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%I #20 Dec 26 2023 12:13:46
%S 1,1,1,1,3,1,-5,7,6,1,-65,-15,25,10,1,-455,-455,0,65,15,1,-1295,-4725,
%T -1715,140,140,21,1
%N In general, let A(n,k,m) denote the (n,k)-th entry of the inverse of the matrix consisting of the (n,k)-th m-restrained Stirling numbers of the second kind (-1)^(n-k) times the number of permutations of an n-set with k disjoint cycles of length less than or equal to m, as the (n+1,k+1)-th entry. The sequence shows A(n,k,3), which is a lower triangular matrix, read by rows.
%C A(n,k,m) also can be expanded for nonpositive integers n and k using the m-restrained Stirling numbers of the first kind.
%H Ji Young Choi, <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.618.3725">Multi-restrained Stirling numbers</a>, Ars Comb. 120 (2015), 113-127.
%H John Engbers, David Galvin, and Cliff Smyth, <a href="https://doi.org/10.1016/j.jcta.2018.08.001">Restricted Stirling and Lah number matrices and their inverses</a>, Journal of Combinatorial Theory, Series A, 161 (2019), 271-298.
%F A(n,k,m) = A(n-1,k-1,m) - Sum_{i=1..m-1} (-1)^{i}(k)...(k+i-1) A(n,k+i,m) A(n,k,m) = A(n-1,k-1,m) + k A(n-1,k,m) + (-1)^m k(k+1)...(k+m-1)A(n,k+m,m).
%e A(1,1,3) = 1, A(1,2,3) = 0, A(1,3,3) = 0, A(1,4,3) = 0, ...
%e A(2,1,3) = 1, A(2,2,3) = 1, A(2,3,3) = 0, A(2,4,3) = 0, ...
%e A(3,1,3) = 1, A(3,2,3) = 3, A(3,3,3) = 1, A(3,4,3) = 0, ...
%e A(4,1,3) = -5, A(4,2,3) = 7, A(4,3,3) = 6, A(4,4,3) = 1, ...
%e In other words, A(n,k,3) is the matrix
%e 1
%e 1 1
%e 1 3 1
%e -5 7 6 1
%e ...
%e with all other entries in each row being 0. - _N. J. A. Sloane_, Dec 21 2019
%Y Cf. A111246, A144633.
%K sign,more,tabl
%O 1,5
%A _Ji Young Choi_, Jan 21 2010
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