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Search: id:A134410
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| A134410 |
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Second-order Lucas numbers; a(n) := (2n+3)*Lucas(n) - n*Lucas(n-1). |
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+0
2
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| 6, 3, 19, 27, 61, 108, 204, 367, 661, 1173, 2069, 3622, 6306, 10923, 18839, 32367, 55421, 94608, 161064, 273527, 463481, 783753, 1322869, 2229002, 3749886, 6299283, 10567579, 17705667, 29630461, 49532148, 82715844, 137997247
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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This sequence is defined by analogy with the sequence of second-order Fibonacci numbers A010049(n) = ((2n+3)*Fib(n)-n*Fib(n-1))/5.
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FORMULA
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Defining equation a(n) := (2n+3)*Lucas(n) - n*Lucas(n-1). Recurrence: a(0) = 6, a(1) = 3, a(n+2) = a(n+1) + a(n) + 5*Lucas(n). O.g.f.: (2-x)*(3-3x+2x^2)/(1-x-x^2)^2. Set A(n) = (a(n-1) + a(n+1))/5, B(n) = a(n+1) - a(n-1). Then A(n+2) = A(n+1) + A(n) + 5*Fibonacci(n) and B(n+2) = B(n+1) + B(n) + 5*Lucas(n). The polynomials L_2(n,-x) = sum {k = 0..n} C(n,k)*a(n-k)*(-x)^k appear to satisfy a Riemann hypothesis; their zeros appear to lie on the vertical line Re x = 1/2 in the complex plane. Compare with the polynomials L(n,-x) defined in A132148.
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CROSSREFS
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Cf. A000032, A010049, A132148.
Adjacent sequences: A134407 A134408 A134409 this_sequence A134411 A134412 A134413
Sequence in context: A067990 A050008 A019069 this_sequence A123153 A088697 A039631
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KEYWORD
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easy,nonn
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AUTHOR
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Peter Bala (pbala(AT)toucansurf.com), Oct 24 2007
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