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A210645
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Area A of the triangles such that A, the sides and one of the altitudes are four consecutive integers of an arithmetic progression d.
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1
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84, 336, 756, 1344, 2100, 3024, 4116, 5376, 6804, 8400, 10164, 12096, 14196, 16464, 18900, 21504, 24276, 27216, 30324, 33600, 37044, 40656, 44436, 48384, 52500, 56784, 61236, 65856, 70644, 75600, 80724, 86016, 91476, 97104, 102900, 108864, 114996, 121296
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OFFSET
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1,1
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COMMENTS
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See the first link for the comments. We use Heron's Formula for a triangle: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.The sides and one of the altitudes are of the form q+d, q, q-d, q-2d.
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LINKS
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FORMULA
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Conjecture: a(n) = 84*n^2. a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). G.f.: 84*x*(1+x)/(1-x)^3. - Colin Barker, Apr 19 2012
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EXAMPLE
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84 is in the sequence because (a, b, c, h) = (15, 14, 13, 12) => A = sqrt((21*(21-15)*(21-14)*(21-13)) = sqrt(7056)= 84 but A = (1/2)*h*b = (1/2)*12*14 = 84.
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MAPLE
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with(numtheory):T:=array(1..1000):k:=0:nn:=800:for q from 1 to nn do: for d from 1 to nn do: a:=q+d:b:=q:c:=q-d:h1:=q-2*d:p:=(a+b+c)/2:s:=p*(p-a)*(p-b)*(p-c):if s>0 then s1:=sqrt(s): h11:=2*s1/a: h22:=2*s1/b:h33:=2*s1/c:if s1=floor(s1) and (h1=h11 or h1=h22 or h1=h33) then k:=k+1:T[k]:=s1:else fi:fi:od:od: L := [seq(T[i], i=1..k)]:L1:=convert(T, set):A:=sort(L1, `<`): print(A):
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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