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Floor of area of triangle with consecutive prime sides.
3

%I #7 Aug 19 2022 14:45:41

%S 0,6,12,38,71,107,158,218,317,436,550,696,817,961,1184,1425,1666,1883,

%T 2134,2377,2635,3008,3437,3931,4351,4645,4887,5199,5778,6548,7484,

%U 7955,8653,9237,10032,10642,11389,12150,12928,13653,14570,15323,16232,16683

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

%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 = 1/2hc. 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).

%e For triangle with sides 3,5,7 area = (1/4)*sqrt(4*9*49 - (9-25+49)^2) = 6.495...

%o (PARI) area(n) = { for(x=1,n, a=prime(x);b=prime(x+1);c=prime(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