A075835 Numbers k such that 13*k^2 + 4 is a square.

0, 3, 33, 360, 3927, 42837, 467280, 5097243, 55602393, 606529080, 6616217487, 72171863277, 787274278560, 8587845200883, 93679022931153, 1021881407041800, 11147016454528647, 121595299592773317

Offset: 1

Author

Gregory V. Richardson, Oct 14 2002

Keywords

• nonn
• easy

Comments

Lim_{n->infinity} a(n)/a(n-1) = (11 + sqrt(13))/2.

References

• A. H. Beiler, "The Pellian", ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
• L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
• Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
• S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234; http://www.scirp.org/journal/am; http://dx.doi.org/10.4236/am.2014.515216

Links

• Tanya Khovanova, Recursive Sequences
• J. J. O'Connor and E. F. Robertson, Pell's Equation
• Eric Weisstein's World of Mathematics, Pell Equation
• Index entries for linear recurrences with constant coefficients, signature (11,-1).

Crossrefs

Cf. A006190.

Programs

Mathematica

LinearRecurrence[{11,-1},{0,3},20] (* Harvey P. Dale, Dec 27 2011 *)
Table[Fibonacci[2n, 3], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)

Formula

a(n) = ((11 + 3*sqrt(13))^n - (11 - 3*sqrt(13))^n) / ((2^n) * sqrt(13)).

From Philippe Deléham, Nov 17 2008: (Start)

a(n) = 11*a(n-1) - a(n-2) with a(1)=0 and a(2)=3.

G.f.: 3x^2/(1-11x+x^2). (End)

a(n) = A006190(2*n). - Vladimir Reshetnikov, Sep 16 2016