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 A096378 Floor of area of triangle with consecutive integer sides. 2

%I

%S 0,2,6,9,14,20,26,34,42,51,61,72,84,96,109,124,139,155,172,190,208,

%T 228,248,269,291,314,338,363,388,415,442,470,499,529,560,591,624,657,

%U 691,727,762,799,837,875,915,955,996,1038,1081,1125,1170,1215,1261,1308

%N Floor of area of triangle with consecutive integer sides.

%H Robert Israel, <a href="/A096378/b096378.txt">Table of n, a(n) for n = 1..10000</a>

%F Given a triangle ABC with sides a, b, base c, height h and x=base of right triangle formed by a and h. Then a^2 = h^2 + x^2, b^2 = h^2 + (c-x)^2, h = sqrt(a^2 - x^2), area = (1/2)hc. Hence x = (a^2 - b^2 + c^2)/2c and so area = (1/4)*sqrt(4*a^2*c^2 - (a^2 - b^2 + c^2)^2).

%F From _Robert Israel_, Jan 16 2018: (Start)

%F a(n) = floor((n+1)*sqrt(3*(n+3)*(n-1))/4).

%F sqrt(3)*(n^2/4 + n/2) - 2 < a(n) < sqrt(3)*(n^2/4 + n/2). (End)

%e For triangle with sides 3,4,5 area = (1/4)*sqrt(4*9*25 - (9-16+25)^2) = 6. This is the 3,4,5 right triangle with base 3 and height 4. (1/2)*3*4 = 6.

%p f:= n -> floor((n+1)*sqrt(3*(n+3)*(n-1))/4):

%p map(f, [\$1..100]); # _Robert Israel_, Jan 16 2018

%o (PARI) area(n) = { for(x=1,n, a=x;b=x+1;c=x+2; z=1/4*sqrt(4*a^2*c^2-(c^2+a^2-b^2)^2); print1(floor(z)",") ) }

%K nonn

%O 1,2

%A _Cino Hilliard_, Aug 04 2004

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Last modified January 21 11:28 EST 2022. Contains 350477 sequences. (Running on oeis4.)