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A283960 a(n) = (Sum_{j=1..h-1} a(n-j) + a(n-1)*a(n-h+1))/a(n-h) with a(1), ..., a(h)=1, where h = 6.

%I A283960 #25 Nov 06 2020 07:48:28
%S A283960 1,1,1,1,1,1,6,16,41,106,276,2101,6026,15976,41901,109726,835906,
%T A283960 2397991,6358066,16676206,43670551,332688201,954394051,2530493951,
%U A283960 6637087801,17380769451,132409067806,379846433966,1007130234091,2641544268306,6917502570826,52698476298301
%N A283960 a(n) = (Sum_{j=1..h-1} a(n-j) + a(n-1)*a(n-h+1))/a(n-h) with a(1), ..., a(h)=1, where h = 6.
%H A283960 Seiichi Manyama, <a href="/A283960/b283960.txt">Table of n, a(n) for n = 1..1929</a>
%H A283960 <a href="/index/Rec#order_15">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,399,0,0,0,0,-399,0,0,0,0,1).
%F A283960 a(5*k-2) = 3*a(5*k-3) - a(5*k-4) - 1,
%F A283960 a(5*k-1) = 3*a(5*k-2) - a(5*k-3) - 1,
%F A283960 a(5*k)   = 3*a(5*k-1) - a(5*k-2) - 1,
%F A283960 a(5*k+1) = 3*a(5*k)   - a(5*k-1) - 1,
%F A283960 a(5*k+2) = 8*a(5*k+1) - a(5*k)   - 1.
%F A283960 From _Colin Barker_, Nov 03 2020: (Start)
%F A283960 G.f.: x*(1 + x + x^2 + x^3 + x^4 - 398*x^5 - 393*x^6 - 383*x^7 - 358*x^8 - 293*x^9 + 276*x^10 + 106*x^11 + 41*x^12 + 16*x^13 + 6*x^14) / ((1 - x)*(1 + x + x^2 + x^3 + x^4)*(1 - 398*x^5 + x^10)).
%F A283960 a(n) = 399*a(n-5) - 399*a(n-10) + a(n-15) for n>15.
%F A283960 (End)
%t A283960 a[n_]:= If[n<7, 1, (Sum[a[n-j] , {j, 5}] +  a[n - 1] a[n - 5])/a[n - 6]]; Table[a[n], {n, 30}] (* _Indranil Ghosh_, Mar 18 2017 *)
%o A283960 (PARI) a(n) = if(n<7, 1, (sum(j=1, 5, a(n - j)) + a(n - 1)*a(n - 5))/a(n - 6));
%o A283960 for(n=1, 30, print1(a(n), ", ")) \\ _Indranil Ghosh_, Mar 18 2017
%Y A283960 Cf. A072881 (h=3), A283958 (h=4), A283959 (h=5), this sequence (h=6).
%K A283960 nonn
%O A283960 1,7
%A A283960 _Seiichi Manyama_, Mar 18 2017