The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

Thanks to everyone who made a donation during our annual appeal!
To see the list of donors, or make a donation, see the OEIS Foundation home page.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A279269 a(n) = floor( (4 + sqrt(11))^n ). 1

%I

%S 1,7,53,391,2865,20967,153413,1122471,8212705,60089287,439650773,

%T 3216759751,23535824145,172202794407,1259943234533,9218531904231,

%U 67448539061185,493495652968327,3610722528440693,26418301962683911,193292803059267825,1414250914660723047

%N a(n) = floor( (4 + sqrt(11))^n ).

%C All numbers are odd.

%H Olimpiada Matemática Española, <a href="http://www.olimpiadamatematica.es/platea.pntic.mec.es/_csanchez/olimpcompendi.htm">Si n es un número natural, demostrar que la parte entera de (4 + sqrt(11))^n es un número impar</a> (in Spanish), Problem 26/3 (1990), page 26.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-13,5).

%F O.g.f.: (1 - 2*x + 3*x^2)/((1 - x)*(1 - 8*x + 5*x^2)). - _Ilya Gutkovskiy_, Dec 13 2016

%F E.g.f.: exp((4 + sqrt(11))*x) + exp((4 - sqrt(11))*x) - exp(x). - _Bruno Berselli_, Dec 14 2016

%F a(n) = 9*a(n-1) - 13*a(n-2) + 5*a(n-3) for n>2.

%F a(n) = 8*a(n-1) - 5*a(n-2) + 2 for n>1.

%F a(n) = (4 + sqrt(11))^n + (4 - sqrt(11))^n - 1. - _Bruno Berselli_, Dec 13 2016

%t Floor[(4+Sqrt[11])^Range[0,30]] (* or *) LinearRecurrence[{9,-13,5},{1,7,53},30] (* _Harvey P. Dale_, Apr 22 2019 *)

%K nonn,easy

%O 0,2

%A _Philippe Deléham_, Dec 13 2016

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified January 18 19:50 EST 2020. Contains 331030 sequences. (Running on oeis4.)