Suppose a triangle with side lengths $\tuple {a - 1, a, a + 1}$ is Heronian. By Heron's Formula, the area of this triangle is: {{begin-eqn}} {{eqn | o =       | r = \sqrt {s \paren {s - a + 1} \paren {s - a} \paren {s - a - 1} } }} {{eqn | r = \sqrt {\frac 3 2 a \paren {\frac 1 2 a + 1} \paren {\frac 1 2 a} \paren {\frac 1 2 a - 1} } }} {{eqn | r = \frac a 4 \sqrt {3 \paren {a + 2} \paren {a - 2} } }} {{eqn | r = \frac a 4 \sqrt {3 a^2 - 12} }} {{end-eqn}} By {{Defof|Heronian Triangle}}, $\dfrac a 4 \sqrt {3 a^2 - 12} \in \N$. Hence it is necessary for $a$ to be even. Writing $a = 2 n$: :$\dfrac a 4 \sqrt {3 a^2 - 12} = \dfrac {2 n} 4 \sqrt {3 \paren {2 n}^2 - 12} = n \sqrt {3 n^2 - 3}$ So we need to find $n$ such that $\sqrt {3 n^2 - 3}$ is an integer. This corresponds to a Pell-like equation $x^2 - 3 n^2 = -3$. This equation has a trivial solution $\tuple {x, n} = \tuple {0, 1}$. The smallest $n$ with integer $x$ solution to $x^2 - 3 n^2 = 1$ are: :$\tuple {n, x} = \tuple {0, 1}$ (trivial), $\tuple {1, 2}$ Then a family of solutions can be found by: :$x^2 - 3 y^2 = 1 = \paren {2^2 - 3 \times 1^2}^m$ for all $m \in \N$ Factorising gives: :$\paren {x - \sqrt 3 y} \paren {x + \sqrt 3 y} = \paren {2 - \sqrt 3}^m \paren {2 + \sqrt 3}^m$ The system of equations: :$x - \sqrt 3 y = \paren {2 - \sqrt 3}^m$ :$x + \sqrt 3 y = \paren {2 + \sqrt 3}^m$ Has solutions: :$x = \dfrac {\paren {2 - \sqrt 3}^m + \paren {2 + \sqrt 3}^m} 2$ :$y = \dfrac {\paren {2 + \sqrt 3}^m - \paren {2 - \sqrt 3}^m} {2 \sqrt 3}$ Suppose that $p^2 - 3 q^2 = -3, r^2 - 3 s^2 = 1$. Then: {{begin-eqn}} {{eqn | o =       | r = \paren {p^2 - 3 q^2} \paren {r^2 - 3 s^2} }} {{eqn | r = p^2 r^2 - 3 q^2 r^2 - 3 p^2 s^2 + 9 q^2 s^2 }} {{eqn | r = p^2 r^2 - 3 q^2 r^2 - 3 p^2 s^2 + 9 q^2 s^2 \pm 6 p q r s \mp 6 p q r s }} {{eqn | r = \paren {p r \pm 3 q s}^2 - 3 \paren {p s \pm q r}^2 }} {{eqn | r = -3}} {{end-eqn}} showing that $\tuple {p r \pm 3 q s, p s \pm q r}$ is a solution for $x^2 - 3 y^2 = -3$. In particular: :$\tuple {\dfrac {\sqrt 3 \paren {\paren {2 + \sqrt 3}^m - \paren {2 - \sqrt 3}^m}} 2, \dfrac {\paren {2 - \sqrt 3}^m + \paren {2 + \sqrt 3}^m} 2}$ is a solution. From our discussion above: :$a = \paren {2 - \sqrt 3}^m + \paren {2 + \sqrt 3}^m$ is a family of solutions for the almost-equilateral Heronian triangles. The sequence above is a subfamily of the solution above, as seen by: {{begin-eqn}} {{eqn | l = m = 1       | o = \implies       | r = \paren {2 - \sqrt 3}^1 + \paren {2 + \sqrt 3}^1 = 4 }} {{eqn | l = \paren {\paren {2 - \sqrt 3}^m + \paren {2 + \sqrt 3}^m}^2 - 2       | r = \paren {2 - \sqrt 3}^{2 m} + \paren {2 + \sqrt 3}^{2 m} + 2 \paren {2 - \sqrt 3}^m \paren {2 + \sqrt 3}^m - 2 }} {{eqn | r = \paren {2 - \sqrt 3}^{2 m} + \paren {2 + \sqrt 3}^{2 m} + 2 \paren {\paren {2 - \sqrt 3} \paren {2 + \sqrt 3} }^m - 2 }} {{eqn | r = \paren {2 - \sqrt 3}^{2 m} + \paren {2 + \sqrt 3}^{2 m} + 2 \paren 1^m - 2 }} {{eqn | r = \paren {2 - \sqrt 3}^{2 m} + \paren {2 + \sqrt 3}^{2 m} }} {{end-eqn}} Some other solutions include: :$m = 3: \tuple {51, 52, 53}$, with area $\sqrt {78 \times 27 \times 26 \times 25} = 1170$ :$m = 5: \tuple {723, 724, 725}$, with area $\sqrt {1086 \times 361 \times 362 \times 363} = 226 \, 974$ {{OEIS|A003500|order = $a$}} {{OEIS|A011945|order = area}}{{qed}}