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A081016
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a(n) = (Lucas(4*n+3) + 1)/5, or Fibonacci(2*n+1)*Fibonacci(2*n+2), or A081015(n)/5.
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16
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1, 6, 40, 273, 1870, 12816, 87841, 602070, 4126648, 28284465, 193864606, 1328767776, 9107509825, 62423800998, 427859097160, 2932589879121, 20100270056686, 137769300517680, 944284833567073, 6472224534451830
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OFFSET
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0,2
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COMMENTS
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a(n-1) is, together with b(n) := A089508(n), n >= 1, the solution to a binomial problem; see A089508.
Numbers k such that 1 - 2*k + 5*k^2 is a square. - Artur Jasinski, Oct 26 2008
Also solution y of Diophantine equation x^2 + 4*y^2 = h^2 for which x = y-1. - Carmine Suriano, Jun 23 2010
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REFERENCES
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Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 26.
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LINKS
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FORMULA
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a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: (1 - 2*x)/((1 - x)*(1 - 7*x + x^2)).
a(n) = F(1) + F(5) + F(9) +...+ F(4*n+1) = F(2*n)*F(2*n+3) + 1, where F(j) = Fibonacci(j).
a(n) = 7*a(n-1) - a(n-2) - 1, n >= 2. - R. J. Mathar, Nov 07 2015
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MAPLE
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luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d, `, (luc(4*n+3)+1)/5) od: # James A. Sellers, Mar 03 2003
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MATHEMATICA
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LinearRecurrence[{8, -8, 1}, {1, 6, 40}, 30] (* Bruno Berselli, Aug 31 2017 *)
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PROG
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(PARI) first(n) = Vec((1-2*x)/((1-x)*(1-7*x+x^2)) + O(x^n)) \\ Iain Fox, Dec 19 2017
(Magma) [(Lucas(4*n+3) +1)/5: n in [0..30]]; // G. C. Greubel, Dec 18 2017
(Sage) [(lucas_number2(4*n+3, 1, -1) +1)/5 for n in (0..30)] # G. C. Greubel, Jul 13 2019
(GAP) List([0..30], n-> (Lucas(1, -1, 4*n+3)[2] +1)/5 ); # G. C. Greubel, Jul 13 2019
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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