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 1-3 of 3 results.

A232461 Integer areas of integer-sided triangles where two sides are of square length.

Original entry on oeis.org

120, 168, 300, 360, 1920, 2016, 2688, 4680, 4800, 5760, 9720, 10140, 13608, 14280, 18720, 19080, 23256, 24300, 29160, 30720, 32760, 34440, 34680, 38640, 42120, 43008, 57720, 74880, 75000, 76800, 92160, 94080, 105000, 128700, 162240, 177072, 187500, 217728
Offset: 1

Views

Author

Michel Lagneau, Nov 24 2013

Keywords

Comments

Subset of A188158.
The areas of the triangles (a,b,c) are given by Heron's formula, A = sqrt(s(s-a)(s-b)(s-c)), where its side lengths are a, b, c and semiperimeter s = (a+b+c)/2.
The areas A of the primitive triangles of sides (a,b,c) are 120, 168, 300, 360, 4680, ...
The areas of the nonprimitive triangle of sides (a*p^2, b*p^2, c*p^2) are in the sequence with the value A*p^4.
It is possible to find integer-sided triangles having two square sides, for example:
a(2) = 168 with sides (25,25,48) and (14,25,25);
a(3) = 300 with sides (25,25,30) and (25,25,40);
a(14) = 14280 with sides (169,169,238), (169,169,240), (100,289,291).
The following table gives the first values (A, a, b, c):
+-------+-----+-----+-----+
| A | a | b | c |
+-------+-----+-----+-----+
| 120 | 16 | 25 | 39 |
| 168 | 14 | 25 | 25 |
| 300 | 25 | 25 | 40 |
| 360 | 25 | 29 | 36 |
| 1920 | 64 | 100 | 156 |
| 2016 | 64 | 225 | 287 |
| 2688 | 100 | 100 | 192 |
| 4680 | 74 | 169 | 225 |
| 4800 | 100 | 100 | 160 |
| 5760 | 100 | 116 | 144 |
| 9720 | 144 | 225 | 351 |
| 10140 | 169 | 169 | 312 |
| 13608 | 225 | 225 | 432 |
+-------+-----+-----+-----+

Examples

			120 is in the sequence because the triangle (4^2, 5^2, 39) has semiperimeter s = (16+25+39)/2 = 40, and A = sqrt(40*(40-16)*(40-25)*(40-39)) = 120.
		

Crossrefs

Programs

  • Mathematica
    nn=1000;lst={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a) (s-b) (s-c);If[0 0 && IntegerQ[ar]]; nn = 80; t = {}; ps = sqr[Range[2, nn]]; mx = 3*ps[[-1]]; Do[If[p <= q && goodQ[p, q, e], aa = area[p, q, e]; If[aa <= mx, AppendTo[t, aa]]], {p, ps}, {q, ps}, {e, q - p + 2, p + q - 2, 2}]; t = Union[t] (* program from T. D. Noe adapted for this sequence - see A229746 *)

A227112 Numbers n such that there exist two primes p and q where the area A of the triangle of sides (n, p, q) is an integer.

Original entry on oeis.org

4, 6, 8, 10, 12, 16, 18, 20, 22, 24, 26, 30, 38, 40, 42, 50, 56, 60, 68, 70, 78, 80, 90, 96, 100, 102, 104, 110, 120, 130, 144, 148, 150, 156, 160, 170, 174, 180, 182, 198, 210, 224, 234, 240, 286, 290, 300, 312, 350, 360, 370, 390, 400, 440, 510, 520, 548
Offset: 1

Views

Author

Michel Lagneau, Oct 02 2013

Keywords

Comments

n is an even composite number because the perimeter of the triangle (n, p, q) is always even. The corresponding areas are {6, 12, 12, 60, 30, 120, 360, 66, ...}
The area is given by Heron's formula A = sqrt(s(s-n)(s-p)(s-q)) where the semiperimeter s = (n + p + q)/2.
The following table gives the first values (n, A, p, q).
+----+-----+----+----+
| n | A | p | q |
+----+-----+----+----+
| 4 | 6 | 3 | 5 |
| 6 | 12 | 5 | 5 |
| 8 | 12 | 5 | 5 |
| 10 | 60 | 13 | 13 |
| 12 | 30 | 5 | 13 |
| 16 | 120 | 17 | 17 |
| 18 | 360 | 41 | 41 |
| 20 | 66 | 11 | 13 |
...

