%I #21 Jan 23 2020 23:38:34
%S 1,1,1,3,2,1,9,7,3,1,31,24,12,4,1,113,89,46,18,5,1,431,342,183,76,25,
%T 6,1,1697,1355,741,323,115,33,7,1,6847,5492,3054,1376,520,164,42,8,1,
%U 28161,22669,12768,5900,2326,786,224,52,9,1,117631,94962,54033,25464
%N Triangle in A071943 with rows reversed.
%C A convolution triangle based on A052709 (with first term omitted). - _Philippe Deléham_, Sep 15 2005
%F G.f.: (1-q)/(z(2 - t + 2z + tq)), where q = sqrt(1 - 4z - 4z^2). - _Emeric Deutsch_, Jun 06 2005
%F T(0, 0) = 1; T(n, k) = 0 if k < 0 or if k > n; T(n, k) = Sum_{j>=0} T(n-1, k-1+j) + Sum_{j>=0} T(n-1, k+1+j). - _Philippe Deléham_, Sep 15 2005
%F T(n,k) = k*Sum_{i=1..(n-k)} C(i,n-k-i)*C(k+2*i-1,i)/(k+i), n > k, T(n,n)=1. - _Vladimir Kruchinin_, Apr 27 2015
%e 1; 1,1; 3,2,1; 9,7,3,1; 31,24,12,4,1; ...
%p q:=sqrt(1-4*z-4*z^2): G:=(1-q)/z/(2-t+2*z+t*q): Gserz:=simplify(series(G,z=0,14)): P[0]:=1: for n from 1 to 10 do P[n]:=coeff(Gserz,z^n) od: for n from 0 to 10 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields sequence in triangular form # _Emeric Deutsch_, Jun 06 2005
%t T[n_, n_] = 1; T[n_, k_] := (k+1)*Sum[Binomial[i, n-k-i] * Binomial[k+2*i, i] / (k+i+1), {i, 1, n-k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Apr 29 2015, after _Vladimir Kruchinin_ *)
%o (Maxima)
%o T(n,k):=if n=k then 1 else k*sum((binomial(i,n-k-i)*binomial(k+2*i-1,i))/(k+i),i,1,n-k); /* _Vladimir Kruchinin_, Apr 27 2015 */
%Y Row sums yield A071356. Column 0 yields A052709.
%K nonn,tabl,easy
%O 0,4
%A _N. J. A. Sloane_, Jun 05 2005
%E More terms from _Emeric Deutsch_, Jun 06 2005