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A233231 a(n) = 10*a(n-3) - a(n-6) + 4 for n>5, a(0)=2, a(1)=3, a(2)=5, a(3)=12, a(4)=29, a(5)=51.

%I A233231 #16 Jun 13 2015 00:54:55
%S A233231 2,3,5,12,29,51,122,291,509,1212,2885,5043,12002,28563,49925,118812,
%T A233231 282749,494211,1176122,2798931,4892189,11642412,27706565,48427683,
%U A233231 115248002,274266723,479384645,1140837612,2714960669,4745418771,11293128122,26875339971
%N A233231 a(n) = 10*a(n-3) - a(n-6) + 4 for n>5, a(0)=2, a(1)=3, a(2)=5, a(3)=12, a(4)=29, a(5)=51.
%C A233231 Apart from a(0), a(n) such that the triple (5,a(n),a(n)+1) forms a Heronian triangle. Equivalently, a(n) such that 6*(a(n)+3)*(a(n)-2) is a square. Note that this sequence generates all Heronian triples with a fixed side of 5 except (5,5,8) which is the only solution to Heronian triples of the form (5,x,x+3).
%H A233231 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,10,-10,0,-1,1 ).
%F A233231 G.f.: (2 + x + 2*x^2 - 13*x^3 + 7*x^4 + 2*x^5 + 3*x^6)/((1 - x)*(1 - 10*x^3 + x^6)). [_Bruno Berselli_, Dec 09 2013]
%F A233231 a(n) = a(n-1) + 10*a(n-3) - 10*a(n-4) - a(n-6) + a(n-7) for n>6. [_Bruno Berselli_, Dec 09 2013]
%e A233231 a(5)=29 as the triangle with sides (5,29,30) has integer area 72.
%t A233231 seq[n_] := seq[n]=Which[n==0, 2, n==1, 3, n==2, 5, n==3, 12, n==4, 29, n==5, 51, True, 10seq[n-3]-seq[n-6]+4]; Table[seq[m], {m, 0, 100}]
%t A233231 LinearRecurrence[{1, 0, 10, -10, 0, -1, 1}, {2, 3, 5, 12, 29, 51, 122}, 30] (* _T. D. Noe_, Dec 09 2013 *)
%Y A233231 Cf. A072221, A230666.
%K A233231 nonn,easy
%O A233231 0,1
%A A233231 _Frank M Jackson_, Dec 06 2013