A241262 Array t(n,k) = binomial(n*k, n+1)/n, where n >= 1 and k >= 2, read by ascending antidiagonals.

1, 2, 3, 5, 10, 6, 14, 42, 28, 10, 42, 198, 165, 60, 15, 132, 1001, 1092, 455, 110, 21, 429, 5304, 7752, 3876, 1020, 182, 28, 1430, 29070, 57684, 35420, 10626, 1995, 280, 36, 4862, 163438, 444015, 339300, 118755, 24570, 3542, 408, 45, 16796, 937365, 3506100, 3362260, 1391280, 324632, 50344, 5850, 570, 55

Offset: 1

Author

Jean-François Alcover, Apr 18 2014

Keywords

• nonn
• tabl
• easy

Comments

About the "root estimation" question asked in MathOverflow, one can check (at least numerically) that, for instance with k = 4 and a = 1/11, the series a^-1 + (k - 1) + Sum_{n>=} (-1)^n*binomial(n*k, n+1)/n*a^n evaluates to the positive solution of x^k = (x+1)^(k-1).

Row 1 is A000217 (triangular numbers),

Row 2 is A006331 (twice the square pyramidal numbers),

Row 3 is A067047(3n) = lcm(3n, 3n+1, 3n+2, 3n+3)/12 (from column r=4 of A067049),

Row 4 is A222715(2n) = (n-1)*n*(2n-1)*(4n-3)*(4n-1)/15,

Row 5 is not in the OEIS.

Column 1 is A000108 (Catalan numbers),

Column 2 is A007226 left shifted 1 place,

Column 4 is A007228 left shifted 1 place,

Column 5 is A124724 left shifted 1 place,

Column 6 is not in the OEIS.

Examples

			Array begins:
    1,    3,     6,     10,      15,      21, ...
    2,   10,    28,     60,     110,     182, ...
    5,   42,   165,    455,    1020,    1995, ...
   14,  198,  1092,   3876,   10626,   24570, ...
   42, 1001,  7752,  35420,  118755,  324632, ...
  132, 5304, 57684, 339300, 1391280, 4496388, ...
  etc.
		

References

• N. S. S. Gu, H. Prodinger, S. Wagner, Bijections for a class of labeled plane trees, Eur. J. Combinat. 31 (2010) 720-732, doi|10.1016/j.ejc.2009.10.007, Theorem 2

Links

• MathOverflow, Root estimation

Crossrefs

Cf. A000217, A006331, A000108, A007226, A007228, A067047, A067049, A124724, A222715.

Programs

Mathematica

t[n_, k_] := Binomial[n*k, n+1]/n; Table[t[n-k+2, k], {n, 1, 10}, {k, 2, n+1}] // Flatten