A106328 Numbers j such that 8*(j^2) + 9 = k^2 for some positive number k.

0, 3, 18, 105, 612, 3567, 20790, 121173, 706248, 4116315, 23991642, 139833537, 815009580, 4750223943, 27686334078, 161367780525, 940520349072, 5481754313907, 31950005534370, 186218278892313, 1085359667819508, 6325939728024735, 36870278700328902, 214895732473948677

Offset: 1

Author

Pierre CAMI, Apr 29 2005

Keywords

• nonn
• easy

Comments

The ratio k(n) /(2*j(n)) tends to sqrt(2) as n increases.

The squares of the numbers in this sequence are one less than a triangular number: a(n)^2 = A164080(n). For example, 18^2 is 324, and 325 is a triangular number. a(n)^2 + 1 = A164055(n). a(n)^2 = A072221(n)(A072221(n)+1)/2 - 1. - Tanya Khovanova & Alexey Radul, Aug 09 2009

For n > 0, a(n+1) is the n-th almost balancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
• Tanya Khovanova, Recursive Sequences
• Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
• Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
• Index entries for linear recurrences with constant coefficients, signature (6,-1).

Crossrefs

Cf. A001109, A005319, A072221, A103328, A164080, A164055.

Programs

Haskell

a106328 n = a106328_list !! (n-1)
a106328_list = 0 : 3 : zipWith (-) (map (* 6) (tail a106328_list)) a106328_list
-- Reinhard Zumkeller, Jan 10 2012

Mathematica

s=0;lst={};Do[s+=n;If[Sqrt[s-1]==Floor[Sqrt[s-1]],AppendTo[lst,Sqrt[s-1]]],{n,8!}];lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
Rest@ CoefficientList[Series[3 x^2/(1 - 6 x + x^2), {x, 0, 24}], x] (* Michael De Vlieger, Nov 02 2020 *)

PARI

concat(0, Vec(3*x^2/(1-6*x+x^2) + O(x^40))) \\ Michel Marcus, Sep 07 2016

PARI

a(n)=([0,1;-1,6]^n*[-3;0])[1,1] \\ Charles R Greathouse IV, Sep 07 2016

Formula

a(1)=0, a(2)=3 then a(n) = 6*a(n-1) - a(n-2).

a(n) = ((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1))*3/4/sqrt(2). - Max Alekseyev, Jan 11 2007

a(n) = 3*A001109(n). - M. F. Hasler, R. J. Mathar, Jun 03 2009

a(n) = (3/4)*A005319(n-1).

G.f.: 3*x^2/(1 - 6*x + x^2). - Philippe Deléham, Nov 17 2008

E.g.f.: 3 - 3*exp(3*x)*(4*cosh(2*sqrt(2)*x) - 3*sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, Nov 25 2022

Extensions

More terms from Max Alekseyev, Jan 11 2007