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.

A305310 Numbers k(n) used for Cassels's Markoff forms MF(n) corresponding to the conjectured unique Markoff triples MT(n) with maximal entry m(n) = A002559(n), for n >= 1.

Original entry on oeis.org

0, 1, 2, 5, 12, 13, 34, 70, 75, 89, 179, 233, 408, 507, 610, 1120, 1597, 2378, 2673, 2923, 3468, 4181, 6089, 10946, 13860, 15571, 16725, 19760, 23763, 28657, 39916, 51709, 80782, 75025, 113922, 162867, 206855, 196418, 249755, 353702
Offset: 1

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Author

Wolfdieter Lang, Jun 26 2018

Keywords

Comments

For these Markoff forms see Cassels, p. 31. A link to the two original Markoff references is given in A305308.
MF(n) = f_{m(n)}(x, y) = m(n)*F_{m(n)}(x, y) = m(n)*x^2 + (3*m(n) - 2*k(n))*x*y + (l(n) - 3*k(n))*y^2, with the Markoff number m = m(n) = A002559(n) and l(n) = (k(n)^2 + 1)/m(n), for n >= 1.
Every m(n) is proved to appear as largest member of a Markoff triple MT(n) = (m_1(n), m_2(n), m(n)), with positive integers m_1(n) < m_2(n) < m(n) for n >= 3 (MT(1) = (1, 1, 1) and MT(2) = (1, 1, 2)) satisfying the Markoff equation m_1(n)^2 + m_2(n)^2 + m(n)^2 = 3*m_1(n)*m_2(n)*m(n). The famous Markoff uniqueness conjecture is that m(n) as largest member determines exactly one ordered triple MT(n). See, e.g., the Aigner reference, pp. 38-39, and Corollary 3.5, p. 48. [In numerating the sequence with n related to A002559(n) this conjecture is assumed to be true. - Wolfdieter Lang, Jul 29 2018]
The nonnegative integers k(n) are defined for the Markoff forms given by Cassels by k(n) = min{k1(n), k2(n)}, where m_1(n)*k1(n) - m_2(n) == 0 (mod m(n)), with 0 <= k1(n) < m(n), and m_2(n)*k2(n) - m_1(n) == 0 (mod m(n)), with 0 <= k2(n) < m(n). The k1 and k2 sequences are k1 = [0, 1, 2, 5, 17, 13, 34, 99, 119, 89, 179, 233, 577, 818, 610, 1777, 1597, 3363, 2673, 2923, 5609, 4181, 6089, 10946, 19601, 22095, 26536, 31881, 38447, 28657, 39916, 51709, 114243, 75025, 113922, 263522, 206855, 196418, 396263, 572063, ...], and k2 = [0, 1, 3, 8, 12, 21, 55, 70, 75, 144, 254, 377, 408, 507, 987, 1120, 2584, 2378, 3793, 4638, 3468, 6765, 8612, 17711, 13860, 15571, 16725, 19760, 23763, 46368, 56641, 83428, 80782, 121393, 180763, 162867, 292538, 317811, 249755, 353702, ...].
The discriminant of the form MF(n) = f_{m(n)}(x, y) is D(n) = 9*m(n)^2 - 4. D(n) = A305312(n), for n >= 1. Because D(n) > 0 (not a square) this is an indefinite binary quadratic form, for n >= 1. See Cassels Fig. 2 on p. 32 for the Markoff tree with these forms.
The quadratic irrational xi, determined by the solution with positive square root of f_{m(n)}(x, 1) = 0, is xi(n) = ((2*k - 3*m) + sqrt(D))/(2*m) (the argument n has been dropped). The regular continued fraction is eventually periodic, but not purely periodic. One can find equivalent Markoff forms determining purely periodic quadratic irrationals. The corresponding k sequence is given in A305311.
For the approximation of xi(n) with infinitely many rationals (in lowest terms) Perron's unimodular invariant M(xi) enters. For quadratic irrationals M(xi) < 3, and the values coincide with the discrete Lagrange spectrum < 3: M(xi(n)) = Lagrange(n) = sqrt{D(n)}/m(n), n >= 1. For n=1..4 see A002163, A010466, A200991 and A305308.

