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a(n) = floor(sqrt(a(n-1)^2 + a(n-2)^2 + a(n-1)*a(n-2))), a(1)=1, a(2)=3.
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%I #13 Nov 03 2018 08:43:40

%S 1,3,3,5,7,10,14,20,29,42,61,89,130,190,278,407,596,873,1279,1874,

%T 2746,4024,5897,8642,12665,18561,27202,39866,58426,85627,125492,

%U 183917,269543,395034,578950,848492,1243525,1822474,2670965,3914489,5736962

%N a(n) = floor(sqrt(a(n-1)^2 + a(n-2)^2 + a(n-1)*a(n-2))), a(1)=1, a(2)=3.

%C For a triangle with sides a(n-1) and a(n-2) and a 120-degree angle between them, a(n) is the floor of the value of the third side.

%C a(n) = A020711(n-4) for 4 <= n <= 41. - _Georg Fischer_, Nov 02 2018

%F Conjectures from _Colin Barker_, Nov 03 2018: (Start)

%F G.f.: x*(1 + x - 2*x^2 + x^3 - 2*x^4 + x^5 - x^6) / ((1 - x)*(1 - x - x^3)).

%F a(n) = 2*a(n-1) - a(n-2) + a(n-3) - a(n-4) for n>7.

%F (End)

%t a[1]=1;a[2]=3;a[n_]:=a[n]=Floor[Sqrt[a[n-1]^2+a[n-2]^2+a[n-1]*a[n-2]]] Table[a[n],{n,45}]

%t RecurrenceTable[{a[1]==1,a[2]==3,a[n]==Floor[Sqrt[a[n-1]^2+a[n-2]^2+ a[n-1]*a[n-2]]]},a,{n,50}] (* _Harvey P. Dale_, Oct 01 2018 *)

%Y Cf. A020711, A104803, A104804, A104805.

%K nonn

%O 1,2

%A _Zak Seidov_, May 04 2007