A136325 a(n) = 8*a(n-1)-a(n-2) with a(0)=0 and a(1)=3.

0, 3, 24, 189, 1488, 11715, 92232, 726141, 5716896, 45009027, 354355320, 2789833533, 21964312944, 172924670019, 1361433047208, 10718539707645, 84386884613952, 664376537203971, 5230625413017816, 41180626766938557

Offset: 0

Author

Lorenz H. Menke, Jr., Mar 26 2008

Keywords

• nonn
• easy

Comments

Nonnegative integers k such that 15*k^2 + 9 is a square.

From the recurrence we have a(n) = sqrt(15)*((4 + sqrt(15))^n - (4 - sqrt(15))^n)/10.

Examples

			G.f. = 3*x + 24*x^2 + 189*x^3 + 1488*x^4 + 11715*x^5 + 92232*x^6 + 726141*x^7 + ...
		

Links

• Index entries for linear recurrences with constant coefficients, signature (8,-1).

Crossrefs

Cf. A001090.

Programs

Mathematica

Do[If[IntegerQ[Sqrt[3 (3 + 5 x^2)]], Print[{x, Sqrt[3 (3 + 5 x^2)]}]], {x, 0, 2000000}]
LinearRecurrence[{8,-1},{0,3},30] (* Harvey P. Dale, Aug 18 2014 *)
a[ n_] := 3 ChebyshevU[ n - 1, 4]; (* Michael Somos, Oct 14 2015 *)
a[ n_] := 3/2 ((4 + Sqrt[15])^n - (4 - Sqrt[15])^n) / Sqrt[15] // Simplify; (* Michael Somos, Oct 14 2015 *)

PARI

{a(n) = subst(poltchebi(n+1) - 4 * poltchebi(n), x, 4) / 5}; /* Michael Somos, Apr 05 2008 */

PARI

{a(n) = 3 * polchebyshev(n-1, 2, 4)}; /* Michael Somos, Oct 14 2015 */

PARI

{a(n) = 3 * imag( (4 + quadgen(60))^n )}; /* Michael Somos, Oct 14 2015 */

Formula

From Colin Barker, Jan 24 2013: (Start)

a(n) = (sqrt(3/5)*(-(4-sqrt(15))^n + (4+sqrt(15))^n))/2.

G.f.: 3*x/(x^2-8*x+1). (End)

a(n) = 3 * A001090(n).

For n > 0, a(n) is the denominator of the continued fraction [2,3,2,3,...,2,3] with n repetitions of 2,3. For the numerators see A070997. - Greg Dresden, Sep 12 2019

Extensions

Definition corrected by Bruno Berselli, Jan 24 2013

Definition, comments, formulas further corrected by Greg Dresden, Sep 13 2019

Exchanged definition and comment, in order to retain offset 0. - N. J. A. Sloane, Sep 23 2019