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A132148
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Triangular array T(n,k) = C(n,k)*Lucas(n-k), 0 <= k <= n.
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3
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2, 1, 2, 3, 2, 2, 4, 9, 3, 2, 7, 16, 18, 4, 2, 11, 35, 40, 30, 5, 2, 18, 66, 105, 80, 45, 6, 2, 29, 126, 231, 245, 140, 63, 7, 2, 47, 232, 504, 616, 490, 224, 84, 8, 2, 76, 423, 1044, 1512, 1386, 882, 336, 108, 9, 2, 123, 760, 2115, 3480, 3780, 2772, 1470, 480, 135, 10, 2
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OFFSET
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0,1
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COMMENTS
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The row polynomials L(n,x) = sum {k = 0 .. n} C(n,k)*Lucas(n-k)*x^k satisfy L(n,x)* F(n,x) = F(2n,x), where F(n,x) = sum {k = 0 .. n} C(n,k)*Fibonacci(n-k)*x^k.
Other identities and formulas include: L(n+1,x)^2 - L(n,x)*L(n+2,x) = -5*(x^2 + x - 1)^n; L(n+1,x) - (x^2 + x - 1)*L(n-1,x) = 5*F(n,x) for n >= 1; L(2n,x) - 2*(x^2 + x - 1)^n = 5*F(n,x)^2; L(n,2x) = sum { k = 0 .. n} C(n,k)*L(n-k,x)*x^k; L(n,3x) = sum { k = 0 .. n} C(n,k)*L(n-k,2x)*x^k etc;
Sum {k = 0 .. n} C(n,k)*L(k,x)*F(n-k,x) = 2^n F(n,x); Row sums: L(n,1) = Lucas(2n); Alternating row sums: L(n,-1) = (-1)^n Lucas(n); L(n,1/phi) = (-1)^n L(n,-phi) = sqrt(5)^n for n >= 1, where phi = (1+sqrt(5))/2.
The polynomials L(n,-x) satisfy a Riemann hypothesis: the zeros of L(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
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LINKS
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Table of n, a(n) for n=0..65.
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FORMULA
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G.f.: (2 - (2x + 1)*t)/(1 - (2x + 1)*t + (x^2 + x - 1)*t^2) = 2 + (1 + 2x)*t + (3 + 2x + 2x^2)*t^2 + (4 + 9x + 3x^2 + 2x^3)*t^3 + ... .
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EXAMPLE
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Triangle starts
2;
1, 2;
3, 2, 2;
4, 9, 3, 2;
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MAPLE
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with(combinat): lucas := n -> fibonacci(n-1) + fibonacci(n+1): T := (n, k) -> binomial(n, k)*lucas(n-k): for n from 0 to 10 do seq( T(n, k), k = 0..n) od;
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MATHEMATICA
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Flatten[Table[Binomial[n, k]LucasL[n-k], {n, 0, 10}, {k, 0, n}]] (* From Harvey P. Dale, Nov 06 2011 *)
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CROSSREFS
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Cf. A000032, A000045, A094440.
Sequence in context: A111725 A112218 A172366 * A159974 A143866 A155002
Adjacent sequences: A132145 A132146 A132147 * A132149 A132150 A132151
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KEYWORD
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easy,nonn,tabl
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AUTHOR
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Peter Bala, Aug 17 2007
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STATUS
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approved
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