

A329950


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


1



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
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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 ngon.


LINKS



FORMULA

The area K of a cyclic quadrilateral with sides a, b, c, d is given by Brahmagupta's formula K = sqrt((sa)(sb)(sc)(sd)) 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+57)(2+35+7)(23+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[(sa)(sb)(sc)(sd)]; AppendTo[lst, IntegerPart@A], {n, 1, 200}]; lst


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



