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

1,1

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.

Links

Table of n, a(n) for n=1..43.

Wikipedia, Cyclic quadrilateral, Bertrand's postulate

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.

Example

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...

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

Crossrefs

Cf. A096377.

Sequence in context: A216193 A139577 A172463 * A183445 A296124 A299130

Adjacent sequences: A329947 A329948 A329949 * A329951 A329952 A329953

Keyword

nonn

Author

Frank M Jackson, Nov 25 2019

Status

approved