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%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
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