Examples

			n = 5: a(5) = k(5) = 12 because m(5) = A002559(5) = 29 with the triple MT(5) = (2, 5, 29). Whence 2*k1(5) - 5 == 0 (mod 29) for k1(5) = 17 < 29, and 5*k2(5) - 2 == 0 (mod 29) leads to k2(5) = 12. The smaller value is k2(5) = k(5) = 12. This leads to the form coefficients MF(5) = [29, 63, -31].
The forms MF(n) = [m(n), 3*m(n) - k(n), l(n) - 3*k(n)] with l(n) := (k(n)^2 + 1)/m(n) begin: [1, 3, 1], [2, 4, -2], [5, 11, -5], [13, 29, -13], [29, 63, -31], [34, 76, -34], [89, 199, -89], [169, 367, -181], [194, 432, -196], [233, 521, -233], [433, 941, -463], [610, 1364, -610], [985, 2139, -1055], [1325, 2961, -1327], [1597, 3571, -1597], [2897, 6451, -2927], [4181, 9349, -4181], [5741, 12467, -6149], [6466, 14052, -6914], [7561, 16837, -7639] ... .
The quadratic irrationals xi(n) = ((2*k(n) - 3*m(n)) + sqrt(D(n)))/(2*m(n)) begin: (-3 + sqrt(5))/2, -1 + sqrt(2), (-11 + sqrt(221))/10, (-29 + sqrt(1517))/26, (-63 + sqrt(7565))/58, (-19 + 5*sqrt(26))/17, (-199 + sqrt(71285))/178, (-367 + sqrt(257045))/338, (-108 + sqrt(21170))/97, (-521 + sqrt(488597))/466, (-941 + sqrt(1687397))/866, (-341 + sqrt(209306))/305, (-2139 + sqrt(8732021))/1970, (-2961 + sqrt(15800621))/2650, (-3571 + sqrt(22953677))/3194, (-6451 + sqrt(75533477))/5794, (-9349 + sqrt(157326845))/8362, (-12467 + 5*sqrt(11865269))/11482, (-3513 + 5*sqrt(940706))/3233, (-16837 + sqrt(514518485))/15122, ... .
The invariant M(xi(n)) = Lagrange(n) numbers begin with n >=1: sqrt(5), 2*sqrt(2), (1/5)*sqrt(221), (1/13)*sqrt(1517), (1/29)*sqrt(7565), (10/17)*sqrt(26), (1/89)*sqrt(71285), (1/169)*sqrt(257045), (2/97)*sqrt(21170), (1/233)*sqrt(488597), (1/433)*sqrt(1687397), (2/305)*sqrt(209306), (1/985)*sqrt(8732021), (1/1325)*sqrt(15800621), (1/1597)*sqrt(22953677), (1/2897)*sqrt(75533477), (1/4181)*sqrt(157326845), (5/5741)*sqrt(11865269), (10/3233)*sqrt(940706), (1/7561)*sqrt(514518485), ... .
		

References

  • Martin Aigner, Markov's Theorem and 100 Years of the Uniqueness Conjecture, Springer, 2013.
  • J. W. S. Cassels, An Introduction to Diophantine Approximation, Cambridge University Press, 1957, Chapter II, The Markoff Chain, pp. 18-44.
  • Julian Havil, The Irrationals, Princeton University Press, Princeton and Oxford, 2012, pp. 172-180 and 222-224.
  • Oskar Perron, Über die Approximation irrationaler Zahlen durch rationale, Sitzungsber. Heidelberger Akademie der Wiss., 1921, 4. Abhandlung, pp. 1-17 , and part II., 8. Abhandlung, pp.1-12, Carl Winters Universitätsbuchhandlung.

Crossrefs

Formula

a(n) = k(n) has been defined in terms of the (conjectured unique) ordered Markoff triple MT(n) = (m_1(n), m_2(n), m(n)) with m(n) = A002559(n) in the comment above as k(n) = min{k1(n), k2(n)}, where m_1(n)*k1(n) - m_2(n) == 0 (mod m(n)), with 0 <= k1(n) < m(n), and m_2(n)*k2(n) - m_1(n) == 0 (mod m(n)), with 0 <= k2(n) < m(n).

A378872 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, 8, 5, 13, 12, 12, 5, 20, 21, 8, 40, 21, 40, 40, 5, 29, 32, 60, 17, 60, 85, 85, 96, 32, 17, 85, 96, 17, 96, 96, 5, 40, 45, 24, 104, 13, 148, 148, 165, 24, 148, 8, 221, 148, 12, 221, 260, 45, 104, 148, 165, 148, 221, 12, 260, 104, 165, 221, 260, 165, 260, 260
Offset: 1

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Author

Pontus von Brömssen, Dec 10 2024

Keywords

Comments

Here, the minimal polynomial is required to have integer coefficients with no common divisors.
If two numbers have eventually periodic continued fraction expansions with the same periodic part, the discriminants of their respective minimal polynomials 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, whose minimal polynomial is 2*x^2-2*x-1 with discriminant a(6) = 12.
		

