OFFSET
1,3
COMMENTS
A167821(n) is the difference between A000918(n), the number of branches of a complete binary tree of n levels, and the number of recursive calls needed to compute the (n+1)-th Fibonacci number F(n+1) as defined in A019274: A167821(n) = A000918(n) - A019274(n+1). - Denis Lorrain, Jan 14 2012
Partial sums of A027934 multiplied term by term by 2 (as shown by the second formula), i.e., partial sums of row sums of A108617. - J. M. Bergot, Oct 02 2012, clarified by R. J. Mathar, Oct 05 2012
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3, -1, -2).
FORMULA
G.f.: (2 x^2)/(1 - 3 x + x^2 + 2 x^3);
a(n) = 2^n - 2*Fibonacci(n+1).
a(n) = 3*a(n-1) - a(n-2) - 2*a(n-3). - G. C. Greubel, Jun 27 2016
MATHEMATICA
CoefficientList[Series[(2 x^2)/(1 - 3 x + x^2 + 2 x^3), {x, 0, 30}], x]
Table[2^n - 2*Fibonacci[n + 1], {n, 1, 31}]
LinearRecurrence[{3, -1, -2}, {0, 0, 2}, 50] (* G. C. Greubel, Jun 27 2016 *)
PROG
(Magma) [2^n-2*Fibonacci(n+1): n in [1..40]]; // Vincenzo Librandi, Jun 28 2016
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
V.J. Pohjola, Nov 13 2009
STATUS
approved