A059426 -id:A059426 - cp's OEIS frontend

A026273 a(n) = least k such that s(k) = n, where s = A026272.

1, 2, 4, 6, 7, 9, 10, 12, 14, 15, 17, 19, 20, 22, 23, 25, 27, 28, 30, 31, 33, 35, 36, 38, 40, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56, 57, 59, 61, 62, 64, 65, 67, 69, 70, 72, 74, 75, 77, 78, 80, 82, 83, 85, 86, 88, 90, 91, 93, 95, 96, 98, 99

Offset: 1

Clark Kimberling

nonn

This is the lower s-Wythoff sequence, where s(n)=n+1.
See A184117 for the definition of lower and upper s-Wythoff sequences.  The first few terms of a and its complement, b=A026274, are obtained generated as follows:
s=(2,3,4,5,6,...);
a=(1,2,4,6,7,...)=A026273;
b=(3,5,8,11,13,...)=A026274.
Briefly:  b=s+a, and a=mex="least missing".
From Michel Dekking, Mar 12 2018: (Start)
One has r*(n-2*r+3) = n*r-2r^2+3*r = (n+1)*r-2.
So  a(n) = (n+1)*r-2, and we see that this sequence is simply the Beatty sequence of the golden ratio, shifted spatially and temporally. In other words: if w = A000201 = 1,3,4,6,8,9,11,12,14,...  is the lower Wythoff sequence, then a(n) = w(n+2) - 2.
(N.B. As so often, there is the 'offset 0 vs 1 argument', w = A000201 has offset 1; it would have been better to give (a(n)) offset 1, too).
This observation also gives an answer to Lenormand's question, and a simple proof of Mathar's conjecture in A059426.
(End)

Jon Asier Bárcena-Petisco, Luis Martínez, María Merino, Juan Manuel Montoya, and Antonio Vera-López, Fibonacci-like partitions and their associated piecewise-defined permutations, arXiv:2503.19696 [math.CO], 2025. See p. 3.

Cf. A184117, A026274.

r=(1+Sqrt[5])/2;
a[n_]:=Floor[r*(n-2r+3)];
b[n_]:=Floor[r*r*(n+2r-3)];
Table[a[n],{n,200}]   (* A026273 *)
Table[b[n],{n,200}]   (* A026274 *)

a(n) = floor[r*(n-2*r+3)], where r=golden ratio.

b(n) = floor[(r^2)*(n+2*r-3)] = floor(n*A104457-A134972+1).

Extended by Clark Kimberling, Jan 14 2011

A082389 a(n) = floor((n+2)phi) - floor((n+1)phi) where phi=(1+sqrt(5))/2.

Original entry on oeis.org

1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2

Offset: 1

Benoit Cloitre, Apr 14 2003

Alternative descriptions (1): unique positive integer sequence taking values in {1,2} satisfying a(1)=1, a(2)=2 and a(a(1)+...+a(n))=a(n) for n >= 3.
(2) Start with 1,2; then for any k>=1, a(a(1)+...+a(k))=a(k), fill in any undefined terms by the rule that a(t) = 1 if a(t-1) = 2 and a(t) = 2 if a(t-1) = 1.
(3) a(1)= 1, a(2)=2, a(a(1)+a(2)+...+a(n))=a(n); a(a(1)+a(2)+...+a(n)+1)=3-a(n).
More generally, the sequence a(n)=floor(r*(n+2))-floor(r*(n+1)), r= (1/2) *(z+sqrt(z^2+4)), z integer >=1, is defined by a(1), a(2) and a(a(1)+a(2)+...+a(n)+f(z))=a(n); a(a(1)+a(2)+...+a(n)+f(z)+1)=(2z+1)-a(n) where f(1)=0, f(z)=z-2 for z>=2.

			a(1)+a(2)=3 and a(a(1)+a(2)) must be a(2) so a(3)=2. Therefore a(a(1)+a(2)+a(3))=a(5)=2 and from the rule the "hole" a(4) is 1. Hence sequence begins 1,2,2,1,2,...

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Same as A014675 without the first term.

