A081016 a(n) = (Lucas(4*n+3) + 1)/5, or Fibonacci(2*n+1)*Fibonacci(2*n+2), or A081015(n)/5.

1, 6, 40, 273, 1870, 12816, 87841, 602070, 4126648, 28284465, 193864606, 1328767776, 9107509825, 62423800998, 427859097160, 2932589879121, 20100270056686, 137769300517680, 944284833567073, 6472224534451830

Offset: 0

Author

R. K. Guy, Mar 01 2003

Keywords

• nonn
• easy

Comments

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

References

• 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.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1190
• Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Crossrefs

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081015.

Partial sums of A033889. Bisection of A001654. Equals A003482 + 1.

Cf. A145995, A178898.

Programs

GAP

List([0..30], n-> (Lucas(1,-1,4*n+3)[2] +1)/5 ); # G. C. Greubel, Jul 13 2019

Magma

[(Lucas(4*n+3) +1)/5: n in [0..30]]; // G. C. Greubel, Dec 18 2017

Maple

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 Sellers, Mar 03 2003

Mathematica

LinearRecurrence[{8,-8,1}, {1,6,40}, 30] (* Bruno Berselli, Aug 31 2017 *)

PARI

a(n)=([0,1,0; 0,0,1; 1,-8,8]^n*[1;6;40])[1,1] \\ Charles R Greathouse IV, Sep 28 2015

PARI

first(n) = Vec((1-2*x)/((1-x)*(1-7*x+x^2)) + O(x^n)) \\ Iain Fox, Dec 19 2017

Sage

[(lucas_number2(4*n+3,1,-1) +1)/5 for n in (0..30)] # G. C. Greubel, Jul 13 2019

Formula

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