Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%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