, 14943, 537, 1}, {1, 1057, 48379, 457711, 1315645, 1315645, 457711, 48379, 1057, 1}, {1, 2090, 153461, 2208437, 9751973, 15743651, 9751973, 2208437, 153461, 2090, 1}
Programs
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Mathematica
<< DiscreteMath`Combinatorica` t[n_, m_, 0] := Binomial[n, m]; t[n_, m_, 1] := Binomial[n, m]*Binomial[n + 1, m]/(m + 1); t[n_, m_, 2] := Eulerian[1 + n, m]; t[n_, m_, q_] := t[n, m, q] = t[n, m, q - 2] + t[n, m, q - 3] - 1; Table[Flatten[Table[Table[t[n, m, q], {m, 0, n}], {n, 0, 10}]], {q, 0, 10}]
Formula
q=5;
t(n,m,0)=Binomial[n,m];
t(n,m,1)=Narayana(n,m);
t(n,m,2)=Eulerian(n+1,m);
t(n,m,q)=t(n,m,g-2)+t(n,m,q-3)
Comments