OFFSET
0,2
COMMENTS
Superseeker suggests a(n+2) - 2*a(n+1) + a(n) = 7*3^n = A005032(n).
Inverse binomial transform gives match with first differences of A026622.
Floretion Algebra Multiplication Program, FAMP Code: kbasefor[(- 'j + 'k - 'ii' - 'ij' - 'ik')], vesfor = A000004, Fortype: 1A, Roktype (leftfactor) is set to:Y[sqa.Findk()] = Y[sqa.Findk()] + Math.signum(Y[sqa.Findk()])*p (internal program code)
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-7,3).
FORMULA
From Rolf Pleisch, Feb 10 2008: (Start)
a(0) = 1; a(n) = 3*a(n-1) + n.
a(n) = (7*3^n - 2*n - 3)/4. (End)
a(0)=1, a(1)=4, a(2)=14, a(n) = 5*a(n-1) - 7*a(n-2) + 3*a(n-3). - Harvey P. Dale, Dec 11 2012
MATHEMATICA
s=1; lst={s}; Do[s+=(s+(n+=s)); AppendTo[lst, s], {n, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Oct 11 2008 *)
CoefficientList[Series[(1-x+x^2)/((1-3x)(x-1)^2), {x, 0, 40}], x] (* or *) LinearRecurrence[{5, -7, 3}, {1, 4, 14}, 40] (* Harvey P. Dale, Dec 11 2012 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Creighton Dement, Jun 24 2005
STATUS
approved