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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A327342 a(n) gives the number of distinct odd prime divisors of m(n) = A002559(n) (Markoff numbers).

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 3, 3, 2, 1, 2, 1, 1, 2, 2, 2, 2, 3, 1, 1, 3, 1, 1, 2, 3, 2, 2, 2, 3, 2, 1, 3, 3, 2, 3, 1, 3, 3, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 1, 3, 2, 1, 4, 1, 2, 2, 4, 2, 1, 3, 3, 4, 3, 1, 1, 2, 2, 2, 1, 3, 2, 3, 2
Offset: 1

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Author

Wolfdieter Lang, Sep 11 2019

Keywords

Comments

These sequence members appear as exponents of 2 in the number of representative parallel primitive forms for binary quadratic forms of discriminant Disc(n) = 9*m(n)^2 - 4 and representation of -m(n)^2. The reduced (primitive) principal form of this discriminant is F_p(n; X, Y) = X^2 + b(n)*X*Y - b(n)*Y^2, written also as F_p(n) = [1, b(n), -b(n)], with b(n) = 3*m(n) - 2 = A324250(n). This form representing -m(n)^2 is important for the determination of Markoff triples MT(n).
For more details see A327343(n) = 2^a(n). The Frobenius-Markoff uniqueness conjecture on ordered triples with largest member m(n) is certainly true for m(n) if a(n) = 0 (so-called singular cases) or 1. See the Aigner reference, p. 59, Corollary 3.20, for n >= 3 (the a(n) = 1 cases).

Examples

			For the examples a(6) = 1 and a(12) = 2 see A327343.
		

References

  • Martin Aigner, Markov's Theorem and 100 Years of the Uniqueness Conjecture, Springer, 2013.

Crossrefs

Formula

a(n) = number of distinct odd prime divisors of m(n) = A002559(n), for n >= 1.
a(n) = A005087(A002559(n)). - Michel Marcus, Sep 18 2023

A327344 9 + 8*a(n) appears in a congruence which determines representative parallel primitive binary quadratic forms for discriminant 9*m(n)^2 - 4 and representation -m(n)^2, where m(n) = A002559(n) (Markoff numbers).

Original entry on oeis.org

0, 0, 39, 273, 1365, 333, 12870, 46410, 10878, 88218, 304668, 107559, 1576614, 2852889, 4144413, 13637988, 28406235, 53558505, 12085458, 92899170, 133886883, 34633998, 351194025, 1334488428, 1819412595, 410100933, 3041210445, 4333538430, 1118696184, 9146719764, 15150288153, 29675764248
Offset: 1

Views

Author

Wolfdieter Lang, Sep 11 2019

Keywords

Comments

See A327343 for the relevance of representative parallel primitive binary quadratic forms (rpapfs) for discriminant Disc(n) = 9*m(n)^2 - 4 and representation -m(n)^2 for the determination of ordered Markoff triples.
These rpapfs FPa(n) = [-m(n)^2, B(n), - C(n)] are determined by the solutions of the congruence z^2 - d(n) == 0 (mod (m(n)/2)^2), where d(n) = 9 + 8*a(n) if m(n) = A002559(n) is even, and the integer z(n) = 1 + 2*J(n) = B(n)/4 is from the set {1, 3, ..., 2*(m(n)/2)^2 - 1}. The members C(n) = 7 + (z^2 - d(n))/(m(n)/2)^2. In the odd m(n) case the congruence is B(n)^2 - d(n) == 0 (mod m(n)^2) , where d(n) = 9 + 8*a(n) and B(n) = 2*j(n) + 1 from the set {1, 3, ..., 2*m(n)^2 -1}. The member C(n) = 1 + (1/4)*(B(n)^2 - d(n))/m(n)^2. The different solutions are then FPa(n;i), for i = 1, 2, ..., #FPa(n), with #FPa(n) = A327343(n) = 2^A327342(n).
The d(n) sequence begins with {9, 9, 321, 2193, 10929, 2673, 102969, 371289, 87033, 705753,...}

Examples

			n = 6: m(6)/2 = 17, M(6) = (17 - 1)/16 = 1, a(6) = 37*1*9 = 333. d(6) = 2673.
n = 7: m(7) = 89, M(7) = 22, a(7) = 13*22*45 = 12870. d(7) = 102969.
The two (#FPa(6) = 2^1) solutions z(6) = B(6)/4 are z(n;1) = 19 and z(n;2) = 1559. They lead to FPa(6;1) = [-34^2, 76, +1] and FPa(6;2) = [-34^2, 2236, -1079].
The two (#FPa(7) = 2^1) solutions B(7) are B(7;1) = 199 and B(7;2) = 15643 (the upper bound was 2*m(7)^2 - 1 = 15841), leading to FPa(7;1) = [-89^2, 199, +1] and FPa(7;2) = [-89^2, 15643, -7721].
In both cases the second solution leads to the ordered Markoff triples MT(6) = (1, 13, 34) and MT(7) = (1, 34, 89). The other solution leads to the unordered triples (1, 34, 13) and (1, 89, 34).
		

Crossrefs

Formula

a(n) = (d(n) - 9)/8 = 37*M(n)*(1 + 8*M(n)) with M(n) = A309376(n) = (m(n)/2 -1)/16 if m(n) is even, and a(n) = (d(n) - 9)/8 = 13*M(n)*(1 + 2*M(n)) with M(n) = A309376(n) = (m(n)-1)/4 if m(n) is odd.
Showing 1-2 of 2 results.