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A row reversed version of A177517
0

%I #7 Jul 10 2012 12:20:30

%S 1,1,1,0,1,1,1,2,0,1,3,2,1,4,5,1,1,5,9,6,1,6,14,15,5,1,7,20,29,20,1,8,

%T 27,49,49,22,1,9,35,76,98,71

%N A row reversed version of A177517

%C The row reversal allows getting rid of the initial zeros in A177517

%C and shows the polynomial structure of the sequence more clearly.

%e {1},

%e {1},

%e {1, 0},

%e {1, 1},

%e {1, 2, 0},

%e {1, 3, 2},

%e {1, 4, 5, 1},

%e {1, 5, 9, 6},

%e {1, 6, 14, 15, 5},

%e {1, 7, 20, 29, 20},

%e {1, 8, 27, 49, 49, 22},

%e {1, 9, 35, 76, 98, 71}

%t Clear[t, n, k]

%t t[n_, 1] := If[n == 1, 1, 0]

%t t[n_, k_] := t[n, k] = Sum[t[n - i, k - 1], {i, 1, k - 1}]

%t a = Table[Reverse[Table[t[n, k], {k, 1, n}]], {n, 1, 12}];

%t Table[Table[a[[n, k]], {k, 1, Floor[(n + 1)/2]}], {n, 1, Length[a]}];

%t Flatten[%]

%Y Cf. A177517

%K nonn,tabf

%O 1,8

%A _Roger L. Bagula_ and Mats Granvik, Dec 12 2010