cp's OEIS Frontend

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

%I A329950 #8 Dec 26 2019 05:44:11
%S A329950 13,30,70,130,214,310,461,627,874,1167,1423,1750,2094,2512,2995,3574,
%T A329950 4137,4603,5237,5829,6526,7522,8507,9478,10390,11014,11650,12932,
%U A329950 14314,16053,17799,19278,20698,22159,23994,25403,27190,29033,30595,32718,34558,36255,38014
%N A329950 Floor of area of quadrilateral with consecutive prime sides configured as a cyclic quadrilateral.
%C A329950 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.
%C A329950 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.
%H A329950 Wikipedia, <a href="https://en.wikipedia.org/wiki/Cyclic_quadrilateral">Cyclic quadrilateral</a>, <a href="https://en.wikipedia.org/wiki/Bertrand%27s_postulate">Bertrand's postulate</a>
%F A329950 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.
%e A329950 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...
%t A329950 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
%Y A329950 Cf. A096377.
%K A329950 nonn
%O A329950 1,1
%A A329950 _Frank M Jackson_, Nov 25 2019