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 1-4 of 4 results.

A195531 Hypotenuses of Pythagorean triples in A195499 and A195503.

Original entry on oeis.org

5, 17, 65, 241, 901, 3361, 12545, 46817, 174725, 652081, 2433601, 9082321, 33895685, 126500417, 472105985, 1761923521, 6575588101, 24540428881, 91586127425, 341804080817, 1275630195845, 4760716702561, 17767236614401, 66308229755041
Offset: 1

Views

Author

Clark Kimberling, Sep 20 2011

Keywords

Comments

See A195500 for discussion and list of related sequences; see A195499 for Mathematica program.
Essentially the same as A120893. - R. J. Mathar, Sep 21 2011

Crossrefs

Formula

Conjecture: a(n) = 3*a(n-1)+3*a(n-2)-a(n-3). G.f.: x*(5+2*x-x^2)/((1+x)*(1-4*x+x^2)). - Colin Barker, Apr 08 2012

A195500 Denominators a(n) of Pythagorean approximations b(n)/a(n) to sqrt(2).

Original entry on oeis.org

3, 228, 308, 5289, 543900, 706180, 1244791, 51146940, 76205040, 114835995824, 106293119818725, 222582887719576, 3520995103197240, 17847666535865852, 18611596834765355, 106620725307595884, 269840171418387336, 357849299891217865
Offset: 1

Views

Author

Clark Kimberling, Sep 20 2011

Keywords

Comments

For each positive real number r, there is a sequence (a(n),b(n),c(n)) of primitive Pythagorean triples such that the limit of b(n)/a(n) is r and
|r-b(n+1)/a(n+1)| < |r-b(n)/a(n)|. Peter Shiu showed how to find (a(n),b(n)) from the continued fraction for r, and Peter J. C. Moses incorporated Shiu's method in the Mathematica program shown below.
Examples:
r...........a(n)..........b(n)..........c(n)
sqrt(2).....A195500.......A195501.......A195502
sqrt(3).....A195499.......A195503.......A195531
sqrt(5).....A195532.......A195533.......A195534
sqrt(6).....A195535.......A195536.......A195537
sqrt(8).....A195538.......A195539.......A195540
sqrt(12)....A195680.......A195681.......A195682
e...........A195541.......A195542.......A195543
pi..........A195544.......A195545.......A195546
tau.........A195687.......A195688.......A195689
1...........A046727.......A084159.......A001653
2...........A195614.......A195615.......A007805
3...........A195616.......A195617.......A097315
4...........A195619.......A195620.......A078988
5...........A195622.......A195623.......A097727
1/2.........A195547.......A195548.......A195549
3/2.........A195550.......A195551.......A195552
5/2.........A195553.......A195554.......A195555
1/3.........A195556.......A195557.......A195558
2/3.........A195559.......A195560.......A195561
1/4.........A195562.......A195563.......A195564
5/4.........A195565.......A195566.......A195567
7/4.........A195568.......A195569.......A195570
1/5.........A195571.......A195572.......A195573
2/5.........A195574.......A195575.......A195576
3/5.........A195577.......A195578.......A195579
4/5.........A195580.......A195611.......A195612
sqrt(1/2)...A195625.......A195626.......A195627
sqrt(1/3)...{1}+A195503...{0}+A195499...{1}+A195531
sqrt(2/3)...A195631.......A195632.......A195633
sqrt(3/4)...A195634.......A195635.......A195636

Examples

			For r=sqrt(2), the first five fractions b(n)/a(n) can be read from the following five primitive Pythagorean triples (a(n), b(n), c(n)) = (A195500, A195501, A195502):
(3,4,5); |r - b(1)/a(1)| = 0.08...
(228,325,397); |r - b(2)/a(2)| = 0.011...
(308,435,533); |r - b(3)/a(3)| = 0.0018...
(5289,7480,9161); |r - b(4)/a(4)| = 0.000042...
(543900,769189,942061); |r - b(5)/a(5)| = 0.0000003...
		

Crossrefs

Programs

  • Maple
    Shiu := proc(r,n)
            t := r+sqrt(1+r^2) ;
            cf := numtheory[cfrac](t,n+1) ;
            mn := numtheory[nthconver](cf,n) ;
            (mn-1/mn)/2 ;
    end proc:
    A195500 := proc(n)
            Shiu(sqrt(2),n) ;
            denom(%) ;
    end proc: # R. J. Mathar, Sep 21 2011
  • Mathematica
    r = Sqrt[2]; z = 18;
    p[{f_, n_}] := (#1[[2]]/#1[[
          1]] &)[({2 #1[[1]] #1[[2]], #1[[1]]^2 - #1[[
             2]]^2} &)[({Numerator[#1], Denominator[#1]} &)[
         Array[FromContinuedFraction[
            ContinuedFraction[(#1 + Sqrt[1 + #1^2] &)[f], #1]] &, {n}]]]];
    {a, b} = ({Denominator[#1], Numerator[#1]} &)[
      p[{r, z}]]  (* A195500, A195501 *)
    Sqrt[a^2 + b^2] (* A195502 *)

A102341 Areas of 'close-to-equilateral' integer triangles.

