

A096378


Floor of area of triangle with consecutive integer sides.


2



0, 2, 6, 9, 14, 20, 26, 34, 42, 51, 61, 72, 84, 96, 109, 124, 139, 155, 172, 190, 208, 228, 248, 269, 291, 314, 338, 363, 388, 415, 442, 470, 499, 529, 560, 591, 624, 657, 691, 727, 762, 799, 837, 875, 915, 955, 996, 1038, 1081, 1125, 1170, 1215, 1261, 1308
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OFFSET

1,2


LINKS

Robert Israel, Table of n, a(n) for n = 1..10000


FORMULA

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 + (cx)^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).
From Robert Israel, Jan 16 2018: (Start)
a(n) = floor((n+1)*sqrt(3*(n+3)*(n1))/4).
sqrt(3)*(n^2/4 + n/2)  2 < a(n) < sqrt(3)*(n^2/4 + n/2). (End)


EXAMPLE

For triangle with sides 3,4,5 area = (1/4)*sqrt(4*9*25  (916+25)^2) = 6. This is the 3,4,5 right triangle with base 3 and height 4. (1/2)*3*4 = 6.


MAPLE

f:= n > floor((n+1)*sqrt(3*(n+3)*(n1))/4):
map(f, [$1..100]); # Robert Israel, Jan 16 2018


PROG

(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^2b^2)^2); print1(floor(z)", ") ) }


CROSSREFS

Sequence in context: A184824 A184836 A327895 * A342426 A217001 A320666
Adjacent sequences: A096375 A096376 A096377 * A096379 A096380 A096381


KEYWORD

nonn


AUTHOR

Cino Hilliard, Aug 04 2004


STATUS

approved



