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A309161 a(n) = (x(n)^2 + 1)/m(n), with m(n) = A002559(n) (Markoff numbers) and x(n)= A324601(n), for n >= 3. The Markoff uniqueness conjecture is assumed to be true.

%I A309161 #10 Aug 01 2019 04:00:49
%S A309161 1,2,5,5,13,29,29,34,74,29,34,25,233,433,202,985,457,1130,541,1597,
%T A309161 2042,4181,5741,145,6466,7561,2957,2378,16501,5,3733,1157,53,62210,
%U A309161 27845,75025,96557,43970,59153,5857,160373,219658,252005,294685,126226,426389,559945,514229,733,514,1278649,706225,3001,1441889,1716469,61913,187045,12994
%N A309161 a(n) = (x(n)^2 + 1)/m(n), with m(n) = A002559(n) (Markoff numbers) and x(n)= A324601(n), for n >= 3. The Markoff uniqueness conjecture is assumed to be true.
%C A309161 See the Aigner reference, Corollary 3.17., p. 58.
%C A309161 If the equation x^2 + 1 = a(n)*m(n), with m(n) = A002559(n) holds for just one integral x = x(n) in the interval [1, floor(m(n)/2)] then the Markoff uniqueness conjecture is true. x(n) = A324601(n) (if the Markoff conjecture holds).
%D A309161 Martin Aigner, Markov's Theorem and 100 Years of the Uniqueness Conjecture, Springer, 2013, p. 58.
%F A309161 a(n) = (A324601(n)^2 + 1)/A002559(n), for n >= 3.
%Y A309161 Cf. A002559, A324601.
%K A309161 nonn
%O A309161 3,2
%A A309161 _Wolfdieter Lang_, Jul 26 2019