A085695 -id:A085695 - cp's OEIS frontend

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

Artur Jasinski, Dec 20 2008

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

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).

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

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

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 *)

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

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

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

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

A215465 a(n) = (Lucas(4n) - Lucas(2n))/4.

Original entry on oeis.org

0, 1, 10, 76, 540, 3751, 25840, 177451, 1217160, 8344876, 57202750, 392089501, 2687463360, 18420257701, 126254611990, 865362736876, 5931286406640, 40653646980451, 278644255208560, 1909856172864751, 13090349042248500

Offset: 0

R. J. Mathar, Aug 11 2012

This is a divisibility sequence, that is, if n | m then a(n) | a(m). However, it is not a strong divisibility sequence. It is the case k = 3 of a 1-parameter family of 4th-order linear divisibility sequences with o.g.f. x*(1 - x^2)/( (1 - k*x + x^2)*(1 - (k^2 - 2)*x + x^2) ). Compare with A000290 (case k = 2) and A085695 (case k = -3). - Peter Bala, Jan 17 2014

In general, for distinct integers r and s with r = s (mod 2), the sequence Lucas(r*n) - Lucas(s*n) is a fourth-order divisibility sequence. See A273622 for the case r = 3, s = 1. - Peter Bala, May 27 2016

			a(3) = 76 because the 12th (4 * 3rd) Lucas number is 22, the 6th (2 * 3rd) Lucas number is 18, and (322 - 18)/4 = 304/4 = 76.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Peter Bala, Lucas sequences and divisibility sequences
E. L. Roettger and H. C. Williams, Appearance of Primes in Fourth-Order Odd Divisibility Sequences, J. Int. Seq., Vol. 24 (2021), Article 21.7.5.
Hugh Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory vol. 7 (5) (2011) 1255-1277
Index entries for linear recurrences with constant coefficients, signature (10,-23,10,-1).

Cf. A085695, A215466, A273622.

[(Lucas(4*n) - Lucas(2*n))/4: n in [0..20]]; // Vincenzo Librandi, Dec 23 2012

A215465 := proc(n)
    (A000032(4*n)-A000032(2*n))/4 ;
end proc:

Table[(LucasL[4n] - LucasL[2n])/4, {n, 0, 19}] (* Alonso del Arte, Aug 11 2012 *)
CoefficientList[Series[-x*(x-1)*(1+x)/((x^2 - 7*x + 1)* (x^2 - 3*x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 23 2012 *)
LinearRecurrence[{10,-23,10,-1},{0,1,10,76},50] (* G. C. Greubel, Jun 02 2016 *)

{a(n) = my(w = quadgen(5)^(2*n)); (2*real(w^2-w) + imag(w^2-w))/4}; /* Michael Somos, Dec 29 2022 */

4*a(n) = A000032(4*n) - A000032(2*n).
a(n) = A056854(n)/4 - A005248(n)/4.
G.f.: -x*(x-1)*(1+x) / ( (x^2-7*x+1)*(x^2-3*x+1) ).
a(n) = 10*a(n-1) - 23*a(n-2) + 10*a(n-3) - a(n-4). - G. C. Greubel, Jun 02 2016
a(n) = 2^(-2-n)*((7-3*sqrt(5))^n-(3-sqrt(5))^n-(3+sqrt(5))^n+(7+3*sqrt(5))^n). - Colin Barker, Jun 02 2016
a(n) = a(-n) for all n in Z. - Michael Somos, Dec 29 2022

cp's OEIS Frontend

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

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A215465 a(n) = (Lucas(4n) - Lucas(2n))/4.

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