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%I #27 Jan 26 2020 21:07:35
%S 1,1,1,1,3,2,1,4,6,3,1,5,10,10,4,1,6,15,20,15,5,1,7,21,35,35,21,6,1,8,
%T 28,56,70,56,28,7,1,9,36,84,126,126,84,36,8,1,10,45,120,210,252,210,
%U 120,45,9,1,11,55,165,330,462,462,330,165,55,10,1,12,66,220,495
%N Triangular array: the fusion of polynomial sequences P and Q given by p(n,x) = x^n + x^(n-1) + ... + x+1 and q(n,x)=(x+1)^n.
%C See A193722 for the definition of fusion of two sequences of polynomials or triangular arrays.
%C Triangle T(n,k), read by rows, given by (1,0,-1,1,0,0,0,0,0,0,0,...) DELTA (1,1,-1,1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Oct 08 2011
%C Row sums are A095121. - _Philippe Deléham_, Nov 24 2011
%H Branko Malesevic, Yue Hu, Cristinel Mortici, <a href="https://arxiv.org/abs/1801.04963">Accurate Estimates of (1+x)^{1/x} Involved in Carleman Inequality and Keller Limit</a>, arXiv:1801.04963 [math.CA], 2018.
%F T(n,k) = A153861(n,n-k). - _Philippe Deléham_, Oct 08 2011
%F G.f.: (1-y*x+y*(y+1)*x^2)/((1-y*x)*(1-(y+1)*x)). - _Philippe Deléham_, Nov 24 2011
%F Sum_{k=0..n} T(n,k)*x^k = (x+1)*((x+1)^n - x^n) + 0^n. - _Philippe Deléham_, Nov 24 2011
%F T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0)=T(1,0)=T(1,1)=T(2,0)=1, T(2,1)=3, T(2,2)=2, T(n,k)=0 if k < 0 or if k > n. - _Philippe Deléham_, Dec 15 2013
%e First six rows:
%e 1;
%e 1, 1;
%e 1, 3, 2;
%e 1, 4, 6, 3;
%e 1, 5, 10, 10, 4;
%e 1, 6, 15, 20, 15, 5;
%t z = 10; c = 1; d = 1;
%t p[0, x_] := 1
%t p[n_, x_] := x*p[n - 1, x] + 1; p[n_, 0] := p[n, x] /. x -> 0;
%t q[n_, x_] := (c*x + d)^n
%t t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
%t w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
%t g[n_] := CoefficientList[w[n, x], {x}]
%t TableForm[Table[Reverse[g[n]], {n, -1, z}]]
%t Flatten[Table[Reverse[g[n]], {n, -1, z}]] (* A193815 *)
%t TableForm[Table[g[n], {n, -1, z}]]
%t Flatten[Table[g[n], {n, -1, z}]] (* A153861 *)
%t t[0, 0] = t[1, 0] = t[1, 1] = t[2, 0] = 1; t[2, 1] = 3; t[2, 2] = 2; t[n_, k_] /; k<0 || k>n = 0; t[n_, k_] := t[n, k] = t[n-1, k]+2*t[n-1, k-1]-t[n-2, k-1]-t[n-2, k-2]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Dec 16 2013, after _Philippe Deléham_ *)
%Y Cf. A193722, A153861, A193818.
%K nonn,tabl
%O 0,5
%A _Clark Kimberling_, Aug 06 2011
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