A154376 a(n) = 25*n^2 - 2*n.

23, 96, 219, 392, 615, 888, 1211, 1584, 2007, 2480, 3003, 3576, 4199, 4872, 5595, 6368, 7191, 8064, 8987, 9960, 10983, 12056, 13179, 14352, 15575, 16848, 18171, 19544, 20967, 22440, 23963, 25536, 27159, 28832, 30555, 32328, 34151, 36024

Offset: 1

Author

Vincenzo Librandi, Jan 08 2009

Keywords

• nonn
• easy

Comments

The identity (1250*n^2 - 100*n + 1)^2 - (25*n^2 - 2*n)*(250*n - 10)^2 = 1 can be written as A154374(n)^2 - a(n)*A154378(n)^2 = 1 (see also the second comment in A154374). - Vincenzo Librandi, Jan 30 2012

The continued fraction expansion of sqrt(a(n)) is [5n-1; {1, 3, 1, 10n-2}]. - Magus K. Chu, Sep 04 2022

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A154374, A154378.

Programs

Mathematica

LinearRecurrence[{3, -3, 1}, {23, 96, 219}, 50] (* Vincenzo Librandi, Jan 30 2012 *)

PARI

a(n)=25*n^2-2*n \\ Charles R Greathouse IV, Dec 26 2011

Formula

From Vincenzo Librandi, Jan 30 2012: (Start)

G.f.: x*(23 + 27*x)/(1-x)^3.

a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). (End)

E.g.f.: (25*x^2 + 23*x)*exp(x). - G. C. Greubel, Sep 15 2016