Original entry on oeis.org

12, 120, 1848, 25080, 351780, 4890480, 68149872, 949077360, 13219419708, 184120982760, 2564481115560, 35718589344360, 497495864091732, 6929223155685600, 96511629630137568, 1344233586759971040, 18722758603319903340
Offset: 1

Views

Author

Johannes Koelman (Joc_Kay(AT)hotmail.com), Feb 20 2005

Keywords

Comments

A close-to-equilateral integer triangle is defined to be a triangle with integer sides and integer area such that the largest and smallest sides differ in length by unity. The first five close-to-equilateral integer triangles have sides (5, 5, 6), (17, 17, 16), (65, 65, 66), (241, 241, 240) and (901, 901, 902).
After these first five triangles, there are two more (namely (3361,3361,3360,4890480) and (12545,12545,12546,68149872)). - NĂ­colas V. Calsavara, Jul 13 2023
Then, the next four terms are {three sides a<=b<=c and area}: {46816, 46817, 46817, 949077360}, {174725, 174725, 174726, 13219419708}, {652080, 652081, 652081, 184120982760}, {2433601, 2433601, 2433602, 2564481115560}. Also, if we allow degenerate triangles (area 0), the first case would be {1,1,2,0}. We have 12 cases and a weak conjecture is that the total number of the 'close-to-equilateral' triangles is finite. - Zak Seidov, Feb 23 2005
This is an infinite series; two sides are equal in length to the hypotenuse of almost 30-60 triangles and the third side alternates between that length +/- 1. - Dan Sanders (dan(AT)ified.ca), Oct 22 2005
Heron's formula: a triangle with side lengths (x,y,z) has area A = sqrt(s*(s-x)*(s-y)*(s-z)) where s = (x+y+z)/2. For this sequence we assume integer side-lengths x = y = z +/- 1. Then for A to also be an integer, x+y+z must be even, so we can assume z = 2k for some positive integer k. Now s = (x+y+z)/2 = 3k +/- 1 and A = sqrt((3*k +/- 1)*k*k*(k +/- 1)) = k*sqrt(3*k^2 +/- 4*k + 1). To determine when this is an integer, set 3*k^2 +/- 4*k + 1 = d^2. If we multiply both sides by 3, it is easier to complete the square: (3*k +/- 2)^2 - 1 = 3*d^2. Now we are looking for solutions to the Pell equation c^2 - 3*d^2 = 1 with c = 3*k +/- 2, for which there are infinitely many solutions: use the upper principal convergents of the continued fraction expansion of sqrt(3) (A001075/A001353). - Danny Rorabaugh, Oct 16 2015

Examples

			a(2) = 120 because 120 is the area of a triangle with side lengths of 16, 17 and 17.
		

Crossrefs

For the continued fraction expansion of sqrt(3), cf. A002530, A002531, A040001.

Formula

(2/3) [ A007655(n+2) - (-1)^n*A001353(n+1) ] (conjectured). - Ralf Stephan, May 17 2007
Empirical g.f.: 12*x / ((x^2-14*x+1)*(x^2+4*x+1)). - Colin Barker, Apr 10 2013
a(n) = A001353(n+1)*A195499(n) = A001353(n+1)*A120892(n+1) - Danny Rorabaugh, Oct 16 2015

Extensions

More terms from Zak Seidov, Feb 23 2005
More terms from Dan Sanders (dan(AT)ified.ca), Oct 22 2005

A195503 Numerators b(n) of Pythagorean approximations b(n)/a(n) to sqrt(3).

Original entry on oeis.org

4, 15, 56, 209, 780, 2911, 10864, 40545, 151316, 564719, 2107560, 7865521, 29354524, 109552575, 408855776, 1525870529, 5694626340, 21252634831, 79315912984, 296011017105, 1104728155436, 4122901604639, 15386878263120, 57424611447841
Offset: 1

Views

Author

Clark Kimberling, Sep 20 2011

Keywords

Comments

See A195500 for discussion and list of related sequences; see A195499 for Mathematica program.
Essentially the same as A001353, A106707, A125905 and likely also A010905. - R. J. Mathar, Sep 21 2011
If started 1, 4, 15, 56, ...., denominators a(n) of Pythagorean approximations b(n)/a(n) to sqrt(1/3). - Clark Kimberling, Sep 22 2011

Crossrefs

Formula

a(n) = A001353(n+1). - R. J. Mathar, Sep 21 2011
Empirical g.f.: x*(4-x)/(1-4*x+x^2). - Colin Barker, Jan 04 2012
Showing 1-4 of 4 results.