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

0,9

Comments

General asymptotic formula for g.f. (1 - 4*x)^(-j/2)*x/(1 - x) and fixed j>0 is a(n) ~ n^(j/2 - 1) * 4^n / (3*Gamma(j/2)). - Vaclav Kotesovec, Jan 29 2019

Links

Table of n, a(n) for n=0..65.

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).

Example

[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;

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;

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}]

Crossrefs

Sums of antidiagonals are A323217. Main diagonal is A323219.

Rows: A057427 (n=0), A006134 (n=1), A002450 (n=2), A277178 (n=3), A014916 (n=4), A323223 (n=5).

Columns: A005408 (k=2), A059993 (k=3), A323218 (k=4).

Similar array based on Catalan numbers is A323224.

Sequence in context: A121314 A119271 A364310 * A125104 A098157 A293617

Adjacent sequences: A323219 A323220 A323221 * A323223 A323224 A323225

Keyword

nonn,tabl

Author

Peter Luschny, Jan 24 2019

Status

approved