cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A229159 Smallest integer areas of integer-sided triangles where at least one side is of length prime(n).

Original entry on oeis.org

0, 6, 6, 42, 66, 24, 36, 114, 966, 60, 930, 114, 126, 1290, 4230, 90, 1770, 330, 2814, 14910, 216, 4740, 1494, 420, 420, 510, 6180, 4494, 840, 570, 8382, 11790, 630, 9174, 210, 4530, 840, 2934, 45090, 3276, 22554, 1260, 24066, 336, 1386, 16716, 26586, 52182
Offset: 1

Views

Author

Michel Lagneau, Sep 17 2013

Keywords

Comments

Conjecture: for all prime p > 2 there exists an integer-sided triangle with integer area where at least one side is of length p.
There exist triangles of integer area and integer side lengths having two sides whose lengths are distinct prime numbers; for example, (3,4,5), (11,13,20), (19, 20,37), (43,61,68), (59,68,109), (11,60,61), (79,241, 312), (41,50,89), (26,73,97), ... corresponding to the areas 6, 66, 114, 1290, 1770, 330, 4740, 420, 420, ...
Observation: there exist some integer-area, integer-sided triangles with two prime sides such that the perimeter equals 4 times the smaller prime. For example:
(3, 4, 5) => 12 = 4*3;
(11, 13, 20) => 44 = 4*11;
(19, 20, 37) => 76 = 4*19;
(43, 61, 68) => 172 = 4*43;
(59, 68, 109) => 236 = 4*59;
(131, 181, 212) => 524 = 4*131;
(139, 157, 260) => 556 = 4*139;
(179, 260, 277) => 716 = 4*179.
The first 25 values (prime(n), smallest area, a, b, c) are:
+---------+-------+-----+-----+-----+
| prime(n)| Area | a | b | c |
+---------+-------+-----+-----+-----+
| 2 | 0 | 0 | 0 | 0 |
| 3 | 6 | 3 | 4 | 5 |
| 5 | 6 | 3 | 4 | 5 |
| 7 | 42 | 7 | 15 | 20 |
| 11 | 66 | 11 | 13 | 20 |
| 13 | 24 | 4 | 13 | 15 |
| 17 | 36 | 9 | 10 | 17 |
| 19 | 114 | 19 | 20 | 37 |
| 23 | 966 | 23 | 140 | 159 |
| 29 | 60 | 6 | 25 | 29 |
| 31 | 930 | 31 | 68 | 87 |
| 37 | 114 | 19 | 20 | 37 |
| 41 | 126 | 15 | 28 | 41 |
| 43 | 1290 | 43 | 61 | 68 |
| 47 | 4230 | 47 | 425 | 468 |
| 53 | 90 | 4 | 51 | 53 |
| 59 | 1770 | 59 | 68 | 109 |
| 61 | 330 | 11 | 60 | 61 |
| 67 | 2814 | 67 | 85 | 116 |
| 71 | 14910 | 71 | 447 | 476 |
| 73 | 216 | 9 | 73 | 80 |
| 79 | 4740 | 79 | 241 | 312 |
| 83 | 1494 | 83 | 85 | 164 |
| 89 | 420 | 41 | 50 | 89 |
| 97 | 420 | 26 | 73 | 97 |

Crossrefs

Cf. A226453.

Programs

  • Maple
    with(numtheory):nn:=500: for m from 2 to 40 do: q:=ithprime(m):ii:=0:for a from 1
      to nn while(ii=0) do: for b from a to nn while(ii=0) do: for c from b to nn while(ii=0) do: p:=(a+b+c)/2 : x:=p*(p-a)*(p-b)*(p-c): if x>0 then x0:= sqrt(x):else fi:if (x0=floor(x0) and a=q) or (x0=floor(x0) and b=q) or (x0=floor(x0) and c=q)then ii:=1: printf ( "%d %d %d %d %d \n",q,x0,a,b,c):
    :else fi:od:od:od:od:

A256579 Integer areas of integer-sided triangles where at least one of the three altitudes is of prime length.