A082389:=n->floor((n+2)*(1+sqrt(5))/2) - floor((n+1)*(1+sqrt(5))/2): seq(A082389(n), n=1..300); # Wesley Ivan Hurt, Jan 16 2017

Rest@Nest[ Flatten[ # /. {1 -> 2, 2 -> {2, 1}}] &, {1}, 11] (* Robert G. Wilson v, Jan 26 2006 *)
#[[2]]-#[[1]]&/@Partition[Table[Floor[GoldenRatio n],{n,0,110}],2,1] (* Harvey P. Dale, Sep 04 2019 *)
Differences[Floor[GoldenRatio Range[2,150]]] (* Harvey P. Dale, Dec 02 2024 *)

from math import isqrt
def A082389(n): return (n+2+isqrt(m:=5*(n+2)**2)>>1)-(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 29 2022

a(n) = A014675(n+1); sum(k = 1, n, a(k)) = A058065(n)
Apparently a(n) = A059426(n).
a(n) = A066096(n+2)-A066096(n+1). - R. J. Mathar, Aug 02 2024

A245977 Limit-reverse of the infinite Fibonacci word A014675 = (s(0),s(1),...) = (2,1,2,2,1,2,1,2, ...) using initial block (s(2),s(3)) = (2,2).

Original entry on oeis.org

2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2

Offset: 0

Clark Kimberling and Peter J. C. Moses, Aug 10 2014

nonn

Suppose, as in A245920, that S = (s(0),s(1),s(2),...) is an infinite sequence such that every finite block of consecutive terms occurs infinitely many times in S.  (It is assumed that A014675 is such a sequence.)  Let B = B(m,k) = (s(m-k),s(m-k+1),...,s(m)) be such a block, where m >= 0 and k >= 0.  Let m(1) be the least i > m such that (s(i-k), s(i-k+1),...,s(i)) = B(m,k), and put B(m(1),k+1) = (s(m(1)-k-1),s(m(1)-k),...,s(m(1))).  Let m(2) be the least i > m(1) such that (s(i-k-1),s(i-k),...,s(i)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)-k-2),s(m(2)-k-1),...,s(m(2))).  Continuing in this manner gives a sequence of blocks B(m(n),k+n).  Let B'(n) = reverse(B(m(n),k+n)), so that for n >= 1, B'(n) comes from B'(n-1) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limit-reverse of S with initial block B(m,k)", denoted by S*(m,k), or simply S*.
...
The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-reversing S with initial block B(m,k)" or simply the index sequence for S*, as in A245921 and A245978.
Apparently a(n) = A082389(n+2) = A059426(n+1). - R. J. Mathar, Sep 01 2014

			S = the infinite Fibonacci word A014675, with B = (s(2), s(3)); that is, (m,k) = (2,3)
S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
B'(0) = (2,2)
B'(1) = (2,2,1)
B'(2) = (2,2,1,2)
B'(3) = (2,2,1,2,1)
B'(4) = (2,2,1,2,1,2)
B'(5) = (2,2,1,2,1,2,2)
S* = (2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,1,...), with index sequence (3,8,11,16,21,29,...)

Clark Kimberling, Table of n, a(n) for n = 0..300

Cf. A245978, A245979, A245920, A014675.

z = 100; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 2]]] &[seqtofind]; x = GoldenRatio; s = Differences[Table[Floor[n*x], {n, 1, z^2}]]; ans = Join[{s[[p[0] = pos = seqPosition2[s, #] - 1]]}, #] &[{s[[3]], s[[4]]}]; (* Initial block is (s(3),s(4)) [OR (s(2),s(3)) if using offset 0] *); cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}]; rcf = Last[Map[Reverse, cfs]] (* A245977 *)

cp's OEIS Frontend

A026273 a(n) = least k such that s(k) = n, where s = A026272.

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A082389 a(n) = floor((n+2)phi) - floor((n+1)phi) where phi=(1+sqrt(5))/2.

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A245977 Limit-reverse of the infinite Fibonacci word A014675 = (s(0),s(1),...) = (2,1,2,2,1,2,1,2, ...) using initial block (s(2),s(3)) = (2,2).

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cp's OEIS Frontend

A026273 a(n) = least k such that s(k) = n, where s = A026272.

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A082389 a(n) = floor((n+2)*phi) - floor((n+1)*phi) where phi=(1+sqrt(5))/2.

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A245977 Limit-reverse of the infinite Fibonacci word A014675 = (s(0),s(1),...) = (2,1,2,2,1,2,1,2, ...) using initial block (s(2),s(3)) = (2,2).

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A082389 a(n) = floor((n+2)phi) - floor((n+1)phi) where phi=(1+sqrt(5))/2.