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.

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A057228 a(n) = u * v = x * y with (u - v) = (x + y) = A009000(n) (u>v, y>0, v>0, x>0, y>0).

Original entry on oeis.org

6, 24, 30, 54, 60, 96, 84, 150, 120, 210, 216, 240, 294, 210, 270, 384, 180, 486, 336, 600, 540, 480, 630, 726, 840, 864, 330, 504, 750, 924, 1014, 960, 1176, 1320, 840, 756, 1350, 1080, 1536, 720, 546, 1386, 1500, 1734, 1890, 1560, 1944, 1470, 2166
Offset: 1

Views

Author

Naohiro Nomoto, Sep 19 2000

Keywords

Comments

Areas of Pythagorean triangles.

Examples

			a(1) = 6 = 6 * 1 = 3 * 2, (6 - 1)=(3 + 2) = 5 = A009000(1).

Links

Crossrefs

Cf. A009000, A009003, A009111, A009112.

Programs

  • PARI
    list(lim) = {my(lh = List()); for(u = 2, sqrtint(lim), for(v = 1, u, if (u^2+v^2 > lim, break); if ((gcd(u, v) == 1) && (0 != (u-v)%2), for (i = 1, lim, if (i*(u^2+v^2) > lim, break); /* if (u^2 - v^2 < 2*u*v, w = [i*(u^2 - v^2), i*2*u*v, i*(u^2+v^2)], w = [i*2*u*v, i*(u^2 - v^2), i*(u^2+v^2)]); */ listput(lh, [i*(u^2+v^2), i^2*(u^2 - v^2)*u*v]); ); ); ); ); lh = vecsort(Vec(lh)); vector(#lh, i, lh[i][2])} \\ David A. Corneth, Apr 10 2021, adapted from A009000

Extensions

Offset changed to 1 by David A. Corneth, Apr 10 2021

A283274 Areas of triples of primitive Pythagorean triangles having the same area.

Original entry on oeis.org

13123110, 2203385574390, 2570042985510, 8943387723270, 826290896699730, 9381843970167926138271390
Offset: 1

Views

Author

Duncan Moore, Mar 04 2017

Keywords

Comments

The generators for the currently known triples are:
a(1): (5,138),(38,77),(55,78): Charles Shedd, 1945
a(2): (55,3422),(61,3306),(266,2035): Randall L Rathbun, 1986
a(3): (143,2622),(869,1610),(1817,2002): Randall L Rathbun, 1986
a(4): (198,3565),(1166,2201),(2035,2438): Randall L Rathbun, 1986
a(5): (731,10434),(1122,9077),(2465,7238): Dan Hoey, May 18 1990
a(6): (352538,2999447),(1931103,2398838),(3063347,3215070): Duncan Moore, Mar 01 2017
The generator (a,b) gives the Pythagorean triangle (b^2+a^2,b^2-a^2,2ab) with area ab(b^2-a^2).

Examples

			The generators of a(1) give the 3 Pythagorean triangles (19069,19019,1380), (7373,4485,5852) and (9109,3059,8580). They have the areas 19019*1380/2 = 4485*5852/2 = 3059*8580/2 = 13123110 = a(1).

References

  • R. K. Guy, Unsolved Problems in Number Theory, D21.

Links

  • Shyam Sunder Gupta, Number Curiosities, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 23, 567-604.

Crossrefs

Cf. A009111.

A057101 Area of a Pythagorean triangle (ordered by the product of the sides).

Original entry on oeis.org

6, 24, 30, 54, 60, 96, 84, 120, 150, 210, 216, 180, 210, 240, 294, 270, 384, 336, 330, 486, 480, 540, 600, 504, 630, 726, 546, 840, 750, 864, 756, 720, 924, 840, 960, 1014, 1176, 1080, 840, 1320, 990, 1350, 1386, 1536, 1500, 1470, 1344, 1560, 1734, 1320
Offset: 1

Views

Author

Henry Bottomley, Aug 01 2000

Keywords

Examples

			a(1)=3*4/2=6 since 3*4*5=60 is smallest possible positive product

Crossrefs

Cf. A009111, A009112, A057096.

Formula

a(n) =A057096(n)/(2*A057100(n)) =A057098(n)*A057099(n)/2

A231328 Integer areas of the reflection triangles of integer-sided triangles.

