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A275176 a(n) = (3 * a(n-3) + a(n-1) * a(n-5)) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

%I A275176 #21 Aug 08 2016 23:50:45
%S A275176 1,1,1,1,1,1,4,7,10,22,43,202,370,547,1264,2521,11881,21781,32221,
%T A275176 74521,148681,700744,1284667,1900450,4395442,8769643,41331982,
%U A275176 75773530,112094287,259256524,517260241,2437886161,4469353561,6611662441,15291739441,30509584561
%N A275176 a(n) = (3 * a(n-3) + a(n-1) * a(n-5)) / a(n-6), a(0) = a(1) = ... = a(5) = 1.
%C A275176 Inspired by A048736.
%H A275176 Colin Barker, <a href="/A275176/b275176.txt">Table of n, a(n) for n = 0..1000</a>
%H A275176 <a href="/index/Rec#order_15">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,60,0,0,0,0,-60,0,0,0,0,1).
%F A275176 G.f.: (1 +x +x^2 +x^3 +x^4 -59*x^5 -56*x^6 -53*x^7 -50*x^8 -38*x^9 +43*x^10 +22*x^11 +10*x^12 +7*x^13 +4*x^14) / ((1 -x)*(1 +x +x^2 +x^3 +x^4)*(1 -59*x^5 +x^10)). - _Colin Barker_, Jul 19 2016
%F A275176 a(n) = 60*a(n-5) - 60*a(n-10) + a(n-15).
%t A275176 RecurrenceTable[{a[n] == (3 a[n - 3] + a[n - 1] a[n - 5])/a[n - 6], a[1] == 1, a[2] == 1, a[3] == 1, a[4] == 1, a[5] == 1, a[6] == 1}, a, {n, 36}] (* _Michael De Vlieger_, Jul 19 2016 *)
%o A275176 (Ruby)
%o A275176 def A(k, l, n)
%o A275176   a = Array.new(k * 2, 1)
%o A275176   ary = [1]
%o A275176   while ary.size < n + 1
%o A275176     break if (a[1] * a[-1] + a[k] * l) % a[0] > 0
%o A275176     a = *a[1..-1], (a[1] * a[-1] + a[k] * l) / a[0]
%o A275176     ary << a[0]
%o A275176   end
%o A275176   ary
%o A275176 end
%o A275176 def A275176(n)
%o A275176   A(3, 3, n)
%o A275176 end
%o A275176 (PARI) Vec((1 +x +x^2 +x^3 +x^4 -59*x^5 -56*x^6 -53*x^7 -50*x^8 -38*x^9 +43*x^10 +22*x^11 +10*x^12 +7*x^13 +4*x^14) / ((1 -x)*(1 +x +x^2 +x^3 +x^4)*(1 -59*x^5 +x^10)) + O(x^50)) \\ _Colin Barker_, Jul 19 2016
%Y A275176 Cf. A048736, A275173, A275175.
%K A275176 nonn,easy
%O A275176 0,7
%A A275176 _Seiichi Manyama_, Jul 19 2016