OFFSET
1,2
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 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.
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
EXAMPLE
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.
MATHEMATICA
t[n_, k_] := Binomial[n*k, n+1]/n; Table[t[n-k+2, k], {n, 1, 10}, {k, 2, n+1}] // Flatten
CROSSREFS
KEYWORD
AUTHOR
Jean-François Alcover, Apr 18 2014
STATUS
approved