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A019274
Number of recursive calls needed to compute the n-th Fibonacci number F(n), starting with F(1) = F(2) = 1.
15
0, 0, 2, 4, 8, 14, 24, 40, 66, 108, 176, 286, 464, 752, 1218, 1972, 3192, 5166, 8360, 13528, 21890, 35420, 57312, 92734, 150048, 242784, 392834, 635620, 1028456, 1664078, 2692536, 4356616, 7049154, 11405772, 18454928, 29860702, 48315632
OFFSET
1,3
COMMENTS
Let g = F(2) + F(3) + ... + F(n) = F(n+2) - 2. Some numbers in the range [0,g] have unique representations of the form Sum_{i=1..n} a(i)*F(i) where each a(i) is 1 or -1. These numbers have the form g-k for k in the sequence. - Louis ten Bosch (louis_ten_bosch(AT)hotmail.com), Jan 01 2003
a(n+2) = Sum_{k=0..n} Fibonacci(n-k) + k*Fibonacci(n-k).
FORMULA
a(n) = a(n-1) + a(n-2) + 2. a(n) = 2*F(n) - 2 = 2*A000071(n).
a(n) = Sum_{k=0..n} (2 - 2*0^(n-k))*F(k). - Paul Barry, Oct 24 2007
a(n) = F(n) + F(n+3) - 2, n>=-1 (where F(n) is the n-th Fibonacci number). - Zerinvary Lajos, Jan 31 2008
G.f.: 2*x^3 / ( (x-1)*(x^2+x-1) ). - R. J. Mathar, Jul 01 2012
a(1)=0, a(2)=0, a(3)=2, a(n) = 2*a(n-1) - a(n-3). - Harvey P. Dale, Oct 16 2012
MAPLE
with(combinat): seq(fibonacci(n-2)+fibonacci(n+1)-2, n=1..35); # Zerinvary Lajos, Jan 31 2008
MATHEMATICA
Fibonacci[Range[5! ]]*2-2 (* Vladimir Joseph Stephan Orlovsky, Mar 19 2010 *)
LinearRecurrence[{2, 0, -1}, {0, 0, 2}, 40] (* Harvey P. Dale, Oct 16 2012 *)
CROSSREFS
Cf. A000045.
Antidiagonal sums of array A017125.
Sequence in context: A271493 A118544 A222037 * A164173 A164162 A164402
KEYWORD
nonn,easy,nice
AUTHOR
Kim Trammell (kim(AT)coc.com) and others
STATUS
approved