A026242 a(n) = j if n is L(j), else a(n) = k if n is U(k), where L = A000201, U = A001950 (lower and upper Wythoff sequences).

1, 1, 2, 3, 2, 4, 3, 5, 6, 4, 7, 8, 5, 9, 6, 10, 11, 7, 12, 8, 13, 14, 9, 15, 16, 10, 17, 11, 18, 19, 12, 20, 21, 13, 22, 14, 23, 24, 15, 25, 16, 26, 27, 17, 28, 29, 18, 30, 19, 31, 32, 20, 33, 21, 34, 35, 22, 36, 37, 23, 38, 24, 39, 40, 25, 41

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• nice

Comments

Every positive integer occurs exactly twice. a(n) is the parent of n in the tree at A074049. - Clark Kimberling, Dec 24 2010

Apparently, if n=F(m) (a Fibonacci number), one of two circumstances arise:

I. a(n)=F(m-1) and a(n-1)=F(m-2). When this happens, a(n) occurs for the first time and a(n-1) occurs for the second time;

II. a(n)=F(m-2) and a(n-1)=F(m-1). When this happens, a(n) occurs for the second time and a(n-1) occurs for the first time. - Bob Selcoe, Sep 18 2014

These are the numerators when all fractions, j/r and k/r^2, are arranged in increasing order (where r = golden ratio and j,k are positive integers). - Clark Kimberling, Mar 02 2015

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..10000 (first 999 terms from M. F. Hasler)
• S. Mneimneh, Fibonacci in The Curriculum: Not Just a Bad Recurrence, in Proceeding SIGCSE '15 Proceedings of the 46th ACM Technical Symposium on Computer Science Education, Pages 253-258. See Figure 2.
• Jeffrey Shallit, Fibonacci automaton for a(n)

Crossrefs

Cf. A026272, A074049, A089910.

Cf. A000045 (Fibonacci numbers).

Programs

Mathematica

mx = 100; gr = GoldenRatio; LW[n_] := Floor[n*gr]; UW[n_] := Floor[n*gr^2]; alw = Array[LW, Ceiling[mx/gr]]; auw = Array[UW, Ceiling[mx/gr^2]]; f[n_] := If[ MemberQ[alw, n], Position[alw, n][[1, 1]], Position[auw, n][[1, 1]]]; Array[f, mx] (* Robert G. Wilson v, Sep 17 2014 *)

PARI

my(A=vector(10^4),i,j=0); while(#A>=i=A000201(j++), A[i]=j; (i=A001950(j))>#A || A[i]=j); A026242=A \\ M. F. Hasler, Sep 16 2014 and Sep 18 2014

PARI

A026242=vector(#A002251,n,abs(A002251[n]-n)) \\ M. F. Hasler, Sep 17 2014

Formula

a(n) = a(m) if a(m) has already occurred exactly once and n = a(m) + m; otherwise, a(n) = least positive integer that has not yet occurred.

a(n) = abs(A002251(n) - n).

n = a(n) + a(n-1) unless n = A089910(m); if n = A089910(m), then n = a(n) + a(n-1) - m. - Bob Selcoe, Sep 20 2014

There is a 17-state automaton that accepts the Zeckendorf (Fibonacci) representation of n and a(n), in parallel. See the file a026242.pdf. - Jeffrey Shallit, Dec 21 2023