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Stirling-like number triangle defined by paired decomposition of C(n+3,3) = A000292.
1

%I #31 Aug 23 2024 08:40:04

%S 1,1,1,1,4,1,1,12,10,1,1,32,67,20,1,1,80,376,252,35,1,1,192,1909,2560,

%T 742,56,1,1,448,9094,22928,12346,1848,84,1,1,1024,41479,189120,177599,

%U 46912,4074,120,1,1,2304,183412,1472704,2318149

%N Stirling-like number triangle defined by paired decomposition of C(n+3,3) = A000292.

%C Note that the Stirling numbers of the second kind are generated in a similar fashion by decomposing the triangular numbers C(n+2,2) as {1}, {1,2}, {1,2,3}, .... The defining sequence A000292 appears as the subdiagonal when the triangle is arranged in lower-triangular form. The second column is A001787.

%C Gives the number of ways to construct pairs of k-block partitions from 1 to n such that the sum of the minima of the i-th block of the first partition and the (k-i+1)th block of the second partition is n+1. - _Ken Joffaniel M Gonzales_, Jun 13 2010

%H R. B. Corcino, K. J. M. Gonzales, M. J. C. Loquias and E. L. Tan, <a href="http://arxiv.org/abs/1302.4694">Dually weighted Stirling-type sequences</a>, arXiv preprint arXiv:1302.4694 [math.CO], 2013-2014.

%H R. B. Corcino, K. J. M. Gonzales, M. J. C. Loquias and E. L. Tan, <a href="http://dx.doi.org/10.1016/j.ejc.2014.06.010">Dually weighted Stirling-type sequences</a>, Europ. J. Combin., 43, 2015, 55-67.

%F Columns are generated as follows: Display C(n+3, 3) as row sums of the triangle A080251, or {1}, {2, 2}, {3, 3, 4}, {4, 4, 6, 6}, {5, 5, 8, 8, 9}, ... The columns are then generated by 1/(1-x), 1/(1-2x)^2, 1/((1-3x)^2*(1-4x)), 1/((1-4x)^2*(1-6x)^2), etc.

%e Rows are

%e {1},

%e {1, 1},

%e {1, 4, 1},

%e {1, 12, 10, 1},

%e {1, 32, 67, 20, 1},

%e ...

%t s[b_, n_, k_] := s[b, n, k] = Which[n==k==0, 1, n==0, 0, k==0, 0, True, s[b+1, n-1, k-1] + k*b*s[b, n-1, k]]

%t Table[s[0, n+2, k+2], {n, 0, 10}, {k, 0, n}] // Flatten

%t (* a specialization of equation (9) in the Corcino et al. paper *)

%t (* _Mikhail Lavrov_, Oct 12 2022 *)

%t T[ n_, k_] := If[n < 0, 0, SeriesCoefficient[x^k / Product[1 + x*(Floor[j/2] + 1)*(Floor[j/2] - k - 1), {j,0,k}], {x,0,n}]]; (* _Michael Somos_, Oct 12 2022 *)

%o (PARI) {T(n, k) = if(n<0, 0, polcoeff(x^k / prod(j=0, k, 1 + x*(j\2 + 1)*(j\2 - k - 1) + x*O(x^n)), n))}; /* _Michael Somos_, Oct 12 2022 */

%Y Cf. A000292, A001787, A008277, A080251.

%K easy,nonn,tabl

%O 0,5

%A _Paul Barry_, Feb 17 2003