A153173 a(n) = L(5*n)/L(n) where L(n) = Lucas number A000204(n).

11, 41, 341, 2161, 15251, 103361, 711491, 4868641, 33391061, 228811001, 1568437211, 10749853441, 73681573691, 505018447961, 3461454668501, 23725145626561, 162614613425891, 1114577020834241, 7639424866266611

Offset: 1

Author

Artur Jasinski, Dec 20 2008

Keywords

• nonn
• easy

Comments

All numbers in this sequence are congruent to 1 mod 10.

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• L. Carlitz, Problem B-185, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 8, No. 3 (1970), p. 325; Lucas Ratio I, Solution to Problem B-185 by C. B. A. Peck, ibid., Vol. 9, No. 1 (1971), p. 109.
• L. Carlitz, Problem B-186, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 8, No. 3 (1970), p. 326; Lucas Ratio II, Solution to Problem B-186 by John Wessner, ibid., Vol. 9, No. 1 (1971), pp. 109-110.
• Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Crossrefs

Cf. A000032, A000045, A000204, A005248, A056854, A085695, A110391.

Programs

Magma

I:=[11, 41, 341, 2161, 15251]; [n le 5 select I[n] else 5*Self(n-1)+15*Self(n-2)-15*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..30]]; // G. C. Greubel, Dec 21 2017

Mathematica

Table[LucasL[5*n]/LucasL[n], {n, 1, 50}]
CoefficientList[Series[x*(11-14*x-29*x^2+6*x^3+x^4)/((1-x)*(x^2-7*x+1)*(x^2+3*x+1)), {x,0,50}], x] (* G. C. Greubel, Dec 21 2017 *)
a[ n_] := 1 + 5*Fibonacci[n]*Fibonacci[3*n]; (* Michael Somos, Apr 23 2022 *)

PARI

{L(n)=fibonacci(n-1)+fibonacci(n+1)}; a(n) = L(5*n)/L(n) \\ Charles R Greathouse IV, Jun 11 2015

PARI

my(x='x+O('x^30)); Vec(x*(11-14*x-29*x^2+6*x^3+x^4 )/((1-x)*(x^2-7*x +1)*(x^2+3*x+1))) \\ G. C. Greubel, Dec 21 2017

PARI

{a(n) = 1 + 5*fibonacci(n)*fibonacci(3*n)}; /* Michael Somos, Apr 23 2022 */

Formula

From R. J. Mathar, Oct 22 2010: (Start)

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5).

G.f.: -x*(11-14*x-29*x^2+6*x^3+x^4)/( (x-1)*(x^2-7*x+1)*(x^2+3*x+1) ).

a(n) = 1 + A056854(n) - (-1)^n*A005248(n). (End)

From Amiram Eldar, Feb 02 2022: (Start)

a(n) = Lucas(2*n)^2 - (-1)^n*Lucas(2*n) - 1 (Carlitz, Problem B-185).

a(n) = (Lucas(2*n) - 3*(-1)^n)^2 + (-1)^n*(5*Fibonacci(n))^2 (Carlitz, Problem B-186). (End)

a(n) = a(-n) = 1 + 10*A085695(n) = 5 + L(n-1)*L(n)^2*L(n+1) for all n in Z. - Michael Somos, Apr 23 2022