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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.)
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%I #9 Feb 19 2015 14:48:41

%S 1,7,29,94,280,765,2023,5116,12710,30715,73381,172026,400036,917497,

%T 2091683,4718584,10594978,23592951,52341409,115343350,253405856

%N 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 For the definition of Q-residue, see A193649.

%F 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 q[n_, k_] := 1;

%t r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}]

%t p[n_, k_] := Floor[1/2 + (n + 1) (n + k + 2)/2]

%t v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]

%t Table[v[n], {n, 0, 20}] (* A193655 *)

%t TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]

%t Table[r[k], {k, 0, 8}]

%t TableForm[Table[p[n, k], {n, 0, 4}, {k, 0, n}]]

%Y Cf. A193649, A193654.

%K nonn

%O 0,2

%A _Clark Kimberling_, Aug 02 2011