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User: Gerry Martens

Gerry Martens has authored 2 sequences.

A358210 Congruent number sequence starting from the Pythagorean triple (3,4,5).

6, 15, 34, 353, 175234, 9045146753, 121609715057619333634, 4138643330264389621194448797227488932353, 27728719906622802548355602700962556264398170527494726660553210068191276023007234

Offset: 1

Gerry Martens, Nov 04 2022

nonn

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

G. Jacob Martens, Rational right triangles and the Congruent Number Problem, arXiv:2112.09553 [math.GM], 2021, see section 8 The unseen recurrence, equations (79,80).

Cf. A081465 (numerators of hypotenuses).

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},
  a = p1/q1; b = 2 n1 q1/p1; c = Sqrt[p1^4 + 4 n1^2 q1^4]/(p1 q1);
  Return [{ a b/2, Numerator[b], Denominator[b]}];]
l = NestList[nxt, {6, 3, 1}, 8] ;
l[[All, 1]]

A258684 a(n) = A041105(4n+1).

Original entry on oeis.org

1, 63, 3905, 242047, 15003009, 929944511, 57641556673, 3572846569215, 221458845734657, 13726875588979519, 850844827670995521, 52738652440012742783, 3268945606453119057025, 202621888947653368792767, 12559288169148055746094529, 778473244598231802889068031

Offset: 0

Gerry Martens, Jun 07 2015

a(n) is a solution to the Pell equation (4*b(n))^2 - 15*a(n)^2 = 1. The corresponding b(n) are A302329(n). - Klaus Purath, Jul 19 2025

Colin Barker, Table of n, a(n) for n = 0..557
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 4, k=1, t=4.
Index entries for linear recurrences with constant coefficients, signature (62,-1).

Cf. A041105 (denominators of continued fraction convergents to sqrt(60)).

I:=[1,63]; [n le 2 select I[n] else 62*Self(n-1)-Self(n-2): n in [1..45]]; // Vincenzo Librandi, Jun 08 2015

a[c_, p_, n_] := Module[{},
   l := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n l]] ;
   t := Table[d[[i + 1]], {i, p, Length[d] - 1, l}]  ;
   Return[t];
   ];
a[60, 1, 20]
CoefficientList[Series[(1 + x)/(x^2 - 62 x + 1), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 08 2015 *)
LinearRecurrence[{62,-1},{1,63},30] (* Harvey P. Dale, Dec 24 2015 *)

Vec((x+1)/(x^2-62*x+1) + O(x^100)) \\ Colin Barker, Jun 07 2015

a(n) = (1/2-2/sqrt(15))*(31-8*sqrt(15))^n+((15+4*sqrt(15))*(31+8*sqrt(15))^n)/30.
From Colin Barker, Jun 07 2015: (Start)
a(n) = 62*a(n-1)-a(n-2).
G.f.: (x+1) / (x^2-62*x+1). (End)
E.g.f.: exp(31*x)*(15*cosh(8*sqrt(15)*x) + 4*sqrt(15)*sinh(8*sqrt(15)*x))/15. - Stefano Spezia, Jul 25 2025

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User: Gerry Martens

A358210 Congruent number sequence starting from the Pythagorean triple (3,4,5).

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A258684 a(n) = A041105(4n+1).

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