A098030 -id:A098030 - cp's OEIS frontend

A335016 Largest side lengths of equable Heronian triangles.

10, 13, 17, 20, 29

Offset: 1

Wesley Ivan Hurt, May 19 2020

Equable Heronian triangles are triangles with integer sides, integer area and whose area is equal to their perimeter. There are exactly five, [6,8,10], [5,12,13], [9,10,17], [7,15,20], [6,25,29].

Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
Wikipedia, Integer Triangle

Cf. A098030 (areas/perimeters), A335013 (middle side lengths), A335015 (smallest side lengths), this sequence (largest side lengths).

A370599 a(n) is the number of distinct triangles with integral side-lengths for which the perimeter 2*n divides the area.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 1, 1, 1, 2, 2, 0, 0, 1, 1, 0, 4, 0, 1, 3, 0, 0, 1, 0, 0, 0, 0, 0, 3, 1, 2, 1, 0, 0, 3, 0, 0, 3, 2, 1, 4, 0, 2, 0, 3, 0, 2, 0, 0, 2, 2, 3, 1, 0, 2, 4, 1, 0, 8, 1, 0, 1

Offset: 1

Felix Huber, Mar 09 2024

If the perimeter 2*n of a triangle with integral edge-lengths divides its area A, then this also applies to a triangle stretched with positive integer k, because A*k^2/(2*n*k) = k*A/(2*n). Therefore a(d) <= a(n) for all positive divisors d of n and a(m) >= a(n) for all positive integer multiples m of n.

With an odd perimeter, according to Heron's formula the area A would have the form A = sqrt((2*k - 1)/8), where k is a positive integer. The area A would be irrational and the integer perimeter would not divide the area A. For this reason, only triangles with an even perimeter are considered in this sequence.

			a(18) = 1, because only the triangle (9, 10, 17) satisfies the condition: A/(2*n) = 36/36 = 1. (9, 10, 17) is one of the five triangles for which the perimeter is equal to the area (see A098030).
a(42) = 4, because exactly the 4 triangles (10, 35, 39) with A/(2*n) = 168/84 = 2, (14, 30, 40) with A/(2*n) = 168/84 = 2, (15, 34, 35) with A/(2*n) = 252/84 = 3 and (26, 28, 30) with A/(2*n) = 336/84 = 4 satisfy the condition.
a(426) = 0, because no triangle satisfies the condition. Therefore, a(n) = 0 for all n for which n*k = 426 for positive integers k.

Felix Huber, Table of n, a(n) for n = 1..500
Wikipedia, Heron's formula

Cf. A010814, A098030, A221837, A221838, A369951.

A370599 := proc(n) local u, v, w, A, q, i; i := 0; for u to floor(2/3*n) do for v from max(u, floor(n - u) + 1) to floor(n - 1/2*u) do w := 2*n - u - v; A := sqrt(n*(n - u)*(n - v)*(n - w)); if A = floor(A) then q := 1/2*A/n; if q = floor(q) then i := i + 1; end if; end if; end do; end do; return i; end proc;
seq(A370599(n), n = 1 .. 87);

a(n*k) >= a(n) for positive integers k.

A348143 Areas of integer-sided cyclic quadrilaterals whose area equals their perimeter.

Original entry on oeis.org

16, 18, 20, 30

Offset: 1

Frank M Jackson, Oct 02 2021

There are no further terms. Note that without the condition "integer-sided" there are other solutions, such as (1, 17/2, 17/2, 16) which has perimeter and area 34.

			The areas or perimeters 16, 18, 20, 30 pertain respectively to cyclic quadrilaterals with sides (4, 4, 4, 4), (3, 3, 6, 6), (2, 5, 5, 8), (5, 5, 6, 14).

Ralph H. Buchholz and James A. MacDougall, Cyclic Polygons with Rational Sides and Area, 2001. JNT 128 (2008) 17-48

Cf. A098030, A290451. First four terms of A161874.

lst={}; Do[s=(a+b+c+d)/2; If[s>a, (K=Sqrt[(s-a)(s-b)(s-c)(s-d)]; If[IntegerQ[K]&&K==2s, AppendTo[lst, Sort@{a,b,c,d}]])], {a, 1, 15}, {b, 1, a}, {c, 1, b}, {d, 1, c}]; lst

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A335016 Largest side lengths of equable Heronian triangles.

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A370599 a(n) is the number of distinct triangles with integral side-lengths for which the perimeter 2*n divides the area.

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A348143 Areas of integer-sided cyclic quadrilaterals whose area equals their perimeter.

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Mathematica