Original entry on oeis.org

6, 12, 30, 60, 84, 168, 330, 546, 660, 1092, 1224, 1710, 2448, 3036, 3420, 6072, 6090, 7440, 12180, 12654, 14880, 17220, 19866, 25308, 25944, 34440, 37206, 39732, 51330, 51888, 56730, 74412, 75174, 89460, 97236, 102660, 113460, 123240, 142926, 150348, 176220
Offset: 1

Views

Author

Michel Lagneau, Apr 02 2015

Keywords

Comments

Subset of A226453.
The corresponding primes are: 3, 3, 5, 5, 7, 7, 11, 13, 11, 13, 17, 19, 17, 23, 19, 23, 29, 31, 29, 37, 31, 41, 43, 37, 47, 41, 53, 43, 59, 47, 61, 53, 67, 71, 73, 59, 61, 79, 83, 67, 89, ...
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 altitudes of a triangle with sides length a, b, c and area A have length given by Ha = 2*A/a, Hb = 2*A/b, Hc = 2*A/c.
Properties of this sequence:
- The sequence is infinite (see the formula below);
- The prime altitude of a triangle is the greatest prime divisor of a(n) (the proof is easy if we observe the formula);
- There exists two subsets of numbers included into a(n):
Case (i): A subset with right triangles (a,b,c) where a^2 + b^2 = c^2 with area a1(n) = {6, 30, 84, 330, 546, 1224, ...}. The lengths of the prime altitudes are Ha or Hb = a = p. The sides are of the form (p, q, q+1) with p = sqrt(2*q+1) => the sides are equal to (p, (p^2-1)/2, (p^2+1)/2) and a(n) = (p^3-p)/4.
Case (ii): A subset with isosceles triangles formed by two right triangles of the sequence. So, the areas are a2(n) = {12, 60, 168, 660, 1092, 2448, ...} = 2*a1(n). The sides are of the form (a, a, 2*(a-1)) = ((p^2+1)/2, (p^2+1)/2, p^2-1) and Ha = sqrt(2*a-1) = p, a2(n) = 2*a1(n) = (p^3-p)/2.
We did not find a class of non-isosceles and non-right triangles (a, b, c) whose three altitudes include one of prime length.
The following table gives the first values (A, a, b, c, Ha, Hb, Hc) where A is the integer area, a, b, c are the sides and Ha <= Hb <= Hc are the altitudes.
+------+-----+-----+-----+----------+----------+---------+
| A | a | b | c | Ha | Hb | Hc |
+------+-----+-----+-----+----------+----------+---------+
| 6 | 3 | 4 | 5 | 12/5 | 3 | 4 |
| 12 | 5 | 5 | 8 | 3 | 24/5 | 24/5 |
| 30 | 5 | 12 | 13 | 5 | 60/13 | 12 |
| 60 | 13 | 13 | 24 | 5 | 120/13 | 120/13 |
| 84 | 7 | 24 | 25 | 168/25 | 7 | 24 |
| 168 | 25 | 25 | 48 | 7 | 336/25 | 336/25 |
| 330 | 11 | 60 | 61 | 660/61 | 11 | 60 |
| 546 | 13 | 84 | 85 | 1092/85 | 13 | 84 |
| 660 | 61 | 61 | 120 | 11 | 1320/61 | 1320/61 |
| 1092 | 85 | 85 | 168 | 13 | 2184/85 | 2184/85 |
| 1224 | 17 | 144 | 145 | 2448/145 | 17 | 144 |
| 1710 | 19 | 180 | 181 | 3420/181 | 19 | 180 |
| 2448 | 145 | 145 | 288 | 4896/145 | 4896/145 | 17 |
+------+-----+-----+-----+----------+----------+---------+

Crossrefs

Programs

  • Maple
    # program using the formula
    lst:={}:for n from 2 to 50 do:p:=ithprime(n):p1:=(p^3-p)/4:p2:=(p^3-p)/2:lst:=lst union {p1} union {p2}:od:print(lst):
  • Mathematica
    nn = 300; lst = {}; Do[s = (a + b + c)/2; area2 = s (s - a) (s - b) (s - c); If[area2>0 && IntegerQ[Sqrt[area2]]&&(PrimeQ[(2*Sqrt[area2])/a]|| PrimeQ[(2*Sqrt[area2])/b]||PrimeQ[(2*Sqrt[area2])/c]), AppendTo[lst, Sqrt[area2]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

Formula

a(n) = (prime(n)^3 - prime(n))/4 for the right triangles;
a(n) = (prime(n)^3 - prime(n))/2 for the isosceles triangles.
Showing 1-2 of 2 results.