A087215 Lucas(6*n): a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

2, 18, 322, 5778, 103682, 1860498, 33385282, 599074578, 10749957122, 192900153618, 3461452808002, 62113250390418, 1114577054219522, 20000273725560978, 358890350005878082, 6440026026380244498, 115561578124838522882, 2073668380220713167378

Offset: 0

Author

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 19 2003

Keywords

• nonn
• easy

Comments

a(n+1)/a(n) converges to 9 + sqrt(80) = 17.9442719... a(0)/a(1) = 2/18; a(1)/a(2) = 18/322; a(2)/a(3) = 322/5778; a(3)/a(4) = 5778/103682; etc.

Lim_{n -> oo} a(n)/a(n+1) = 0.05572809000084... = 1/(9 + sqrt(80)) = 9 - sqrt(80).

From Peter Bala, Oct 13 2019: (Start)

Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = (1/2)*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^6) = 1.0555459720... = 1 + 1/(18 + 1/(322 + 1/(5778 + ...))).

Also F(-Phi^6) = 0.9444348576... has the continued fraction representation 1 - 1/(18 - 1/(322 - 1/(5788 - ...))) and the simple continued fraction expansion 1/(1 + 1/((18 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/((5788 - 2) + 1/(1 + ...))))))).

F(Phi^6)*F(-Phi^6) = 0.9968944099... has the simple continued fraction expansion 1/(1 + 1/((18^2 - 4) + 1/(1 + 1/((322^2 - 4) + 1/(1 + 1/((5788^2 - 4) + 1/(1 + ...))))))).

1/2 + (1/2)*F(Phi^6)/F(-Phi^6) = 1.0588241282... has the simple continued fraction expansion 1 + 1/((18 - 2) + 1/(1 + 1/((5778 - 2) + 1/(1 + 1/(1860498 - 2) + 1/(1 + ...))))). (End)

Examples

			a(4) = 103682 = 18*a(3) - a(2) = 18*5778 - 322 = (9 + sqrt(80))^4 + (9 - sqrt(80))^4 = 103681.99999035512... + 0.00000964487... = 103682.
		

References

• J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
• R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Links

• Colin Barker, Table of n, a(n) for n = 0..750
• Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
• P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence R_4.
• Tanya Khovanova, Recursive Sequences
• A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
• Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
• Index entries for linear recurrences with constant coefficients, signature (18,-1).

Crossrefs

Cf. A074919.

Row 2 * 2 of array A188645.

Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

Programs

Magma

[ Lucas(6*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

Mathematica

a[0] = 2; a[1] = 18; a[n_] := 18a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
Table[LucasL[6n], {n, 0, 18}]  (* or *) CoefficientList[Series[2*(1 - 9*x)/(1 - 18*x + x^2), {x, 0, 17}], x] (* Indranil Ghosh, Mar 15 2017 *)

PARI

Vec(2*(1-9*x)/(1-18*x+x^2) + O(x^20)) \\ Colin Barker, Jan 24 2016

PARI

a(n) = if(n<2, 17^n + 1, 18*a(n - 1) - a(n - 2));
for(n=0, 17, print1(a(n),", ")) \\ Indranil Ghosh, Mar 15 2017

Formula

a(n) = A000032(6*n).

a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

a(n) = (9 + sqrt(80))^n + (9 - sqrt(80))^n.

G.f.: 2*(1-9*x)/(1-18*x+x^2). - Philippe Deléham, Nov 17 2008

a(n) = 2*A023039(n). - R. J. Mathar, Oct 22 2010

From Peter Bala, Oct 13 2019: (Start)

a(n) = F(6*n+6)/F(6) - F(6*n-6)/F(6) = A049660(n+1) - A049660(n-1).

a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^6 = [5, 8; 8, 13].

Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).

16*Sum_{n >= 1} 1/(a(n) - 20/a(n)) = 1: (20 = Lucas(6) + 2 and 16 = Lucas(6) - 2)

20*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 16/a(n)) = 1.

Series acceleration formulas for sum of reciprocals:

Sum_{n >= 1} 1/a(n) = 1/16 - 20*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 20)).

Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/20 + 16*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 16)).

Sum_{n >= 1} 1/a(n) = ( (theta_3(9-4*sqrt(5)))^2 - 1 )/4 and

Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(4*sqrt(5)-9))^2 )/4,

where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.

x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 18*x^2 + 323*x^3 + ... is the o.g.f. for A049660. (End)

E.g.f.: 2*exp(9*x)*cosh(4*sqrt(5)*x). - Stefano Spezia, Oct 18 2019

a(n) = L(2n-1)^2 * F(2n+1) + L(2n+1)^2 * F(2n-1), where F(n) = A000045(n) and L(n) = A000032(n). - Diego Rattaggi, Nov 12 2020

From Peter Bala, Apr 16 2025: (Start)

a(n) = Lucas(2*n)^3 - 3*Lucas(2*n) = 2*T(3, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind; more generally, for k >= 0, Lucas(2*k*n) = 2*T(k, Lucas(2*n)/2).

Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3(9 - sqrt(80))^2 - 1) and

Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3(sqrt(80) - 9)^2), where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)