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A276308 a(n) = (a(n-1)+1)*(a(n-3)+1)/a(n-4) for n > 3, a(0) = a(1) = a(2) = a(3) = 1.

%I A276308 #35 Apr 24 2021 03:31:39
%S A276308 1,1,1,1,4,10,22,115,319,736,3886,10816,24991,131989,367405,848947,
%T A276308 4483720,12480934,28839196,152314471,423984331,979683706,5174208274,
%U A276308 14402986300,33280406797,175770766825,489277549849,1130554147381,5971031863756,16621033708546
%N A276308 a(n) = (a(n-1)+1)*(a(n-3)+1)/a(n-4) for n > 3, a(0) = a(1) = a(2) = a(3) = 1.
%H A276308 Colin Barker, <a href="/A276308/b276308.txt">Table of n, a(n) for n = 0..1000</a>
%H A276308 S. Fomin and A. Zelevinsky, <a href="http://dx.doi.org/10.1006/aama.2001.0770">The Laurent Phenomenon</a>, Advances in Applied Mathematics, 28 (2002), 119-144.
%H A276308 Matthew Christopher Russell, <a href="https://www.semanticscholar.org/paper/Using-experimental-mathematics-to-conjecture-and-in-Russell/fdebe20954dacb7ec7a24afe2cf491b951c5a28d">Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences</a>, PhD Dissertation, Mathematics Department, Rutgers University, May 2016. See Eq. (6.137).
%H A276308 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,35,0,0,-35,0,0,1).
%F A276308 From _Colin Barker_, Aug 29 2016: (Start)
%F A276308 a(n) = 35*a(n-3)-35*a(n-6)+a(n-9) for n>8.
%F A276308 G.f.: (1+x+x^2-34*x^3-31*x^4-25*x^5+22*x^6+10*x^7+4*x^8) / ((1-x)*(1+x+x^2)*(1-34*x^3+x^6)).
%F A276308 (End)
%o A276308 (Ruby)
%o A276308 def A(m, n)
%o A276308   a = Array.new(m, 1)
%o A276308   ary = [1]
%o A276308   while ary.size < n + 1
%o A276308     i = (a[1] + 1) * (a[-1] + 1)
%o A276308     break if i % a[0] > 0
%o A276308     a = *a[1..-1], i / a[0]
%o A276308     ary << a[0]
%o A276308   end
%o A276308   ary
%o A276308 end
%o A276308 def A276308(n)
%o A276308   A(4, n)
%o A276308 end
%o A276308 (PARI) Vec((1+x+x^2-34*x^3-31*x^4-25*x^5+22*x^6+10*x^7+4*x^8)/((1-x)*(1+x+x^2)*(1-34*x^3+x^6)) + O(x^35)) \\ _Colin Barker_, Aug 29 2016
%o A276308 (PARI) a276308(maxn) = {a=vector(maxn); a[1]=a[2]=a[3]=a[4]=1; for(n=5, maxn, a[n]=(a[n-1]+1)*(a[n-3]+1)/a[n-4]); a} \\ _Colin Barker_, Aug 30 2016
%Y A276308 Cf. A276123, A276175.
%K A276308 nonn,easy
%O A276308 0,5
%A A276308 _Seiichi Manyama_, Aug 29 2016