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%I #40 Sep 08 2022 08:46:07
%S 0,1,14,211,3164,47461,711914,10678711,160180664,2402709961,
%T 36040649414,540609741211,8109146118164,121637191772461,
%U 1824557876586914,27368368148803711,410525522232055664,6157882833480834961,92368242502212524414,1385523637533187866211
%N a(n) = (15^n - (-1)^n)/16.
%C Let k and t be positive integers and consider a(n) = k*a(n-1)+t*a(n-2) for n>=2, with a(0)=0, a(1)=1.
%C The roots of its characteristic equation are r1 = (k+sqrt(k^2+4t))/2 and r2 =(k-sqrt(k^2+4t))/2. Hence, the solution to the recurrence relation is the sequence {a(n)} where a(n) = alpha1*r1^n + alpha2*r2^n. It can be shown that alpha1 = 1/sqrt(k^2+4t) and alpha2 = -alpha1. It can be shown also that |r2/r1|< 1. Thus, the ratio a(n+1)/a(n) converges to r as n approaches infinity.
%C Note that limit a(n+1)/a(n) = 15 as n approaches infinity with k=14 and t=15.
%C If n > 15 then | a(n+1)/a(n) - 15 | < 10^(-16).
%H G. C. Greubel, <a href="/A239284/b239284.txt">Table of n, a(n) for n = 0..849</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (14,15).
%F G.f.: x/(1 -14*x - 15*x^2).
%F a(n) = 14*a(n-1) + 15*a(n-2) for n > 1, a(0)=0, a(1)=1.
%F a(n) = (1/16)*(15^n - (-1)^n).
%F a(n) = (1/16)*( A001024(n) - A033999(n) ).
%F E.g.f.: (exp(15*x) - exp(-x))/16. - _G. C. Greubel_, May 26 2018
%t CoefficientList[Series[x/(1-14*x-15*x^2), {x,0,50}], x] (* or *) Table[ (15^n - (-1)^n)/16, {n,0,30}] (* or *) LinearRecurrence[{14,15}, {0,1}, 30] (* _G. C. Greubel_, May 26 2018 *)
%o (PARI) a(n) = (15^n - (-1)^n)/16; \\ _Michel Marcus_, Mar 16 2014
%o (PARI) x='x+O('x^30); concat([0], Vec(x/(1 -14*x - 15*x^2))) \\ _G. C. Greubel_, May 26 2018
%o (Magma) [(15^n - (-1)^n)/16: n in [0..30]]; // _G. C. Greubel_, May 26 2018
%K nonn,easy
%O 0,3
%A _Felix P. Muga II_, Mar 14 2014
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