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Array t(n,k) = binomial(n*k, n+1)/n, where n >= 1 and k >= 2, read by ascending antidiagonals.
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%I #19 Jun 15 2018 09:10:18

%S 1,2,3,5,10,6,14,42,28,10,42,198,165,60,15,132,1001,1092,455,110,21,

%T 429,5304,7752,3876,1020,182,28,1430,29070,57684,35420,10626,1995,280,

%U 36,4862,163438,444015,339300,118755,24570,3542,408,45,16796,937365,3506100,3362260,1391280,324632,50344,5850,570,55

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

%C 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).

%C Row 1 is A000217 (triangular numbers),

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

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

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

%C Row 5 is not in the OEIS.

%C Column 1 is A000108 (Catalan numbers),

%C Column 2 is A007226 left shifted 1 place,

%C Column 4 is A007228 left shifted 1 place,

%C Column 5 is A124724 left shifted 1 place,

%C Column 6 is not in the OEIS.

%D 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

%H MathOverflow, <a href="http://mathoverflow.net/questions/60943">Root estimation</a>

%e Array begins:

%e 1, 3, 6, 10, 15, 21, ...

%e 2, 10, 28, 60, 110, 182, ...

%e 5, 42, 165, 455, 1020, 1995, ...

%e 14, 198, 1092, 3876, 10626, 24570, ...

%e 42, 1001, 7752, 35420, 118755, 324632, ...

%e 132, 5304, 57684, 339300, 1391280, 4496388, ...

%e etc.

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

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

%K nonn,tabl,easy

%O 1,2

%A _Jean-François Alcover_, Apr 18 2014