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A209573
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Triangle of coefficients of polynomials u(n,x) jointly generated with A209574; see the Formula section.
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3
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1, 1, 1, 2, 4, 1, 3, 9, 9, 1, 4, 17, 29, 16, 1, 5, 28, 69, 74, 25, 1, 6, 42, 138, 224, 160, 36, 1, 7, 59, 245, 541, 613, 307, 49, 1, 8, 79, 399, 1127, 1781, 1469, 539, 64, 1, 9, 102, 609, 2111, 4331, 5103, 3171, 884, 81, 1, 10, 128, 884, 3649, 9281, 14419
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OFFSET
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1,4
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COMMENTS
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For n>1, let r(n,k) be the k-th number in row n. Then
r(n,1)=n-1, r(n,n-1)=(n-1)^2, and r(n,n)=1. For a
discussion and guide to related arrays, see A208510.
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LINKS
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FORMULA
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u(n,x)=x*u(n-1,x)+v(n-1,x),
v(n,x)=2x*u(n-1,x)+(x+1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
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EXAMPLE
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First five rows:
1
1...1
2...4....1
3...9....9....1
4...17...29...16...1
First three polynomials v(n,x): 1, 1 + x, 2 + 4x + x^2.
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MATHEMATICA
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u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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