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A241262 Array t(n,k) = binomial(n*k, n+1)/n, where n >= 1 and k >= 2, read by ascending antidiagonals. 0
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 (list; table; graph; refs; listen; history; text; internal format)
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 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.
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
Sequence in context: A250745 A109204 A250747 * A244489 A367297 A286144
KEYWORD
nonn,tabl,easy
AUTHOR
STATUS
approved

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Last modified February 27 16:21 EST 2024. Contains 370378 sequences. (Running on oeis4.)