%I #24 Nov 12 2019 03:35:11
%S 1,0,1,0,1,2,0,3,7,5,0,15,39,37,14,0,105,296,326,176,42,0,945,2838,
%T 3458,2228,794,132,0,10395,32859,43191,31235,13553,3473,429,0,135135,
%U 445767,622259,489899,241225,76417,14893,1430
%N Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,2,3,4,5,6,...] DELTA [1,1,1,1,1,1,1,1,...] where DELTA is the operator defined in A084938.
%C This triangle enumerates fixed-point-free involutions in S_n by number of left-to-right maxima. For instance there are 15 fixed point free involutions on 6 elements: 3 have 1 left to right maxima, namely (1,6)(2,3)(4,5), (1,6)(2,4)(3,5) and (3,6)(2,5)(3,4); 7 have 2 left-to right maxima and 5 have 3 left to right maxima. - Robert Cori (rcori(AT)cs.brown.edu), Apr 25 2008
%C A053979*A130595 as infinite lower triangular matrices. - _Philippe Deléham_, Jan 06 2012
%F Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000108(n), A001147(n), A128709(n) for x = -1,0,1,2 respectively.
%F From _Peter Bala_, Dec 22 2011: (Start)
%F The o.g.f. in the form G(x,t) = x/(1 - t*x^2/(1 - (t+1)*x^2/(1 - (t+2)*x^2/(1 - (t+3)*x^2/(1 - ... ))))) = x + t*x^3 + (t+2*t^2)*x^5 + ... satisfies the Riccati differential equation (1 - (t-1)*x*G)*G = x + x^3*dG/dx. Writing G(x,t) = sum {n = 1..inf} R(n,t)*x^(2*n-1), the row generating polynomials R(n,t) satisfy the recurrence R(n+1,t) = (2*n-1)*R(n,t) + (t-1)*sum {k = 1..n} R(k,t)*R(n+1-k,t) with initial condition R(1,t) = 1.
%F G(x,t+1) = x + (1+t)*x^3 + (3+5*t+2*t^2)*x^5 + ... is an o.g.f. for A053979.
%F (End)
%e Triangle begins:
%e 1;
%e 0, 1;
%e 0, 1, 2;
%e 0, 3, 7, 5;
%e 0, 15, 39, 37, 14;
%e 0, 105, 296, 326, 176, 42;
%e 0, 945, 2838, 3458, 2228, 794, 132;
%e 0, 10395, 32859, 43191, 31235, 13553, 3473, 429;
%e 0, 135135, 445767, 622259, 489899, 241225, 76417, 14893, 1430;
%t nmax = 8;
%t DELTA[r_, s_, m_] := Module[{p, q, t, x, y}, q[k_] := x r[[k + 1]] + y s[[k + 1]]; p[0, _] = 1; p[_, -1] = 0; p[n_ /; n >= 1, k_ /; k >= 0] := p[n, k] = p[n, k-1] + q[k] p[n-1, k+1] // Expand; t[n_, k_] := Coefficient[ p[n, 0], x^(n - k) y^k]; t[0, 0] = p[0, 0]; Table[t[n, k], {n, 0, m}, {k, 0, n}]];
%t DELTA[Range[0, nmax], Table[1, {nmax+1}], nmax] // Flatten (* _Jean-François Alcover_, Nov 12 2019 *)
%Y A053979.
%K nonn,tabl
%O 0,6
%A _Philippe Deléham_, Mar 25 2007
%E Corrected and extended by _Peter Bala_, Dec 20 2011