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 A102341 Areas of 'close-to-equilateral' integer triangles. 6
 12, 120, 1848, 25080, 351780, 4890480, 68149872, 949077360, 13219419708, 184120982760, 2564481115560, 35718589344360, 497495864091732, 6929223155685600, 96511629630137568, 1344233586759971040, 18722758603319903340 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS A close-to-equilateral integer triangle is defined to be a triangle with integer sides and integer area such that the largest and smallest sides differ in length by unity. The first five close-to-equilateral integer triangles have sides (5, 5, 6), (17, 17, 16), (65, 65, 66), (241, 241, 240) and (901, 901, 902). Next four terms are {three sides a<=b<=c and area}: {46816, 46817, 46817, 949077360}, {174725, 174725, 174726, 13219419708}, {652080, 652081, 652081, 184120982760}, {2433601, 2433601, 2433602, 2564481115560}. Also, if we allow degenerate triangles (area 0), the first case would be {1,1,2,0}. We have 12 cases and a weak conjecture is that the total number of the 'close-to-equilateral' triangles is finite. - Zak Seidov, Feb 23 2005 This is an infinite series; two sides are equal in length to the hypotenuse of almost 30-60 triangles and the third side alternates between that length +/- 1. - Dan Sanders (dan(AT)ified.ca), Oct 22 2005 Heron's formula: a triangle with side lengths (x,y,z) has area A = sqrt(s*(s-x)*(s-y)*(s-z)) where s = (x+y+z)/2. For this sequence we assume integer side-lengths x = y = z +/- 1. Then for A to also be an integer, x+y+z must be even, so we can assume z = 2k for some positive integer k. Now s = (x+y+z)/2 = 3k +/- 1 and A = sqrt((3*k +/- 1)*k*k*(k +/- 1)) = k*sqrt(3*k^2 +/- 4*k + 1). To determine when this is an integer, set 3*k^2 +/- 4*k + 1 = d^2. If we multiply both sides by 3, it is easier to complete the square: (3*k +/- 2)^2 - 1 = 3*d^2. Now we are looking for solutions to the Pell equation c^2 - 3*d^2 = 1 with c = 3*k +/- 2, for which there are infinitely many solutions: use the upper principal convergents of the continued fraction expansion of sqrt(3) (A001075/A001353). - Danny Rorabaugh, Oct 16 2015 LINKS Danny Rorabaugh, Table of n, a(n) for n = 1..875 Steven Dutch, Almost 30-60 Triples ProjectEuler, Problem 94: Almost equilateral triangles Eric Weisstein's World of Mathematics, Heronian Triangle and Pell Equation FORMULA (2/3) [ A007655(n+2) - (-1)^n*A001353(n+1) ] (conjectured). - Ralf Stephan, May 17 2007 Empirical g.f.: 12*x / ((x^2-14*x+1)*(x^2+4*x+1)). - Colin Barker, Apr 10 2013 a(n) = A001353(n+1)*A195499(n) = A001353(n+1)*A120892(n+1) - Danny Rorabaugh, Oct 16 2015 EXAMPLE a(2) = 120 because 120 is the area of a triangle with side lengths of 16, 17 and 17. CROSSREFS For the continued fraction expansion of sqrt(3), cf. A002530, A002531, A040001. Sequence in context: A200163 A012565 A012621 * A174561 A009078 A221493 Adjacent sequences:  A102338 A102339 A102340 * A102342 A102343 A102344 KEYWORD easy,nonn AUTHOR Johannes Koelman (Joc_Kay(AT)hotmail.com), Feb 20 2005 EXTENSIONS More terms from Zak Seidov, Feb 23 2005 More terms from Dan Sanders (dan(AT)ified.ca), Oct 22 2005 STATUS approved

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Last modified January 20 07:59 EST 2020. Contains 331081 sequences. (Running on oeis4.)