%I #37 Mar 05 2018 04:56:05
%S 1,1,1,2,3,1,4,8,5,1,8,20,18,7,1,16,48,56,32,9,1,32,112,160,120,50,11,
%T 1,64,256,432,400,220,72,13,1,128,576,1120,1232,840,364,98,15,1,256,
%U 1280,2816,3584,2912,1568,560,128,17,1,512,2816,6912,9984,9408,6048,2688,816,162,19,1
%N Triangle T(n,k), read by rows, given by (1,1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.
%C Riordan array ((1-x)/(1-2x),x/(1-2x)).
%C Product A097805*A007318 as infinite lower triangular arrays.
%C Product A193723*A130595 as infinite lower triangular arrays.
%C T(n,k) is the number of ways to place n unlabeled objects into any number of labeled bins (with at least one object in each bin) and then designate k of the bins. - _Geoffrey Critzer_, Nov 18 2012
%C Apparently, rows of this array are unsigned diagonals of A028297. - _Tom Copeland_, Oct 11 2014
%C Unsigned A118800, so my conjecture above is true. - _Tom Copeland_, Nov 14 2016
%F T(n,k) = 2*T(n-1,k)+T(n-1,k-1) with T(0,0)=T(1,0)=T(1,1)=1 and T(n,k)=0 for k<0 or for n<k.
%F T(n,k) = A011782(n-k)*A135226(n,k) = 2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2.
%F Sum_{k, 0<=k<=n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for n=-1,0,1,2,3,4,5,6,7,8,9,10 respectively.
%F G.f.: (1-x)/(1-(2+y)*x).
%F T(n,k) = Sum_j>=0 T(n-1-j,k-1)*2^j.
%F T = A007318*A059260, so the row polynomials of this entry are given umbrally by p_n(x) = (1 + q.(x))^n, where q_n(x) are the row polynomials of A059260 and (q.(x))^k = q_k(x). Consequently, the e.g.f. is exp[tp.(x)] = exp[t(1+q.(x))] = e^t exp(tq.(x)) = [1 + (x+1)e^((x+2)t)]/(x+2), and p_n(x) = (x+1)(x+2)^(n-1) for n > 0. - _Tom Copeland_, Nov 15 2016
%F T^(-1) = A130595*(padded A130595), differently signed A118801. Cf. A097805. - _Tom Copeland_, Nov 17 2016
%F The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 + x)/(1 + 2*x) * (1 + 2*x)^n about 0. For example, for n = 4, (1 + x)/(1 + 2*x) * (1 + 2*x)^4 = (8*x^4 + 20*x*3 + 18*x^2 + 7*x + 1) + O(x^5). - _Peter Bala_, Feb 24 2018
%e Triangle begins:
%e 1
%e 1, 1
%e 2, 3, 1
%e 4, 8, 5, 1
%e 8, 20, 18, 7, 1
%e 16, 48, 56, 32, 9, 1
%e 32, 112, 160, 120, 50, 11, 1
%t nn=15;f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[(1-x)/(1-2x-y x) ,{x,0,nn}],{x,y}]]//Grid (* _Geoffrey Critzer_, Nov 18 2012 *)
%Y Cf. A118800 (signed version), A081277, A039991, A001333 (antidiagonal sums), A025192 (row sums); diagonals: A000012, A005408, A001105, A002492, A072819l; columns: A011782, A001792, A001793, A001794, A006974, A006975, A006976.
%Y Cf. A007318, A028297, A059260, A097805, A118801, A130595.
%K nonn,tabl,easy
%O 0,4
%A _Philippe Deléham_, Nov 13 2011