A279269 a(n) = floor( (4 + sqrt(11))^n ).

1, 7, 53, 391, 2865, 20967, 153413, 1122471, 8212705, 60089287, 439650773, 3216759751, 23535824145, 172202794407, 1259943234533, 9218531904231, 67448539061185, 493495652968327, 3610722528440693, 26418301962683911, 193292803059267825, 1414250914660723047

Offset: 0

Author

Philippe Deléham, Dec 13 2016

Keywords

• nonn
• easy

Comments

All numbers are odd.

Links

• Olimpiada Matemática Española, Si n es un número natural, demostrar que la parte entera de (4 + sqrt(11))^n es un número impar (in Spanish), Problem 26/3 (1990), page 26.
• Index entries for linear recurrences with constant coefficients, signature (9,-13,5).

Programs

Mathematica

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

Formula

O.g.f.: (1 - 2*x + 3*x^2)/((1 - x)*(1 - 8*x + 5*x^2)). - Ilya Gutkovskiy, Dec 13 2016

E.g.f.: exp((4 + sqrt(11))*x) + exp((4 - sqrt(11))*x) - exp(x). - Bruno Berselli, Dec 14 2016

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

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

a(n) = (4 + sqrt(11))^n + (4 - sqrt(11))^n - 1. - Bruno Berselli, Dec 13 2016