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A193655 Q-residue of the triangle p(n,k)=floor(1/2+(n+1)/(n+k+2)/2), 0<=k<=n, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

%I A193655 #9 Feb 19 2015 14:48:41
%S A193655 1,7,29,94,280,765,2023,5116,12710,30715,73381,172026,400036,917497,
%T A193655 2091683,4718584,10594978,23592951,52341409,115343350,253405856
%N A193655 Q-residue of the triangle p(n,k)=floor(1/2+(n+1)/(n+k+2)/2), 0<=k<=n, where Q is the triangular array (t(i,j)) given by t(i,j)=1.  (See Comments.)
%C A193655 For the definition of Q-residue, see A193649.
%F A193655 Conjecture: G.f.: ( -1-2*x+3*x^2+9*x^3-8*x^4-4*x^5 ) / ( (1+x)*(2*x+1)*(x-1)^2*(2*x-1)^3 ). - _R. J. Mathar_, Feb 19 2015
%t A193655 q[n_, k_] := 1;
%t A193655 r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}]
%t A193655 p[n_, k_] := Floor[1/2 + (n + 1) (n + k + 2)/2]
%t A193655 v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
%t A193655 Table[v[n], {n, 0, 20}]    (* A193655 *)
%t A193655 TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
%t A193655 Table[r[k], {k, 0, 8}]
%t A193655 TableForm[Table[p[n, k], {n, 0, 4}, {k, 0, n}]]
%Y A193655 Cf. A193649, A193654.
%K A193655 nonn
%O A193655 0,2
%A A193655 _Clark Kimberling_, Aug 02 2011