%I #20 Aug 08 2016 23:50:29
%S 1,1,1,1,1,1,1,1,2,3,4,5,7,10,14,33,53,74,96,141,209,300,714,1151,
%T 1611,2094,3083,4578,6579,15665,25257,35355,45959,67673,100497,144431,
%U 343906,554491,776186,1008991,1485711,2206346,3170896,7550257,12173533,17040724
%N a(n) = (a(n-4) + a(n-1) * a(n-7)) / a(n-8), a(0) = a(1) = ... = a(7) = 1.
%C Inspired by A048736.
%H Colin Barker, <a href="/A275174/b275174.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_21">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,23,0,0,0,0,0,0,-23,0,0,0,0,0,0,1).
%F G.f.: (1 +x +x^2 +x^3 +x^4 +x^5 +x^6 -22*x^7 -21*x^8 -20*x^9 -19*x^10 -18*x^11 -16*x^12 -13*x^13 +14*x^14 +10*x^15 +7*x^16 +5*x^17 +4*x^18 +3*x^19 +2*x^20) / ((1 -x)*(1 +x +x^2 +x^3 +x^4 +x^5 +x^6)*(1 -22*x^7 +x^14)). - _Colin Barker_, Jul 19 2016
%F a(n) = 23*a(n-7) - 23*a(n-14) + a(n-21).
%t RecurrenceTable[{a[n] == (a[n - 4] + a[n - 1] a[n - 7])/a[n - 8], a[1] == 1, a[2] == 1, a[3] == 1, a[4] == 1, a[5] == 1, a[6] == 1, a[7] == 1, a[8] == 1}, a, {n, 42}] (* _Michael De Vlieger_, Jul 19 2016 *)
%o (Ruby)
%o def A(k, l, n)
%o a = Array.new(k * 2, 1)
%o ary = [1]
%o while ary.size < n + 1
%o break if (a[1] * a[-1] + a[k] * l) % a[0] > 0
%o a = *a[1..-1], (a[1] * a[-1] + a[k] * l) / a[0]
%o ary << a[0]
%o end
%o ary
%o end
%o def A275174(n)
%o A(4, 1, n)
%o end
%o (PARI) Vec((1 +x +x^2 +x^3 +x^4 +x^5 +x^6 -22*x^7 -21*x^8 -20*x^9 -19*x^10 -18*x^11 -16*x^12 -13*x^13 +14*x^14 +10*x^15 +7*x^16 +5*x^17 +4*x^18 +3*x^19 +2*x^20) / ((1 -x)*(1 +x +x^2 +x^3 +x^4 +x^5 +x^6)*(1 -22*x^7 +x^14)) + O(x^20)) \\ _Colin Barker_, Jul 19 2016
%Y Cf. A101879, A048736, A275173.
%K nonn,easy
%O 0,9
%A _Seiichi Manyama_, Jul 19 2016