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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A246921 Irregular triangular array: row n gives numbers D, each being the discriminant of the minimal polynomial of a quadratic irrational represented by a continued fraction with period an n-tuple of 1s and 3s.

Original entry on oeis.org

5, 13, 5, 13, 21, 5, 13, 17, 65, 5, 13, 21, 165, 725, 2805, 5, 13, 445, 1525, 1853, 5933, 7925, 30629, 5, 13, 17, 21, 65, 136, 288, 960, 1260, 4224, 16128, 21608, 83520, 5, 13, 3029, 10205, 11029, 12773, 28157, 34973, 42853, 47965, 53365, 136165, 184045
Offset: 1

Views

Author

Clark Kimberling, Sep 07 2014

Keywords

Examples

			First 5 rows:
  5 ... 13
  5 ... 13 ... 21
  5 ... 13 ... 17 .. 65
  5 ... 13 ... 21 .. 165 .. 725 .. 2805
  5 ... 13 ... 445 . 1525 . 1853 . 5933 . 7925 . 30629
The following list shows for n = 3 the purely periodic continued fractions (with period an n-tuple of 1s and 2s), each followed by the number r it represents, the minimal polynomial a*x^2 + b*x + c of r, and the discriminant, D = b^2 - 4*a*c.
[(1,1,1)] = (1+sqrt(5))/2, -1 - x + x^2, D = 5
[(1,1,3)] = (-1 + sqrt(17))/2, -4 + x + x^2, D = 17
[(1,3,1)] = (3 + sqrt(17))/4, -1 - 3 x + 2 x^2, D = 17
[(3,1,1)] = (1 + sqrt(17))/4, -2 - x + 2 x^2, D = 17
[(1,3,3)] = (-1 + sqrt(65))/4, -8 + x + 2 x^2, D = 65
[(3,1,3)] = (-3 + sqrt(65))/4, -7 + 3 x + 2 x^2, D = 65
[(3,3,1)] = (5 + sqrt(65))/10, -2 - 5 x + 5 x^2, D = 65
[(3,3,3)] = (-1 + sqrt(13))/2, -3 + x + x^2, D = 13
The distinct values of D are 5, 13, 17, 65, as in row 3.
		

Crossrefs

Programs

  • Mathematica
    z = 7; u[n_] := u[n] = Table[MinimalPolynomial[Map[FromContinuedFraction[{1, #}] &, Tuples[{1, 3}, k]], x], {k, 1, n}]; d = Discriminant[u[z], x];
    t = Table[Union[d[[n]]], {n, 1, z}]; TableForm[t] (* A246921 array *)
    Flatten[t] (* A246921 sequence *)

Extensions

Edited by Clark Kimberling, Dec 05 2024

A378873 Squarefree part of A378872(n) (the discriminant of the minimal polynomial of a number whose continued fraction expansion has periodic part given by the n-th composition (in standard order)).

Original entry on oeis.org

5, 2, 5, 13, 3, 3, 5, 5, 21, 2, 10, 21, 10, 10, 5, 29, 2, 15, 17, 15, 85, 85, 6, 2, 17, 85, 6, 17, 6, 6, 5, 10, 5, 6, 26, 13, 37, 37, 165, 6, 37, 2, 221, 37, 3, 221, 65, 5, 26, 37, 165, 37, 221, 3, 65, 26, 165, 221, 65, 165, 65, 65, 5, 53, 15, 35, 37, 3, 229
Offset: 1

Views

Author

Pontus von Brömssen, Dec 10 2024

Keywords

Comments

Any number x whose continued fraction expansion is eventually periodic can be written uniquely as x = (c+f*sqrt(d))/b, where b, c, f, d are integers, b > 0, d > 0 is squarefree, and GCD(b,c,f) = 1. a(n) is equal to d when the periodic part of the continued fraction of x is given by the n-th composition. If two numbers have eventually periodic continued fraction expansions with the same periodic part, their respective values of d are the same.

Examples

			For n = 6, the 5th composition is (1,2). The value of the continued fraction 1+1/(2+1/(1+1/(2+...))) is (1+sqrt(3))/2, so a(6) = 3.
		

Crossrefs

Cf. A007913, A066099 (compositions in standard order), A246904, A246922, A259911, A259912, A305311, A378872, A378874.

Formula

a(n) = A007913(A378872(n)) = A378872(n)/A378874(n)^2.
a(2^n) = A259912(n+1) if a(2^n) == 1 (mod 4), a(2^n) = A259912(n+1)/4 otherwise.
For n > k >= 0, a(2^n+2^k) = A259911(n,k+1) if a(2^n+2^k) == 1 (mod 4), a(2^n+2^k) = A259911(n,k+1)/4 otherwise.
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