%I #35 Jan 03 2024 06:16:45

%S 1,2,1,3,4,1,4,9,7,1,5,16,22,11,1,6,25,50,46,16,1,7,36,95,130,86,22,1,

%T 8,49,161,295,296,148,29,1,9,64,252,581,791,610,239,37,1,10,81,372,

%U 1036,1792,1897,1163,367,46,1,11,100,525,1716,3612,4900,4166,2083,541,56,1

%N Triangle of coefficients of polynomials u(n,x) jointly generated with A210220; see the Formula section.

%C First two terms in row n: n,(n-1)^2; last term: 1.

%C Period of alternating row sums: (1,1,0).

%C For a discussion and guide to related arrays, see A208510.

%C Apparently this is A071920 without the marginal zeros, read by downwards antidiagonals, or T(n,k) = A071922(n,k). - _R. J. Mathar_, May 17 2014

%H Andrew Howroyd, <a href="/A210219/b210219.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50)

%F u(n,x) = x*u(n-1,x) + v(n-1,x) + 1, v(n,x) = x*u(n-1,x) + (x+1)*v(n-1,x) + 1, where u(1,x)=1, v(1,x)=1.

%F T(n,k) = Sum_{j=0..floor(k/2)} binomial(k,2*j)*binomial(n-j,k). - _Detlef Meya_, Dec 05 2023

%e First five rows:

%e 1

%e 2...1

%e 3...4....1

%e 4...9....7....1

%e 5...16...22...11...1

%e First three polynomials u(n,x): 1, 2 + x, 3 + 4x + x^2.

%t u[1, x_] := 1; v[1, x_] := 1; z = 16;

%t u[n_, x_] := x*u[n - 1, x] + v[n - 1, x] + 1;

%t v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;

%t Table[Expand[u[n, x]], {n, 1, z/2}]

%t Table[Expand[v[n, x]], {n, 1, z/2}]

%t cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];

%t TableForm[cu]

%t Flatten[%] (* A210219 *)

%t Table[Expand[v[n, x]], {n, 1, z}]

%t cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];

%t TableForm[cv]

%t Flatten[%] (* A210220 *)

%t (* alternative program *)

%t T[n_,k_] := Sum[Binomial[k, 2*j]*Binomial[n-j, k], {j, 0, Floor[k/2]}]; Flatten[Table[T[n, k],{n, 1, 11}, {k, 1, n}]] (* _Detlef Meya_, Dec 05 2023 *)

%o (PARI) T(n,k) = sum(j=0, k\2, binomial(k,2*j)*binomial(n-j,k)) \\ _Andrew Howroyd_, Jan 01 2024

%Y Cf. A210220, A208510, A001906 (row sums).

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Mar 19 2012