OFFSET
2,1
COMMENTS
The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula : A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.
The corresponding areas are 6, 6, 84, 330, 84, 36, 1710, 3036, 210,...
The following table gives the first values of n, the sides (prime(n), k, k+1) and the area A of each triangle.
+-----+---------+------+------+-------+
| n | prime(n)| k | k+1 | A |
+-----+---------+------+------+-------+
| 2 | 3 | 4 | 5 | 6 |
| 3 | 5 | 3 | 4 | 6 |
| 4 | 7 | 24 | 25 | 84 |
| 5 | 11 | 60 | 61 | 330 |
| 6 | 13 | 14 | 15 | 84 |
| 7 | 17 | 9 | 10 | 36 |
| 8 | 19 | 180 | 181 | 1710 |
| 9 | 23 | 264 | 265 | 3036 |
| 10 | 29 | 20 | 21 | 210 |
.......................................
We observe triangles of sides (prime(m), prime(m)+ 1, prime(m)+ 2)) = (3, 4, 5), (13, 14, 15), (193, 194, 195), (37633, 37634, 37635), ... with the corresponding areas 6, 84, 16296, 613283664, ... (subsequence of A011945).
We observe Pythagorean triangles for n = 2, 3, 4, 5, 8, 9, 10, ....
In this case, if prime(n) < k, the numbers k of the sequence such that prime(n) = sqrt(2k+1) are given by the numbers {4, 24, 60, 180, 264, ...}, subsequence of {A084921} = {4, 12, 24, 60, 84, 144, 180, 264, ...}. If prime(n) > k, the numbers k of the sequence such that prime(n) = sqrt(2k^2+2k+1) are given by the numbers 3, 20, 4059, 23660, ....
From Chai Wah Wu, May 15 2017: (Start)
Assumes triangle has positive area.
Let p = prime(n). Then
(p+1)/2 <= a(n) <= (p^2-1)/2.
a(n) = (p+1)/2 if n > 1 is a term in A062325, i.e. p is of the form m^2+1 (A002496); otherwise, a(n) > (p+1)/2.
a(n) is the smallest k >= (p+1)/2 such that sum_{i=(p+1)/2}^{k} i*(p^2-1)/2 is a square.
These statements follow from the fact that the area of a triangle with sides of length p, k and k+1 is equal to (p^2-1)*((2k+1)^2-p^2)/16.
(End)
LINKS
Chai Wah Wu, Table of n, a(n) for n = 2..1000
EXAMPLE
a(4) = 24 because the area of the triangle (prime(4), 24, 25) = (7, 24, 25) = sqrt(28*(28-7)*(28-24)*(28-25)) = 84, where the semiperimeter 28 = (7+24+25)/2.
MAPLE
nn:=10^7:
for n from 2 to 50 do:
a:=ithprime(n):ii:=0:
for k from 1 to nn while(ii=0) do:
p:=(a+2*k+1)/2:q:=p*(p-a)*(p-k)*(p-k-1):
if q>0 and floor(sqrt(q))=sqrt(q) then
ii:=1: printf(`%d, `, k):
else
fi:
od:
od:
MATHEMATICA
Do[kk=0; Do[s=(Prime[n]+2k+1)/2; If[IntegerQ[s], area2=s(s-Prime[n])(s-k)(s-k-1); If[area2>0&&kk==0&&IntegerQ[Sqrt[area2]], Print[n, " ", k]; kk=1]], {k, 1, 3*10^4}], {n, 2, 10}] (* or *)
a[n_] := Block[{p = Prime@n, k}, k = (p + 1)/2; While[! IntegerQ@ Sqrt[(4 k^2 - p^2 + 4 k + 1) (p^2 - 1)/16], k++]; k]; a /@ Range[2, 50] (* Giovanni Resta, May 07 2017 *)
PROG
(Python)
from __future__ import division
from sympy import prime
from gmpy2 import is_square
def A286328(n): # assumes n >= 2
p, area = prime(n), 0
k, q, kq = (p + 1)//2, (p**2 - 1)//2, (p - 1)*(p + 1)**2//4
while True:
area += kq
if is_square(area):
return k
k += 1
kq += q # Chai Wah Wu, May 15 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, May 07 2017
STATUS
approved