Original entry on oeis.org

18, 72, 90, 162, 180, 252, 288, 360, 450, 540, 630, 648, 720, 810, 882, 990, 1008, 1152, 1440, 1458, 1512, 1620, 1638, 1800, 1890, 2160, 2178, 2250, 2268, 2520, 2592, 2772, 2880, 2970, 3042, 3240, 3528, 3672, 3960, 4032, 4050, 4158, 4410, 4500, 4608, 4680, 4860
Offset: 1

Views

Author

Michel Lagneau, Nov 07 2013

Keywords

Comments

The triangle A'B'C' obtained by reflecting the vertices of a reference triangle ABC about the opposite sides is called the reflection triangle (Grinberg 2003).
The area of the reflection triangle is given by
A' = A*t/(a^2*b^2*c^2) where A is the area of the reference triangle of sides (a, b, c) and
t=-(a^6-b^2*a^4-c^2*a^4-b^4*a^2-c^4*a^2-b^2*c^2*a^2+b^6+c^6-b^2*c^4-b^4*c^2)/(a^2*b^2*c^2).
See the link for the side lengths of the reflection triangles.
Properties of this sequence:
The areas corresponding to the primitive reflection triangles are 18, 90, 180, 252, 540,...
The non-primitive triangles of areas 4*a(n),9*a(n),...,p^2*a(n),... are in the sequence.
It appears that one of the side of the reflection triangles equals the greatest side of the initial triangle (see the table below), and the initial triangles are Pythagorean triangles => a(n) = 3*A009112(n).
The following table gives the first values (A, A', a, b, c, a', b', c') where A' is the area of the reflection triangles, A is the area of the initial triangles, a, b, c are the integer sides of the initial triangles, and a', b', c' are the sides of the reflection triangles.
-------------------------------------------------------------------------
| A' | A | a | b | c | a' | b' | c'|
-------------------------------------------------------------------------
| 18 | 6 | 3 | 4 | 5 | 9*sqrt(17)/5 | 4*sqrt(97)/5 | 5 |
| 72 | 24 | 6 | 8 | 10 | 18*sqrt(17)/5 | 8*sqrt(97)/5 | 10 |
| 90 | 30 | 5 | 12 | 13 | 5*sqrt(1321)/13 | 36*sqrt(41)/13 | 13 |
| 162 | 54 | 9 | 12 | 15 | 27*sqrt(17)/5 | 12*sqrt(97)/5 | 15 |
| 180 | 60 | 8 | 15 | 17 | 8*sqrt(2089)/17 | 45*sqrt(89)/17 | 17 |
| 252 | 84 | 7 | 24 | 25 | 7*sqrt(5233)/25 | 72*sqrt(113)/25 | 25 |
| 288 | 96 | 12 | 16 | 20 | 36*sqrt(17)/5 | 16*sqrt(97)/5 | 20 |
| 360 | 120 | 10 | 24 | 26 | 10*sqrt(1321)/13 | 72*sqrt(41)/13 | 26 |
| 450 | 150 | 15 | 20 | 25 | 9*sqrt(17) | 4*sqrt(97) | 25 |
| 540 | 180 | 9 | 40 | 41 | 27*sqrt(1609)/41 | 40*sqrt(2329)/41 | 41 |
| 630 | 210 | 12 | 35 | 37 | 36*sqrt(1241)/37 | 35*sqrt(2521)/37 | 37 |
| 648 | 216 | 18 | 24 | 30 | 54*sqrt(17)/5 | 24*sqrt(97)/5 | 30 |
.......................................................................

Examples

			18 is in the sequence. We use two ways:
First way: with the triangle (3, 4, 5) the formula A' = A*t/(a^2*b^2*c^2) gives directly the result: A'= 18 where the area A = 6 is obtained by Heron's formula A =sqrt(s*(s-a)*(s-b)*(s-c))= sqrt(6*(6-3)*(6-4)*(6-5)) = 6, where s is the semiperimeter.
Second way: by calculation of the sides a', b', c' and by using Heron's formula. We obtain from the formulas given in the link:
a' = 9*sqrt(17)/5;
b' = 4*sqrt(97/5);
c' = 5.
Now, we use Heron's formula with (a',b',c'). We find A'=sqrt(s1*(s1-a')*(s1-b')*(s1-c')) with:
s1 =(a'+b'+c')/2 = (9*sqrt(17)/5+ 4*sqrt(97/5)+ 5)/2. We find A'= 18.

References

  • D. Grinberg, On the Kosnita Point and the Reflection Triangle, Forum Geom. 3, 105-111, 2003.

Links

Crossrefs

Cf. A188158, A009111, A009112.

Programs

  • Mathematica
    nn = 300 ; lst = {}; Do[s = (a + b + c)/2 ; If[IntegerQ[s],area2 = s (s-a)(s-b) (s-c); If[area2 > 0 && IntegerQ[Sqrt[area2] + (a^2 + b^2 + c^2)/8], AppendTo[lst, Sqrt[area2] + (a^2 + b^2 + c^2)/8]]],{a,nn},{b,a},{c,b}] ; Union[lst]
Previous Showing 11-14 of 14 results.

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