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Triangle of coefficients of polynomials u(n,x) jointly generated with A208337; see the Formula section.
5

%I #15 Jun 26 2023 07:43:07

%S 1,1,1,1,2,2,1,3,5,3,1,4,9,10,5,1,5,14,22,20,8,1,6,20,40,51,38,13,1,7,

%T 27,65,105,111,71,21,1,8,35,98,190,256,233,130,34,1,9,44,140,315,511,

%U 594,474,235,55,1,10,54,192,490,924,1295,1324,942,420,89,1,11

%N Triangle of coefficients of polynomials u(n,x) jointly generated with A208337; see the Formula section.

%C coef. of x^(n-1) in u(n,x): A000045(n), Fibonacci numbers

%C coef. of x^(n-1) in v(n,x): A000045(n+1)

%C row sums, u(n,1): A000129

%C row sums, v(n,1): A001333

%C alternating row sums, u(n,-1): 1,0,1,0,1,0,1,0,1,0,...

%C alternating row sums, v(n,-1): 1,-1,1,-1,1,-1,1,-1,...

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

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

%F where u(1,x)=1, v(1,x)=1.

%F T(n,k) = A038137(n-1,k). - _Philippe Deléham_, Apr 05 2012

%e First five rows:

%e 1

%e 1...1

%e 1...2...2

%e 1...3...5...3

%e 1...4...9...10...5

%e First five polynomials u(n,x):

%e 1

%e 1 + x

%e 1 + 2x + 2x^2

%e 1 + 3x + 5x^2 + 3x^3

%e 1 + 4x + 9x^2 + 10x^3 + 5x^4

%t u[1, x_] := 1; v[1, x_] := 1; z = 13;

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

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

%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[%] (* A208336 *)

%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[%] (* A208337 *)

%t Table[u[n, x] /. x -> 1, {n, 1, z}] (* u row sums *)

%t Table[v[n, x] /. x -> 1, {n, 1, z}] (* v row sums *)

%t Table[u[n, x] /. x -> -1, {n, 1, z}](* u alt. row sums *)

%t Table[v[n, x] /. x -> -1, {n, 1, z}](* v alt. row sums *)

%Y Apart from offsets the same as A038137.

%Y Cf. A208337.

%K nonn,tabl

%O 1,5

%A _Clark Kimberling_, Feb 26 2012