A329950 Floor of area of quadrilateral with consecutive prime sides configured as a cyclic quadrilateral.

13, 30, 70, 130, 214, 310, 461, 627, 874, 1167, 1423, 1750, 2094, 2512, 2995, 3574, 4137, 4603, 5237, 5829, 6526, 7522, 8507, 9478, 10390, 11014, 11650, 12932, 14314, 16053, 17799, 19278, 20698, 22159, 23994, 25403, 27190, 29033, 30595, 32718, 34558, 36255, 38014

Offset: 1

Author

Frank M Jackson, Nov 25 2019

Keywords

• nonn

Comments

Because it is possible to generate triangles using three consecutive odd primes (see A096377) any four consecutive primes will form a quadrilateral. If such quadrilaterals are configured to be cyclic they will have maximal area. This sequence comprises the integer part of these maximal areas.

Proof: Given 4 consecutive odd primes a, b, c, d with a < b < c < d we only have to prove that a+b+c > d for a quadrilateral to exist. However we know that 3 consecutive odd primes will form a triangle hence a+b > c and a+b+c > 2c and by Bertrand's postulate there exists a prime d such that c < d < 2c so a+b+c > d. By induction this can be extended such that n consecutive primes will always form an n-gon.

Examples

			a(1)=13 because the area of the cyclic quadrilateral with sides 2,3,5,7 is (1/4)*sqrt((2+3+5-7)(2+3-5+7)(2-3+5+7)(-2+3+5+7)) = 13.699...
		

Links

• Wikipedia, Cyclic quadrilateral, Bertrand's postulate

Crossrefs

Cf. A096377.

Programs

Mathematica

lst = {}; Do[{a, b, c, d} = {Prime[n], Prime[n+1], Prime[n+2], Prime[n+3]}; s=(a+b+c+d)/2; A=Sqrt[(s-a)(s-b)(s-c)(s-d)]; AppendTo[lst, IntegerPart@A], {n, 1, 200}]; lst

Formula

The area K of a cyclic quadrilateral with sides a, b, c, d is given by Brahmagupta's formula K = sqrt((s-a)(s-b)(s-c)(s-d)) where s = (a+b+c+d)/2.