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A275175 a(n) = (2 * a(n-3) + a(n-1) * a(n-5)) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

%I A275175 #21 Aug 08 2016 23:50:38
%S A275175 1,1,1,1,1,1,3,5,7,13,23,83,147,215,423,771,2801,4971,7281,14351,
%T A275175 26181,95133,168845,247317,487493,889373,3231703,5735737,8401475,
%U A275175 16560393,30212491,109782751,194846191,285402811,562565851,1026335311,3729381813,6619034735,9695294077,19110678523
%N A275175 a(n) = (2 * a(n-3) + a(n-1) * a(n-5)) / a(n-6), a(0) = a(1) = ... = a(5) = 1.
%C A275175 Inspired by A048736.
%H A275175 Colin Barker, <a href="/A275175/b275175.txt">Table of n, a(n) for n = 0..1000</a>
%H A275175 <a href="/index/Rec#order_15">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,35,0,0,0,0,-35,0,0,0,0,1).
%F A275175 G.f.: (1 +x +x^2 +x^3 +x^4 -34*x^5 -32*x^6 -30*x^7 -28*x^8 -22*x^9 +23*x^10 +13*x^11 +7*x^12 +5*x^13 +3*x^14) / ((1 -x)*(1 +x +x^2 +x^3 +x^4)*(1 -34*x^5 +x^10)). - _Colin Barker_, Jul 19 2016
%F A275175 a(n) = 35*a(n-5) - 35*a(n-10) + a(n-15). - _G. C. Greubel_, Jul 20 2016
%t A275175 RecurrenceTable[{a[n] == (2 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, 40}] (* _Michael De Vlieger_, Jul 19 2016 *)
%o A275175 (Ruby)
%o A275175 def A(k, l, n)
%o A275175   a = Array.new(k * 2, 1)
%o A275175   ary = [1]
%o A275175   while ary.size < n + 1
%o A275175     break if (a[1] * a[-1] + a[k] * l) % a[0] > 0
%o A275175     a = *a[1..-1], (a[1] * a[-1] + a[k] * l) / a[0]
%o A275175     ary << a[0]
%o A275175   end
%o A275175   ary
%o A275175 end
%o A275175 def A275175(n)
%o A275175   A(3, 2, n)
%o A275175 end
%o A275175 (PARI) Vec((1 +x +x^2 +x^3 +x^4 -34*x^5 -32*x^6 -30*x^7 -28*x^8 -22*x^9 +23*x^10 +13*x^11 +7*x^12 +5*x^13 +3*x^14) / ((1 -x)*(1 +x +x^2 +x^3 +x^4)*(1 -34*x^5 +x^10)) + O(x^50)) \\ _Colin Barker_, Jul 19 2016
%Y A275175 Cf. A048736, A275173, A275176.
%K A275175 nonn,easy
%O A275175 0,7
%A A275175 _Seiichi Manyama_, Jul 19 2016