A157664 a(n) = 80000*n^2 + 800*n + 1.

80801, 321601, 722401, 1283201, 2004001, 2884801, 3925601, 5126401, 6487201, 8008001, 9688801, 11529601, 13530401, 15691201, 18012001, 20492801, 23133601, 25934401, 28895201, 32016001, 35296801, 38737601, 42338401, 46099201

Offset

1,1

Comments

The identity (80000n^2 + 800n + 1)^2 - (100n^2 + n)*(8000n + 40)^2 = 1 can be written as a(n)^2 - A055438(n)*A157663(n)^2 = 1. - Vincenzo Librandi, Feb 04 2012

This is the case s=10 of the identity (8*n^2*s^4 + 8*n*s^2 + 1)^2 - (n^2*s^2 + n)*(8*n*s^3 + 4*s)^2 = 1. - Bruno Berselli, Feb 04 2012

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Vincenzo Librandi, X^2-AY^2=1

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

G.f.: x*(80801 + 79198*x + x^2)/(1-x)^3. - Vincenzo Librandi, Feb 04 2012

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Feb 04 2012

E.g.f.: (1 + 80800*x + 80000*x^2)*exp(x) - 1. - G. C. Greubel, Nov 17 2018

Mathematica

LinearRecurrence[{3, -3, 1}, {80801, 321601, 722401}, 50] (* Vincenzo Librandi, Feb 04 2012 *)

Prog

(Magma) I:=[80801, 321601, 722401]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 04 2012
(PARI) for(n=1, 40, print1(80000*n^2 + 800*n + 1", ")); \\ Vincenzo Librandi, Feb 04 2012
(Sage) [80000*n^2+800*n+1 for n in (1..40)] # G. C. Greubel, Nov 17 2018
(GAP) List([1..40], n -> 80000*n^2+800*n+1); # G. C. Greubel, Nov 17 2018

Crossrefs

Cf. A055438, A157663.

Sequence in context: A264805 A050517 A069304 * A218106 A064001 A252625

Adjacent sequences: A157661 A157662 A157663 * A157665 A157666 A157667

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 04 2009

Status

approved