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%I #41 Apr 29 2019 09:45:07
%S 0,1,12,189,2808,42201,632772,9492309,142382448,2135743281,
%T 32036129532,480542002029,7208129853288,108121948330761,
%U 1621829223367092,24327438355289349,364911575314991328,5473673629767916641,82105104446389609452
%N a(n) = (15^n - (-3)^n)/18.
%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, a(n+1)/a(n) converges to r1 as n approaches infinity.
%C Note that limit a(n+1)/a(n) = 15 with k=12 and t=45. Thus, we use this as the name of the sequence.
%C If n > 25, then |a(n+1)/a(n) - 15| < 10^(-16).
%C a(n+2) is the number of strings of length n containing the 12-ary digits 0,1,...,9,A,B or any of the 45 two-consecutive digits C0,C1,...,C9,CA,...,CZ,Ca,...,Ci where A corresponds to 10, B to 11, ..., Z to 35, a to 36, ..., i to 44.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (12,45).
%F G.f.: x/((1-15*x)*(1+3*x)).
%F a(n) = 12*a(n-1)+45*a(n-2), for n>=2, a(0)=0, a(1)=1.
%F a(n) = (1/18)*(A001024(n) - (-1)^n * A000244(n)).
%F a(n) = A000244(n-1) * Sum_{k=0...n-1} (A000351(k) * A033999(n-1-k)).
%t Table[(15^n-(-3)^n)/18,{n,0,20}] (* or *) LinearRecurrence[{12,45},{0,1},20] (* _Harvey P. Dale_, Apr 29 2019 *)
%o (PARI) a(n) = (15^n - (-3)^n)/18; \\ _Michel Marcus_, Mar 16 2014
%K nonn,easy
%O 0,3
%A _Felix P. Muga II_, Mar 14 2014
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