OFFSET
1,1
COMMENTS
Apparently a(n) = A094958(n+4)-5.
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1, 2, -2).
FORMULA
a(n) = 2*a(n-2) + 5 for n > 2; a(1) = 3, a(2) = 5.
a(n) = (13 - 3*(-1)^n)*2^(1/4*(2*n -1 +(-1)^n))/2 - 5.
G.f.: x*(3+2*x)/(1-x-2*x^2+2*x^3).
a(1)=3, a(2)=5, a(3)=11, a(n)=a(n-1)+2*a(n-2)-2*a(n-3). - Harvey P. Dale, Aug 28 2012
MATHEMATICA
Accumulate[LinearRecurrence[{0, 2}, {3, 2}, 50]] (* or *) LinearRecurrence[ {1, 2, -2}, {3, 5, 11}, 50] (* Harvey P. Dale, Aug 28 2012 *)
PROG
(Magma) T:=[ n le 2 select 4-n else 2*Self(n-2): n in [1..39] ]; [ n eq 1 select T[1] else Self(n-1)+T[n]: n in [1..#T]];
(PARI) x='x+O('x^50); Vec(x*(3+2*x)/(1-x-2*x^2+2*x^3)) \\ G. C. Greubel, Sep 09 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Klaus Brockhaus, Aug 08 2009
STATUS
approved