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, 44361286907595736, 304056783818718321, 2084036199823432510

Offset

0,2

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

Arthur T. Benjamin and Jennifer. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 26.

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1190

Ke Feng, The Effects of Costly Information Processing on Shareholder Voting, Ph. D. dissertation, Univ. Alberta (Canada 2025). See p. 25.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Formula

a(n) = A003482(n) + 1.

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

Sum_{n>=0} 1/a(n) = (A094214 + A290565)/2. - Amiram Eldar, Dec 11 2025

E.g.f.: exp(x)*(1 + 2*exp(5*x/2)*(2*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2)))/5. - Stefano Spezia, Dec 12 2025

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 *)

Prog

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

Crossrefs

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

Partial sums of A033889.

Bisection of A001654.

Cf. A003482, A089508, A094214, A145995, A178898, A290565.

Sequence in context: A289208 A244253 A123357 * A390456 A388138 A083426

Adjacent sequences: A081013 A081014 A081015 * A081017 A081018 A081019

Keyword

nonn,easy

Author

R. K. Guy, Mar 01 2003

Status

approved