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A096378
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Floor of area of triangle with consecutive integer sides.
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3
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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
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1,2
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LINKS
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FORMULA
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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).
a(n) = floor((n+1)*sqrt(3*(n+3)*(n-1))/4).
sqrt(3)*(n^2/4 + n/2) - 2 < a(n) < sqrt(3)*(n^2/4 + n/2). (End)
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EXAMPLE
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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.
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MAPLE
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f:= n -> floor((n+1)*sqrt(3*(n+3)*(n-1))/4):
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PROG
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(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)", ") ) }
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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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