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

12, 120, 1848, 25080, 351780, 4890480, 68149872, 949077360, 13219419708, 184120982760, 2564481115560, 35718589344360, 497495864091732, 6929223155685600, 96511629630137568, 1344233586759971040, 18722758603319903340

Offset: 1

Author

Johannes Koelman (Joc_Kay(AT)hotmail.com), Feb 20 2005

Keywords

• easy
• nonn

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).

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

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

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

Examples

			a(2) = 120 because 120 is the area of a triangle with side lengths of 16, 17 and 17.
		

Links

• Danny Rorabaugh, Table of n, a(n) for n = 1..875
• Steven Dutch, Almost 30-60 Triples
• Project Euler, Problem 94: Almost equilateral triangles
• Eric Weisstein's World of Mathematics, Heronian Triangle and Pell Equation

Crossrefs

For the continued fraction expansion of sqrt(3), cf. A002530, A002531, A040001.

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

Extensions

More terms from Zak Seidov, Feb 23 2005

More terms from Dan Sanders (dan(AT)ified.ca), Oct 22 2005