A210645 Area A of the triangles such that A, the sides and one of the altitudes are four consecutive integers of an arithmetic progression d.

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

Offset: 1

Author

Michel Lagneau, Mar 26 2012

Keywords

• nonn

Comments

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.

Examples

			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.
		

Links

• Eric Weisstein, Altitude

Crossrefs

Cf. A188158, A210643.

Programs

Maple

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):

Formula

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