OFFSET
0,1
COMMENTS
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-4).
FORMULA
a(n) = 4^(n+1) + 7.
From Alexander R. Povolotsky, Sep 19 2011: (Start)
G.f.: (11 - 32*x)/(1 - 5*x + 4*x^2).
a(n+1) = 4*a(n) - 21. (End)
a(n) = A188165(2n+2) - 2. - Bruno Berselli, Sep 26 2011
PROG
(PARI) a(n)=4^(n+1)+7 \\ Charles R Greathouse IV, Sep 19 2011
(Magma) [4^(n+1) + 7: n in [0..30]]; // Vincenzo Librandi, Sep 30 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Brad Clardy, Sep 19 2011
STATUS
approved