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.

A081461 Consider the mapping f(a/b) = (a^2+b^3)/(a^3+b^2) from rationals to rationals. Starting with 1/2 (a=1, b=2) and applying the mapping to each new (reduced) rational number gives 1/2, 9/5, 103/377, ... . Sequence gives values of the numerators.

Original entry on oeis.org

1, 9, 103, 26796621, 236092315725004393, 3561970421302126514421966146019939188025056477849165490630219227287
Offset: 1

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Author

Amarnath Murthy, Mar 22 2003

Keywords

Comments

For the mapping g(a/b) = (a^2+b)/(a+b^2), starting with 1/2 the same procedure leads to the periodic sequence 1/2, 3/5, 1/2, 3/5, ...

Crossrefs

Programs

  • Mathematica
    nxt[{a_,b_}]:=Module[{frac=(a^2+b^3)/(a^3+b^2)},{Numerator[frac], Denominator[ frac]}]; Transpose[NestList[nxt,{1,2},5]][[1]] (* Harvey P. Dale, Nov 09 2011 *)
  • PARI
    {r=1/2; for(n=1,7,a=numerator(r); b=denominator(r); print1(a,","); r=(a^2+b^3)/(a^3+b^2))}

Extensions

Edited and extended by Klaus Brockhaus, Mar 28 2003

A081466 Consider the mapping f(a/b) = (a^2+b^2)/(a^2-b^2) from rationals to rationals. Starting with 2/1 (a=2, b=1) and applying the mapping to each new (reduced) rational number gives 2/1, 5/3, 17/8, 353/225, ... Sequence gives values of the denominators.

Original entry on oeis.org

1, 3, 8, 225, 36992, 6308330625, 21009822254496776192, 3255818067933293622186199316985612890625, 3264008661830516310447364816658205121507617681188862393654856638929469798612992
Offset: 1

Views

Author

Amarnath Murthy, Mar 22 2003

Keywords

Crossrefs

Programs

  • PARI
    {r=2; for(n=1,9,a=numerator(r); b=denominator(r); print1(b,","); r=(a^2+b^2)/(a^2-b^2))}

Extensions

Edited and extended by Klaus Brockhaus, Mar 24 2003

A358210 Congruent number sequence starting from the Pythagorean triple (3,4,5).

Original entry on oeis.org

6, 15, 34, 353, 175234, 9045146753, 121609715057619333634, 4138643330264389621194448797227488932353, 27728719906622802548355602700962556264398170527494726660553210068191276023007234
Offset: 1

Views

Author

Gerry Martens, Nov 04 2022

Keywords

Examples

			Starting with the triple (3,4,5) and choosing the b side we obtain by the recurrence the right triangles: (15/2, 4, 17/2), (136/15, 15/2, 353/30), (5295/136, 272/15, 87617/2040), (47663648/79425, 79425/136, 9045146753/10801800), ...
So a(4) = (5295/136) * (272/15) / 2 = 353.
		

Crossrefs

Cf. A081465 (numerators of hypotenuses).

Programs

  • Mathematica
    nxt[{n_, p_, q_}] := Module[{n1 = Sqrt[p^4 + 4 n^2 q^4], p1 = p Sqrt[p^4 + 4 n^2 q^4], q1 = q^2 n},
      a = p1/q1; b = 2 n1 q1/p1; c = Sqrt[p1^4 + 4 n1^2 q1^4]/(p1 q1);
      Return [{ a b/2, Numerator[b], Denominator[b]}];]
    l = NestList[nxt, {6, 3, 1}, 8] ;
    l[[All, 1]]
Showing 1-3 of 3 results.