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%I #39 Mar 04 2024 09:01:37
%S 2,123,15127,1860498,228826127,28143753123,3461452808002,
%T 425730551631123,52361396397820127,6440026026380244498,
%U 792070839848372253127,97418273275323406890123,11981655542024930675232002,1473646213395791149646646123,181246502592140286475862241127
%N a(n) = Lucas(10*n).
%C Lim_{n->infinity} a(n+1)/a(n) = (123 + sqrt(15125))/2 = 122.9918693812...
%C Lim_{n->infinity} a(n)/a(n+1) = (123 - sqrt(15125))/2 = 0.00813061875578...
%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^10) = 1.0081300769... = 1 + 1/(123 + 1/(15127 + 1/(1860498 + ...))).
%C Also F(-Phi^10) = 0.9918699143... has the continued fraction representation 1 - 1/(123 - 1/(15127 - 1/(1860498 - ...))) and the simple continued fraction expansion 1/(1 + 1/((123 - 2) + 1/(1 + 1/((15127 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + ...))))))).
%C F(Phi^10)*F(-Phi^10) = 0.9999338930... has the simple continued fraction expansion 1/(1 + 1/((123^2 - 4) + 1/(1 + 1/((15127^2 - 4) + 1/(1 + 1/((1860498^2 - 4) + 1/(1 + ...))))))).
%C 1/2 + (1/2)*F(Phi^10)/F(-Phi^10) = 1.0081967213... has the simple continued fraction expansion 1 + 1/((123 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + 1/(28143753123 - 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="/A065705/b065705.txt">Table of n, a(n) for n = 0..477</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 (123,-1).
%F a(n) = 123*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 123.
%F a(n) = ((123 + sqrt(15125))/2)^n + ((123 - sqrt(15125))/2)^n.
%F a(n)^2 = a(2*n) + 2.
%F G.f.: (2 - 123*x)/(1 - 123*x + x^2). - _Philippe Deléham_, Nov 18 2008
%F From _Peter Bala_, Oct 14 2019: (Start)
%F a(n) = F(10*n+10)/F(10) - F(10*n-10)/F(10) = A049670(n+1) - A049670(n-1).
%F a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^10 = [34, 55; 55, 89].
%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 121*Sum_{n >= 1} 1/(a(n) - 125/a(n)) = 1: (125 = Lucas(10) + 2 and 121 = Lucas(10) - 2)
%F 125*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 121/a(n)) = 1.
%F x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 123*x^2 + 15128*x^3 + ... is the o.g.f. for A049670. (End)
%F E.g.f.: exp((1/2)*(123 - 55*sqrt(5))*x)*(1 + exp(55*sqrt(5)*x). - _Stefano Spezia_, Oct 18 2019
%e a(4) = 228826127 = 123*a(3) - a(2) = 123*1860498 - 15127=((123+sqrt(15125))/2)^4 + ( (123-sqrt(15125))/2)^4 =228826126.99999999562986 + 0.00000000437013 = 228826127.
%e a(4) = L(10 * 4) = L(40) = 228826127. - _Indranil Ghosh_, Feb 08 2017
%t LucasL[10*Range[0, 20]] (* _Paolo Xausa_, Mar 04 2024 *)
%o (Magma) [Lucas(10*n): n in [0..90]]; // _Vincenzo Librandi_, Apr 14 2011
%Y Cf. A000032: a(n) = A000032(10*n).
%Y Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).
%K nonn,easy
%O 0,1
%A Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 25 2003
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