A095903 Lexical ordering of the lazy Fibonacci representations.

1, 2, 3, 4, 5, 7, 6, 8, 9, 12, 10, 13, 15, 20, 11, 14, 16, 21, 17, 22, 25, 33, 18, 23, 26, 34, 28, 36, 41, 54, 19, 24, 27, 35, 29, 37, 42, 55, 30, 38, 43, 56, 46, 59, 67, 88, 31, 39, 44, 57, 47, 60, 68, 89, 49, 62, 70, 91, 75, 96, 109, 143, 32, 40, 45, 58, 48, 61, 69, 90, 50, 63

Offset

1,2

Comments

A permutation of the natural numbers. As suggested by the example, the numbers can be generated as a graph consisting of two components, each being a tree. One tree has root 1 and consists of the numbers in the lower Wythoff sequence, A000201; the other has root 2 and consists of the numbers in the upper Wythoff sequence, A001950. (One could start with 0 and have a single tree instead of two components.)

Regard generations g(n) of the graph as rows of an array (see Example); then |g(n)| = 2^n. Every row includes exactly two Fibonacci numbers; specifically, row n includes F(2n) and F(2n+1). - Clark Kimberling, Mar 11 2015

Links

Clark Kimberling, Table of n, a(n) for n = 1..4000

Example

Start with 1,2. Suffix the next two Fibonacci numbers, getting 1+2, 1+3; 2+3, 2+5. Suffix the next two Fibonacci numbers, getting 1+2+3, 1+2+5, 1+3+5, 1+3+8; 2+3+5, 2+3+8, 2+5+8, 2+5+13. Continue, obtaining
row 1:  1,2
row 2:  3,4,5,7
row 3:  6,8,9,12,10,13,15,20
row 4:  11,14,16,21,17,22,25,33,18,23,26,34,28,36,41,54

Mathematica

Map[Total, Fibonacci[Flatten[NestList[Flatten[Map[{Join[#, {Last[#]+1}], Join[#, {Last[#]+2}]}&, #], 1]&, {{2}, {3}}, 7], 1]]]  (* Peter J. C. Moses, Mar 06 2015 *)

Prog

(PARI) a(n) = n++; my(x=0, y=0); for(i=0, logint(n, 2)-1, y++; [x, y]=[y, x+y]; if(bittest(n, i), [x, y]=[y, x+y])); y; \\ Kevin Ryde, Jun 19 2021

Crossrefs

Cf. A000045, A095791, A000201, A001950, A255773 (the 1-tree), A255774 (the 2-tree).

Sequence in context: A267111 A161919 A269391 * A267112 A269392 A166277

Adjacent sequences: A095900 A095901 A095902 * A095904 A095905 A095906

Keyword

easy,nonn

Author

Clark Kimberling, Jun 12 2004

Status

approved