%I #40 Sep 01 2024 09:35:12
%S 1,2,3,1,5,3,8,8,13,18,1,21,38,5,34,76,18,55,147,53,1,89,277,139,7,
%T 144,512,336,32,233,932,766,116,1,377,1676,1670,364,9,610,2984,3516,
%U 1032,50,987,5269,7198,2714,215,1,1597,9239,14402,6734,785,11
%N Triangular array read by rows. T(n,k) is the number of length n words on alphabet {0,1} with k maximal runs of 0's having length 2 or more, n>=0, 0<=k<=nearest integer to n/3.
%H Sergi Elizalde, Johnny Rivera Jr., and Yan Zhuang, <a href="https://arxiv.org/abs/2408.15111">Counting pattern-avoiding permutations by big descents</a>, arXiv:2408.15111 [math.CO], 2024. See p. 6.
%H P. Flajolet and R. Sedgewick, <a href="http://algo.inria.fr/flajolet/Publications/books.html">Analytic Combinatorics</a>, 2009; see page 304.
%F O.g.f.: ((u x^2)/(1 - x) + (1 - x^2)/(1 - x))/(1 - x ((u x^2)/(1 - x) + (1 - x^2)/(1 - x))).
%F Generally, the o.g.f. for such words having maximal runs of length at least r is: ((u x^r)/(1 - x) + (1 - x^r)/(1 - x))/(1 - x ((u x^r)/(1 - x) + (1 - x^r)/(1 - x))).
%e 1,
%e 2,
%e 3, 1,
%e 5, 3,
%e 8, 8,
%e 13, 18, 1,
%e 21, 38, 5,
%e 34, 76, 18,
%e 55, 147, 53, 1
%e T(6,2) = 5 because we have: 000100, 001000, 001001, 001100, 100100.
%t nn = 15; c[z_, u_] := ((1 - z^r)/(1 - z) + u z^r/(1 - z))*1/(1 - z ((1 - z^r)/(1 - z) + u z^r/(1 - z))) /. r -> 2; Map[Select[#, # > 0 &] &, CoefficientList[Series[c[z, u], {z, 0, nn}], {z, u}]] // Grid
%Y Cf. A000045 (column k=0) A006478 (column k=1).
%Y Row sums give A000079.
%K nonn,tabf
%O 0,2
%A _Geoffrey Critzer_, Jul 25 2020