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Lucas numbers L(8*n).
10

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

%S 2,47,2207,103682,4870847,228826127,10749957122,505019158607,

%T 23725150497407,1114577054219522,52361396397820127,

%U 2459871053643326447,115561578124838522882,5428934300813767249007,255044350560122222180447

%N Lucas numbers L(8*n).

%C a(n+1)/a(n) converges to (47+sqrt(2205))/2 = 46.9787137... a(0)/a(1)=2/47; a(1)/a(2)=47/2207; a(2)/a(3)=2207/103682; a(3)/a(4)=103682/4870847; etc. Lim_{n->infinity} a(n)/a(n+1) = 0.02128623625... = 2/(47+sqrt(2205)) = (47-sqrt(2205))/2.

%C a(n) = a(-n). - _Alois P. Heinz_, Aug 07 2008

%C From _Peter Bala_, Oct 14 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^8) = 1.0212763906... = 1 + 1/(47 + 1/(2207 + 1/(103682 + ...))).

%C Also F(-Phi^8) = 0.9787231991... has the continued fraction representation 1 - 1/(47 - 1/(2207 - 1/(103682 - ...))) and the simple continued fraction expansion 1/(1 + 1/((47 - 2) + 1/(1 + 1/((2207 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + ...))))))).

%C F(Phi^8)*F(-Phi^8) = 0.9995468962... has the simple continued fraction expansion 1/(1 + 1/((47^2 - 4) + 1/(1 + 1/((2207^2 - 4) + 1/(1 + 1/((103682^2 - 4) + 1/(1 + ...))))))).

%C 1/2 + 1/2*F(Phi^8)/F(-Phi^8) = 1.0217391349... has the simple continued fraction expansion 1 + 1/((47 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + 1/(228826127 - 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 Indranil Ghosh, <a href="/A087265/b087265.txt">Table of n, a(n) for n = 0..596</a>

%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 (47,-1).

%F a(n) = 47*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 47.

%F a(n) = ((47+sqrt(2205))/2)^n + ((47-sqrt(2205))/2)^n

%F (a(n))^2 = a(2n)+2.

%F G.f.: (2-47*x)/(1-47*x+x^2). - _Alois P. Heinz_, Aug 07 2008

%F From _Peter Bala_, Oct 14 2019: (Start)

%F a(n) = F(8*n+8)/F(8) - F(8*n-8)/F(8) = A049668(n+1) - A049668(n-1).

%F a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^8 = [13, 21; 21, 34].

%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 45*Sum_{n >= 1} 1/(a(n) - 49/a(n)) = 1: (49 = Lucas(8) + 2 and 45 = Lucas(8) - 2)

%F 49*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 45/a(n)) = 1.

%F x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 47*x^2 + 2208*x^3 + ... is the o.g.f. for A049668. (End)

%F E.g.f.: 2*exp(47*x/2)*cosh(21*sqrt(5)*x/2). - _Stefano Spezia_, Oct 18 2019

%e a(4) = 4870847 = 47*a(3) - a(2) = 47*103682 - 2207=((47+sqrt(2205))/2)^4 + ( (47-sqrt(2205))/2)^4 =4870846.999999794696 + 0.000000205303 = 4870847.

%p a:= n-> (Matrix([[2,47]]). Matrix([[47,1],[ -1,0]])^(n))[1,1]:

%p seq(a(n), n=0..14); # _Alois P. Heinz_, Aug 07 2008

%t LucasL[8*Range[0,20]] (* or *) LinearRecurrence[{47,-1},{2,47},20] (* _Harvey P. Dale_, Oct 23 2017 *)

%o (Magma) [ Lucas(8*n) : n in [0..100]]; // _Vincenzo Librandi_, Apr 14 2011

%Y Cf. A000032. Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

%Y a(n) = A000032(8n).

%K easy,nonn

%O 0,1

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

%E Terms a(22)-a(27) from _John W. Layman_, Jun 14 2004