A323222 A(n, k) = [x^k] (1 - 4*x)^(-n/2)*x/(1 - x), square array read by ascending antidiagonals with n >= 0 and k >= 0.
0, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 5, 9, 1, 0, 1, 7, 21, 29, 1, 0, 1, 9, 37, 85, 99, 1, 0, 1, 11, 57, 177, 341, 351, 1, 0, 1, 13, 81, 313, 807, 1365, 1275, 1, 0, 1, 15, 109, 501, 1593, 3579, 5461, 4707, 1, 0, 1, 17, 141, 749, 2811, 7737, 15591, 21845, 17577, 1
Offset: 0
Examples
[n\k] 0 1 2 3 4 5 6 7 8 9 ------------------------------------------------------------------- [0] 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, ... A057427 [1] 0, 1, 3, 9, 29, 99, 351, 1275, 4707, 17577, ... A006134 [2] 0, 1, 5, 21, 85, 341, 1365, 5461, 21845, 87381, ... A002450 [3] 0, 1, 7, 37, 177, 807, 3579, 15591, 67071, 285861, ... A277178 [4] 0, 1, 9, 57, 313, 1593, 7737, 36409, 167481, 757305, ... A014916 [5] 0, 1, 11, 81, 501, 2811, 14823, 74883, 366603, 1752273, ... A323223 [6] 0, 1, 13, 109, 749, 4589, 26093, 140781, 730605, 3679725, ... [7] 0, 1, 15, 141, 1065, 7071, 43107, 247311, 1355847, 7175661, ... [8] 0, 1, 17, 177, 1457, 10417, 67761, 411825, 2377905, 13191345, ... [9] 0, 1, 19, 217, 1933, 14803, 102319, 656587, 3982195, 23104441, ... Triangle given by antidiagonals: 0; 0, 1; 0, 1, 1; 0, 1, 3, 1; 0, 1, 5, 9, 1; 0, 1, 7, 21, 29, 1; 0, 1, 9, 37, 85, 99, 1; 0, 1, 11, 57, 177, 341, 351, 1; 0, 1, 13, 81, 313, 807, 1365, 1275, 1; 0, 1, 15, 109, 501, 1593, 3579, 5461, 4707, 1;
Crossrefs
Programs
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Maple
Row := proc(n, len) local ogf, ser; ogf := (1 - 4*x)^(-n/2)*x/(1 - x); ser := series(ogf, x, (n+1)*len+1); seq(coeff(ser, x, j), j=0..len) end: for n from 0 to 9 do Row(n, 9) od;
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Mathematica
BF[N_, K_] := Module[{}, r[n_, k_] := FrobeniusSolve[ConstantArray[1, n], k]; X[n_] := Flatten[Table[r[N, j], {j, 0, n - 1}], 1]; CentralBinomial[n_] := Binomial[2 n, n]; Sum[Product[CentralBinomial[m[[i]]], {i, 1, N}], {m , X[K]}]]; Trow[n_] := Table[BF[n, k], {k, 0, 9}]; Table[Trow[n], {n, 1, 9}]
Formula
For n>0 and k>0 let X(n, k) denote the set of all tuples of length n with elements from {0, ..., k-1} with sum < k. Let b(m) = binomial(2*m, m). Then A(n, k) = Sum_{(j1,...,jn) in X(n, k)} b(j1)*b(j2)*...*b(jn).
Comments