OFFSET
1,2
COMMENTS
If the Fibonacci pairs are kept in the natural order (F(n),F(n+1)), it appears that the first term of the pair occurs in A002487 at the index given by A061547(n).
Equals row sums of triangle A177954. - Gary W. Adamson, May 15 2010
Starting at n=3, begin subtracting from (2^(n-1)-1)/2^(n-1): 3/4 - 1/2 = 1/4 with 1+4=5=a(3); 7/8 - 1/4 = 5/8 with 5+8=13=a(4); 15/16 - 5/8 = 5/16 with 5+16=21= a(5); 31/32 - 5/16 = 21/32 with 21+32=53=a(6); 63/64 - 21/32 = 21/64 with 21+64=85=a(7) and so on. For n odd in the first fraction (2^(n-1)-1)/2^(n-1), the result approaches 1/3, and for n even in the first fraction, the result approaches 2/3. - J. M. Bergot, May 08 2015
Also, the decimal representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 678", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. See A283641. - Robert Price, Mar 12 2017
LINKS
FORMULA
It appears that a(n)=(4^((n+1)/2)-1)/3 if n is odd and a(n)=(a(n-1)+a(n+1))/2 if n is even.
G.f.: (1+2*x-2*x^2)/((1-x)*(1-4*x^2)); a(n) = 2^(n-1)(3-(-1)^n/3)-1/3 (offset 0); a(n) = Sum{k=0..n+1, 4^floor(k/2)/2} (offset 0); a(2n) = A002450(n+1) (offset 0); a(2n+1) = A072197(n) (offset 0). - Paul Barry, May 21 2004
a(n+2) = 4*a(n) + 1, a(1) = 1, a(2) = 3, n > 0. - Yosu Yurramendi, Mar 07 2017
a(n+1) = a(n) + A158302(n), a(1) = 1, n > 0. - Yosu Yurramendi, Mar 07 2017
EXAMPLE
MATHEMATICA
f[n_] := Module[{a = 1, b = 0, m = n}, While[m > 0, If[OddQ@ m, b = a + b, a = a + b]; m = Floor[m/2]]; b]; a = Table[f[n], {n, 0, 10^6}]; b = Reverse /@ Partition[Map[Fibonacci, Range[Ceiling@ Log[GoldenRatio, Max@ a] + 1]], 2, 1]; Map[If[Length@ # > 0, #[[1, 1]] - 1, 0] &@ SequencePosition[a, #] &, b] (* Michael De Vlieger, Mar 15 2017, Version 10.1, after Jean-François Alcover at A002487 *)
PROG
(PARI) a(n)=if(n%2, 2^(n+1), 2^(n+1)+2^(n-1))\3 \\ Charles R Greathouse IV, May 08 2015
(Magma) [2^(n-1)*(3-(-1)^n/3)-1/3: n in [0..35]]; // Vincenzo Librandi, May 09 2015
(Python)
def A086893(n): return (1<<n+1 if n&1 else 5<<n-1)//3 # Chai Wah Wu, Apr 29 2024
KEYWORD
nonn,easy
AUTHOR
John W. Layman, Sep 18 2003
EXTENSIONS
More terms from Paul Barry, May 21 2004
STATUS
approved