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.

A246904 Irregular triangular array: every periodic simple continued fraction CF represents a quadratic irrational (c + f*sqrt(d))/b, where b,c,f,d are integers and d is squarefree. Row n of this array shows the distinct values of d as CF ranges through the periodic continued fractions having period an n-tuple of 1s and 2s.

Original entry on oeis.org

2, 5, 2, 3, 5, 2, 5, 10, 85, 2, 3, 5, 6, 30, 221, 2, 5, 13, 65, 145, 290, 533, 2813, 2, 3, 5, 10, 42, 85, 87, 105, 210, 455, 1365, 1517, 7565, 2, 5, 29, 58, 74, 149, 185, 442, 565, 901, 2026, 2117, 2210, 3973, 10001, 11026, 16133, 18229, 2, 3, 5, 6, 26, 30
Offset: 1

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Author

Clark Kimberling, Sep 06 2014

Keywords

Comments

Is every squarefree positive integer in this array?

Examples

			First 5 rows:
  2 ... 5
  2 ... 3 ... 5
  2 ... 5 ... 10 .. 85
  2 ... 3 ... 5 ... 6 ... 30 ... 221
  2 ... 5 ... 13 .. 65 .. 145 .. 290 .. 533 .. 2813
The following list shows for n = 3 the 2^n purely periodic continued fractions, each followed by the number r it represents, the minimal polynomial a*x^2 + b*x + c of r, the discriminant, D = b^2 - 4*a*c, and the squarefree factor, d, of D.
[(1,1,1)] = (1+sqrt(5))/2, -1 - x + x^2, D = 5
[(1,1,2)] = sqrt(5/2), -5 + 2 x^2, D = 40
[(1,2,1)] = (2 + sqrt(10))/3, -2 - 4 x + 3 x^2, D = 40
[(2,1,1)] = (1 + sqrt(10))/3, -3 - 2 x + 3 x^2, D = 40
[(1,2,2)] = (1 + sqrt(85))/6, -7 - x + 3 x^2, D = 85
[(2,1,2)] = (-1 + sqrt(85))/6, -7 + x + 3 x^2, D = 85
[(2,2,1)] = (5 + sqrt(85))/10, -3 - 5 x + 5 x^2, D = 85
[(2,2,2)] = sqrt(2), -2 + x^2, D = 8
The distinct values of d are 2, 5, 10, 85, as in row 3.
		

Crossrefs

Programs

  • Mathematica
    z = 6;
    t[n_] := t[n] = Map[FromContinuedFraction[{1, #}] &, Tuples[{1, 2}, n]]
    u[n_] := u[n] = Table[MinimalPolynomial[t[k], x], {k, 1, n}]
    d = Discriminant[u[z], x]; v[n_] := Table[{p, m} =
    Transpose[FactorInteger[k]]; Times @@ (p^Mod[m, 2]), {k, d[[n]]}];
    w = Table[Union[Table[v[n], {n, 1, z}][[n]]], {n, 1, z}];
    TableForm[w] (* A246904 array *)
    Flatten[w]   (* A246904 sequence *)

Extensions

Edited by Clark Kimberling, Dec 05 2024