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%I #34 May 05 2019 07:41:03
%S 1,0,1,0,1,1,0,2,2,1,0,4,6,3,1,0,9,20,12,4,1,0,21,72,54,20,5,1,0,51,
%T 272,261,112,30,6,1,0,127,1064,1323,672,200,42,7,1,0,323,4272,6939,
%U 4224,1425,324,56,8,1,0,835,17504,37341,27456,10625,2664,490,72,9,1
%N A square array of Motzkin related transforms, read by antidiagonals.
%C Rows are transforms of k^n, k>=0, under the matrix A107131. As a number triangle, with T(n,k)=if(k<=n,sum{j=0..n-k, (1/(j+1))C(j+1,n-k-j+1)C(n-k,j)k^j},0), row sums are A107268 and diagonal sums are A107269. Rows are series reversions of x/(1+kx+kx^2), k>=0. Conjecture: rows count weighted Motzkin paths.
%C Row k counts colored Motzkin paths, where H(1,0) and U(1,1) each have k colors and D(1,-1) one color. - _Paul Barry_, May 16 2005
%H Seiichi Manyama, <a href="/A107267/b107267.txt">Antidiagonals n = 0..139, flattened</a>
%F Number array T(n,k) = Sum_{j=0..k} n^j * binomial(k,j) * binomial(j+1,k-j+1)/(j+1).
%F G.f. of row k: 1/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - ...))))), a continued fraction. - _Ilya Gutkovskiy_, Sep 21 2017
%F From _Seiichi Manyama_, May 05 2019: (Start)
%F T(n,k) = Sum_{j=0..floor(k/2)} n^(k-j) * binomial(k,2*j) * binomial(2*j,j)/(j+1) = Sum_{j=0..floor(k/2)} n^(k-j) * binomial(k,2*j) * A000108(j).
%F (k+2) * T(n,k) = n * (2*k+1) * T(n,k-1) - n * (n-4) * (k-1) * T(n,k-2). (End)
%e Array begins
%e 1, 0, 0, 0, 0, 0, 0, ...
%e 1, 1, 2, 4, 9, 21, 51, ...
%e 1, 2, 6, 20, 72, 272, 1064, ...
%e 1, 3, 12, 54, 261, 1323, 6939, ...
%e 1, 4, 20, 112, 672, 4224, 27456, ...
%e 1, 5, 30, 200, 1425, 10625, 81875, ...
%e 1, 6, 42, 324, 2664, 22896, 203256, ...
%Y Rows n=0..6 give A000007, A001006, A071356, A107264, A003645, A107265, A107266.
%Y Main diagonal gives A292716.
%Y Cf. A000108.
%K easy,nonn,tabl
%O 0,8
%A _Paul Barry_, May 15 2005
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