A087215 - OEIS

%I #81 Sep 08 2022 08:45:11

%S 2,18,322,5778,103682,1860498,33385282,599074578,10749957122,

%T 192900153618,3461452808002,62113250390418,1114577054219522,

%U 20000273725560978,358890350005878082,6440026026380244498,115561578124838522882,2073668380220713167378

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

%C 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.

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

%C From _Peter Bala_, Oct 13 2019: (Start)

%C 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 + ...))).

%C 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 + ...))))))).

%C 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 + ...))))))).

%C 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)

%D R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

%H Colin Barker, <a href="/A087215/b087215.txt">Table of n, a(n) for n = 0..750</a>

%H Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Nemeth/nemeth7.html">Ellipse Chains and Associated Sequences</a>, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.

%H P. Bhadouria, D. Jhala, and B. Singh, <a href="http://dx.doi.org/10.22436/jmcs.08.01.07">Binomial Transforms of the k-Lucas Sequences and its Properties</a>, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence R_4.

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H A. V. Zarelua, <a href="https://doi.org/10.1007/s11006-006-0090-y">On Matrix Analogs of Fermat's Little Theorem</a>, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.

%H <a href="/index/Rea#recur1">Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (18,-1).

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

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

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

%F G.f.: 2*(1-9*x)/(1-18*x+x^2). - _Philippe Deléham_, Nov 17 2008

%F a(n) = 2*A023039(n). - _R. J. Mathar_, Oct 22 2010

%F From _Peter Bala_, Oct 13 2019: (Start)

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

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

%F 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).

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

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

%F Series acceleration formulas for sum of reciprocals:

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

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

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

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

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

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

%F E.g.f.: 2*exp(9*x)*cosh(4*sqrt(5)*x). - _Stefano Spezia_, Oct 18 2019

%F 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

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

%t 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 *)

%t 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 *)

%o (Magma) [ Lucas(6*n) : n in [0..100]]; // _Vincenzo Librandi_, Apr 14 2011

%o (PARI) Vec(2*(1-9*x)/(1-18*x+x^2) + O(x^20)) \\ _Colin Barker_, Jan 24 2016

%o (PARI) a(n) = if(n<2, 17^n + 1, 18*a(n - 1) - a(n - 2));

%o for(n=0, 17, print1(a(n),", ")) \\ _Indranil Ghosh_, Mar 15 2017

%Y Cf. A074919.

%Y Row 2 * 2 of array A188645.

%Y 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).

%K easy,nonn

%O 0,1

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