cp's OEIS Frontend

A102341 Areas of 'close-to-equilateral' integer triangles.

%I A102341 #40 Feb 16 2025 08:32:55
%S A102341 12,120,1848,25080,351780,4890480,68149872,949077360,13219419708,
%T A102341 184120982760,2564481115560,35718589344360,497495864091732,
%U A102341 6929223155685600,96511629630137568,1344233586759971040,18722758603319903340
%N A102341 Areas of 'close-to-equilateral' integer triangles.
%C A102341 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).
%C A102341 After these first five triangles, there are two more (namely (3361,3361,3360,4890480) and (12545,12545,12546,68149872)). - _Nícolas V. Calsavara_, Jul 13 2023
%C A102341 Then, the 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
%C A102341 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
%C A102341 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
%H A102341 Danny Rorabaugh, <a href="/A102341/b102341.txt">Table of n, a(n) for n = 1..875</a>
%H A102341 Steven Dutch, <a href="http://web.archive.org/web/20060830154134/https://www.uwgb.edu/dutchs/RECMATH/rmpowers.htm/#almost30">Almost 30-60 Triples</a>
%H A102341 Project Euler, <a href="https://projecteuler.net/problem=94">Problem 94: Almost equilateral triangles</a>
%H A102341 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/HeronianTriangle.html">Heronian Triangle</a> and <a href="https://mathworld.wolfram.com/PellEquation.html">Pell Equation</a>
%F A102341 (2/3) [ A007655(n+2) - (-1)^n*A001353(n+1) ] (conjectured). - _Ralf Stephan_, May 17 2007
%F A102341 Empirical g.f.: 12*x / ((x^2-14*x+1)*(x^2+4*x+1)). - _Colin Barker_, Apr 10 2013
%F A102341 a(n) = A001353(n+1)*A195499(n) = A001353(n+1)*A120892(n+1) - _Danny Rorabaugh_, Oct 16 2015
%e A102341 a(2) = 120 because 120 is the area of a triangle with side lengths of 16, 17 and 17.
%Y A102341 For the continued fraction expansion of sqrt(3), cf. A002530, A002531, A040001.
%K A102341 easy,nonn
%O A102341 1,1
%A A102341 Johannes Koelman (Joc_Kay(AT)hotmail.com), Feb 20 2005
%E A102341 More terms from _Zak Seidov_, Feb 23 2005
%E A102341 More terms from Dan Sanders (dan(AT)ified.ca), Oct 22 2005