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a(n) = a(n-1) + a(n-3) - a(n-4) for n>4, with a(1)=1, a(2)=4, a(3)=10, a(4)=4.
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%I #11 Oct 05 2017 20:25:06

%S 1,4,10,4,7,13,7,10,16,10,13,19,13,16,22,16,19,25,19,22,28,22,25,31,

%T 25,28,34,28,31,37,31,34,40,34,37,43,37,40,46,40,43,49,43,46,52,46,49,

%U 55,49,52,58,52,55,61,55,58,64,58,61,67,61,64,70,64,67,73

%N a(n) = a(n-1) + a(n-3) - a(n-4) for n>4, with a(1)=1, a(2)=4, a(3)=10, a(4)=4.

%C This sequence gives a linearly increasing triangle shaped form on plotting.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,-1).

%F G.f.: -x*(-1-3*x-6*x^2+7*x^3)/((1+x+x^2)*(x-1)^2). [Oct 14 2009]

%F a(n) = n+3 + (-1)^n * A130806(n+1). [Oct 14 2009]

%F a(n) = (24*n + 75 - 18*cos(2*(n+1)*Pi/3) + 9*cos(2*Pi*sin(2*n*Pi/3)/sqrt(3)) + 50*sqrt(3)*sin(2*(n+1)*Pi/3))/24. - _Wesley Ivan Hurt_, Oct 05 2017

%t M = {{0, 1, 0}, {0, 0, 1}, {1, 0, 0}} v[1] = {1, 4, 10} v[n_] := v[n] = M.v[n - 1] + {0, 0, 3} a = Table[v[n][[1]], {n, 1, 50}]

%Y Cf. A005448, A130806.

%K nonn,easy

%O 1,2

%A _Roger L. Bagula_, Sep 07 2006

%E Definition replaced by recurrence - The Assoc. Editors of the OEIS, Oct 14 2009