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%I #18 Mar 11 2024 08:29:25
%S 1,2,1,5,4,1,14,15,6,1,41,56,30,8,1,122,205,140,50,10,1,365,732,615,
%T 280,75,12,1,1094,2555,2562,1435,490,105,14,1,3281,8752,10220,6832,
%U 2870,784,140,16,1,9842,29529,39384,30660,15372,5166,1176,180,18,1,29525
%N Triangle given by p(n,k)=(coefficient of x^(n-k) in (1/2) ((x+3)^n+(x+1)^n)), 0<=k<=n.
%F From _Mélika Tebni_, Dec 09 2023: (Start)
%F T(n,k) = binomial(n,k)*(1 + 3^(n-k)) / 2.
%F E.g.f. of column k: exp(2*x)*cosh(x)*x^k / k!. (End)
%F From _Peter Bala_, Mar 07 2024: (Start)
%F Exponential Riordan array (exp(2*x)*cosh(x), x).
%F The zeros of the n-th row polynomial R(n,x) = ((1 + x)^n + (3 + x)^n)/2 lie on the vertical line Re(x) = -2 in the complex plane.
%F Triangle equals P * (I + P^2)/2 = P * A119468 = P^2 * A119467, where P denotes Pascal's triangle A007318. (End)
%e First five rows:
%e 1
%e 2 1
%e 5 4 1
%e 14 15 6 1
%e 41 56 30 8 1
%t q[n_, k_] := 1; r[0] = 1;
%t r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}]
%t p[n_, k_] := Coefficient[(1/2) ((x + 3)^n + (x + 1)^n), x, k] (* A193673 *)
%t v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
%t Table[v[n], {n, 0, 20}] (* A193661 *)
%t TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
%t Table[r[k], {k, 0, 8}] (* 2^k *)
%t TableForm[Table[p[n, k], {n, 0, 10}, {k, 0, n}]] (* A193673 as a triangle *)
%t Flatten[%] (* A193673 as a sequence *)
%Y Cf. A119467, A119468, A193661.
%K nonn,tabl,easy
%O 0,2
%A _Clark Kimberling_, Aug 02 2011
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