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Congruent number sequence starting from the Pythagorean triple (3,4,5).
0

%I #16 Dec 21 2022 21:59:03

%S 6,15,34,353,175234,9045146753,121609715057619333634,

%T 4138643330264389621194448797227488932353,

%U 27728719906622802548355602700962556264398170527494726660553210068191276023007234

%N Congruent number sequence starting from the Pythagorean triple (3,4,5).

%H G. Jacob Martens, <a href="https://arxiv.org/abs/2112.09553">Rational right triangles and the Congruent Number Problem</a>, arXiv:2112.09553 [math.GM], 2021, see section 8 The unseen recurrence, equations (79,80).

%e Starting with the triple (3,4,5) and choosing the b side we obtain by the recurrence the right triangles: (15/2, 4, 17/2), (136/15, 15/2, 353/30), (5295/136, 272/15, 87617/2040), (47663648/79425, 79425/136, 9045146753/10801800), ...

%e So a(4) = (5295/136) * (272/15) / 2 = 353.

%t nxt[{n_, p_, q_}] := Module[{n1 = Sqrt[p^4 + 4 n^2 q^4], p1 = p Sqrt[p^4 + 4 n^2 q^4], q1 = q^2 n},

%t a = p1/q1; b = 2 n1 q1/p1; c = Sqrt[p1^4 + 4 n1^2 q1^4]/(p1 q1);

%t Return [{ a b/2, Numerator[b], Denominator[b]}];]

%t l = NestList[nxt, {6, 3, 1}, 8] ;

%t l[[All, 1]]

%Y Cf. A081465 (numerators of hypotenuses).

%K nonn

%O 1,1

%A _Gerry Martens_, Nov 04 2022