A086893 a(n) is the index of F(n+1) at the unique occurrence of the ordered pair of reversed consecutive terms (F(n+1),F(n)) in Stern's diatomic sequence A002487, where F(k) denotes the k-th term of the Fibonacci sequence A000045.

1, 3, 5, 13, 21, 53, 85, 213, 341, 853, 1365, 3413, 5461, 13653, 21845, 54613, 87381, 218453, 349525, 873813, 1398101, 3495253, 5592405, 13981013, 22369621, 55924053, 89478485, 223696213, 357913941, 894784853, 1431655765, 3579139413

Offset: 1

Author

John W. Layman, Sep 18 2003

Keywords

• nonn
• easy

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

Examples

			A002487 begins 0,1,1,2,1,3,2,... with offset 0. Thus a(1)=1 since (F(2),F(1)) = (1,1) occurs at term 1 of A002487. Similarly, a(2)=3 and a(3)=5, since (F(3),F(2))=(2,1) occurs at term 3 and (F(4),F(3))=(3,2) at term 5 of A002487.
		

Links

• A. J. Macfarlane, Generating functions for integer sequences defined by the evolution of cellular automata..., Fig. 12.

Crossrefs

Cf. A000045, A002487, A061547, A077954, A283641, A372286, A372352.

Interleaving of A002450\{0} and A072197.

Positive terms of A096773 in ascending order.

Partial sums of A158302.

Programs

Magma

[2^(n-1)*(3-(-1)^n/3)-1/3: n in [0..35]]; // Vincenzo Librandi, May 09 2015

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 *)

PARI

a(n)=if(n%2,2^(n+1),2^(n+1)+2^(n-1))\3 \\ Charles R Greathouse IV, May 08 2015

Python

def A086893(n): return (1<Chai Wah Wu, Apr 29 2024

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

Extensions

More terms from Paul Barry, May 21 2004