Examples

			12 is in the sequence because the triangle (12, 5, 13) => semiperimeter s = (12+5+13)/2 = 15, and A = sqrt(15*(15-12)*(15-5)*(15-13))= 30, with 5 and 13 prime numbers.
		

Crossrefs

Programs

  • Mathematica
    area[a_, b_, c_] := Module[{s = (a + b + c)/2, a2}, a2 = s (s - a) (s - b) (s - c); If[a2 < 0, 0, Sqrt[a2]]]; goodQ[a_, b_, c_] := Module[{ar = area[a, b, c]}, ar > 0 && IntegerQ[ar]]; nn = 300; t = {}; ps = Prime[Range[2, nn]]; mx = 3*ps[[-1]]; Do[If[p <= q && goodQ[p, q, e], aa = area[p, q, e]; If[aa <= mx, AppendTo[t, e]]], {p, ps}, {q, ps}, {e, q - p + 2, p + q - 2, 2}]; t = Union[t] (* program from T. D. Noe adapted for this sequence - see A229746 *)

A283446 Integer areas of triangles with side lengths A001223(m), A001223(m+1) and A001223(m+2) for some m.

Original entry on oeis.org

24, 96, 120, 144, 168, 216, 240, 264, 336, 360, 384, 432, 456, 480, 504, 528, 576, 600, 624, 672, 720, 792, 816, 840, 864, 936, 960, 1008, 1056, 1080, 1176, 1200, 1224, 1296, 1320, 1344, 1440, 1512, 1536, 1560, 1584, 1680, 1728, 1824, 1848, 1920, 1944, 2016
Offset: 1

Views

Author

Michel Lagneau, Mar 08 2017

Keywords

Comments

A001223(n) = A000040(n+1) - A000040(n) = prime(n+1) - prime(n).
The Mathematica program examines all triangles with n <= 10^8.
The sequence a(n) is a subsequence of A188158, and the lengths of the sides are even.
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.
a(n) == 0 mod 24 => {b(n)} = {a(n)/24} = {1, 4, 5, 6, 7, 9, 10, 11, 14, 15, 16, 18, 19, 20, 21, 22, 24, 25, 26, 28, 30, 33, 34, 35, 36, 39, 40, 42, 44, 45, 49, 51, 54, 55, 56, 60, 63, 64, 65, 70, 72, ...}. It seems that the primes > 19 are not in {b(n)}.
For the same area, the number of distinct triangles is not always unique; for example, the area 336 can be obtained with triangle (30, 28, 26) starting from prime 461983 and also from triangle (34, 20, 42) starting from prime 2473663 (Giovanni Resta, Mar 08 2017).
The following table gives the first values (A, m, sides of the triangles) where A is the area of the triangles and m is the smallest value generating A.
+-----+--------+-----------+-------------+-------------+
| A | m | A001223(m)| A001223(m+1)| A001223(m+2)|
+-----+--------+-----------+-------------+-------------+
| 24 | 123 | 6 | 8 | 10 |
| 96 | 3935 | 16 | 20 | 12 |
| 120 | 8101 | 10 | 26 | 24 |
| 144 | 13097 | 34 | 18 | 20 |
| 168 | 12226 | 30 | 40 | 14 |
| 216 | 9864 | 24 | 18 | 30 |
| 240 | 102715 | 58 | 50 | 12 |
| 264 | 98259 | 22 | 26 | 40 |
| 336 | 38604 | 30 | 28 | 26 |
+-----+--------+-----------+-------------+-------------+

Examples

			24 is in the sequence because, for the smallest value m = 123, we obtain the triangle of sides (A001223(123), A001223(124), A001223(125)) = (6, 8, 10) and the area is given by Heron's formula with s = 12 and A = sqrt(12(12-6)(12-8)(12-10)) = 24.
The set of the others values m > 123 giving the same area A = 24 starts with 127, 192, 269, 304, 417, 420, ...
		

Crossrefs

Programs

  • Mathematica
    nn=10^5;lst={};Do[u=Prime[a+1]-Prime[a];v=Prime[a+2]-Prime[a+1]; w=Prime[a+3]-Prime[a+2];s=(u+v+w)/2;If[IntegerQ[s],area2=s (s-u)(s-v)(s-w);If[area2>0&&IntegerQ[Sqrt[area2]],AppendTo[lst,Sqrt[area2]]]],{a,nn}];Union[lst]

Extensions

Missing terms 1200 and 1584 from Giovanni Resta, Mar 08 2017
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