Crossrefs

Cf. A059893, A066099 (compositions in standard order), A139706, A246903, A246921, A305311, A378873, A378874.

Formula

a(n) = A378873(n)*A378874(n)^2.
a(A059893(n)) = a(n), since reversing the periodic part of a continued fraction leaves the discriminant unchanged.
a(A139706(n)) = a(n), since a circular shift of the periodic part of a continued fraction leaves the discriminant unchanged.

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.

A305317 a(n) gives the length of the period of the regular continued fraction of the quadratic irrational of any Markoff form representative Mf(n), n >= 1 (assuming the uniqueness conjecture).

Original entry on oeis.org

1, 1, 4, 6, 6, 8, 10, 8, 10, 12, 10, 14, 10, 14, 16, 14, 18, 12, 14, 16, 18, 20, 14, 22, 14, 16, 18, 20, 22, 24, 18, 22, 16, 26, 22, 26, 18, 28, 22, 26
Offset: 1

Views

Author

Wolfdieter Lang, Jul 30 2018

Keywords

Comments

The index n enumerates the Markoff triples with largest member m from A002559 in increasing order. If the Markoff-Frobenius uniqueness conjecture (see, e.g. the book of Aigner) is true then the triples can be numbered by n if the largest member is m(n) = A002559(n). In the other (unlikely) case there may be more than one triple (hence forms) for some Markoff numbers m from A002559, and then one orders these triples lexicographically.
The indefinite binary quadratic Markoff form Mf(n) = Mf(n;x,y) for the given Markoff number m(n) = A002559(n), n >= 1, (assuming that the mentioned uniqueness conjecture is true) is m(n)*x^2 + (3*m(n) - 2*k(n))*x*y + (l(n) - 3*k(n))y^2 with l(n) = (k(n)^2 +1)/m(n), and k(n) is defined for the representative form (of the unimodualar equvivalence class), e.g., in Cassels as k(n) = k_C(n) = A305310(n). The qudadratic irrational xi(n) is the solution of Mf(n;x,1) = 0 with the positive root. For the representative forms used by Cassels the regular continued fractions for xi(n) = xi_C(n) are not purely periodic. The smallest preperiod is -1 for n = 1 and 0 for n >= 2.
For the representative Mf(n) with k(n) = A305311(n) = k_C(n) + 2*m(n) one obtains purely periodic regular continued fractions for the quadratic irrationals xi(n). They were considered by Perron, pp. 5-6, for n=1..11. See the examples below, and in the W. Lang link, Table 2.

Examples

			The periods for the representative form Mf(n) with k(n) = A305311(n) are given for n=1..40 in the W. Lang link in Table 2.
The first 11 examples (given by Perron) are:
  n     periods             length  quadratic irrationals xi  Markoff form coeffs.
  1:    (1)                    1    (1 + sqrt(5))/2           [1, -1, -1]
  2:    (2)                    1     1 + sqrt(2)              [2, -4, -2]
  3:    (2_2, 1_2)             4    (9 + sqrt(221))/10        [5, -9, -7]
  4:    (2_2, 1_4)             6    (23 + sqrt(1517))/26      [13, -23,-19]
  5:    (2_4, 1_2)             6    (53 + sqrt(7565))/58      [29, -53, -4]
  6:    (2_2, 1_6)             8    (15 + 5*sqrt(26))/17      [34, -60, -50]
  7:    (2_2, 1_8)            10    (157 + sqrt(71285))/178   [89, -157, -131]
  8:    (2_6, 1_2)             8    (309 + sqrt(257045))/338  [169, -309, -239]
  9:    (2_2, 1_2, 2_2, 1_4)  10    (86 + sqrt(21170))/97     [194, -344, -284]
  10:   (2_2, 1_10)           12    (411 + sqrt(488597))/466  [233, -411, -343]
  11:   (2_4, 1_2, 2_2, 1_2)  10    (791 + sqrt(1687397))/866 [433, -791, -613]
  ...
		

References

  • Aigner, Martin. Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings. Springer, 2013.
  • Oskar Perron, Über die Approximation irrationaler Zahlen durch rationale, II, pp. 1-12, Sitzungsber. Heidelberger Akademie der Wiss., 1921, 8. Abhandlung, Carl Winters Universitätsbuchhandlung.

Crossrefs

Showing 1-4 of 4 results.