A004957 -id:A004957 - cp's OEIS frontend

A026352 a(n) = floor(n*tau) + n + 1 where tau is the golden ratio A001622.

1, 3, 6, 8, 11, 14, 16, 19, 21, 24, 27, 29, 32, 35, 37, 40, 42, 45, 48, 50, 53, 55, 58, 61, 63, 66, 69, 71, 74, 76, 79, 82, 84, 87, 90, 92, 95, 97, 100, 103, 105, 108, 110, 113, 116, 118, 121, 124, 126, 129, 131, 134, 137, 139, 142, 144

Offset: 0

Clark Kimberling, Dec 11 1999

nonn

a(n) = greatest k such that s(k) = n+1, where s = A026350.
Indices at which blocks (0;1) occur in infinite Fibonacci word; i.e., n such that A005614(n)=0 and A005614(n+1)=1. - Benoit Cloitre, Nov 15 2003
Except for the first term, these are the numbers whose lazy Fibonacci representation (see A095791) includes both 1 and 2; thus this sequence is a subsequence of the lower Wythoff sequence, A000201. - Clark Kimberling, Jun 10 2004 [A-number typo corrected by Nathan Fox, May 03 2014]
a(n) = n-th number k whose lazy Fibonacci representation (as in A095791) has more summands than that of k-1. - Clark Kimberling, Jun 12 2004
a(n) = position of n-th 0 in A096270. - Clark Kimberling, Apr 22 2011
Maximum number of chips in a pile created at each step in the game described by Roland Schroeder in his comment at A000201. (From Allan C. Wechsler via Seqfan.)
In the Fokkink-Joshi paper, this sequence is the Cloitre (1,1,2,3)-hiccup sequence, i.e., a(1) = 1; for m < n, a(n) = a(n-1)+2 if a(m) = n-1, else a(n) = a(n-1)+3. - Michael De Vlieger, Jul 28 2025

Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 9.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See pp. 8, 17.
Eric Friedman, Scott M. Garrabrant, Ilona K. Phipps-Morgan, A. S. Landsberg and Urban Larsson, Geometric analysis of a generalized Wythoff game, in Games of no Chance 5, MSRI publ. Cambridge University Press, 2019.
Urban Larsson and Nathan Fox, An Aperiodic Subtraction Game of Nim-Dimension Two, Journal of Integer Sequences, 2015, Vol. 18, #15.7.4.
Ali Sada, Should A000201 and A026352 be cross-referenced?, Seqfan thread, Jun 2023.

Essentially same as A004957.
Cf. A001622, A005614, A026350, A095791, A096270.
Subsequence of A000201.
Complement of A026351.

Table[Floor[GoldenRatio*n]+n+1,{n,0,60}] (* Harvey P. Dale, Aug 24 2021 *)
cloitreH[j_, x_, y_, z_, w_ : 120] := Block[{c, k}, c[] := False; k = x; c[x] = True; {x}~Join~Reap[Do[If[c[n - j], k += y, k += z]; c[k] = True; Sow[k], {n, 2, w}] ][[-1, 1]] ]; cloitreH[1, 1, 2, 3] (* _Michael De Vlieger, Jul 28 2025 *)

a(n) = floor(n*(sqrt(5)+1)/2) + n + 1; \\ Michel Marcus, Sep 15 2016

from math import isqrt
def A026352(n): return (n+isqrt(5*n**2)>>1)+n+1 # Chai Wah Wu, Aug 25 2022

a(n) = A026351(n)+n. - R. J. Mathar, Jun 24 2025

A060142 Ordered set S defined by these rules: 0 is in S and if x is in S then 2x+1 and 4x are in S.

Original entry on oeis.org

0, 1, 3, 4, 7, 9, 12, 15, 16, 19, 25, 28, 31, 33, 36, 39, 48, 51, 57, 60, 63, 64, 67, 73, 76, 79, 97, 100, 103, 112, 115, 121, 124, 127, 129, 132, 135, 144, 147, 153, 156, 159, 192, 195, 201, 204, 207, 225, 228, 231, 240, 243, 249, 252, 255, 256, 259, 265, 268, 271

Offset: 0

Clark Kimberling, Mar 05 2001

nonn

After expelling 0 and 1, the numbers 4x occupy same positions in S that 1 occupies in the infinite Fibonacci word (A003849).
a(A026351(n)) = A219608(n); a(A004957(n)) = 4 * a(n). - Reinhard Zumkeller, Nov 26 2012
Apart from the initial term, this lists the indices of the 1's in A086747. - N. J. A. Sloane, Dec 05 2019
From Gus Wiseman, Jun 10 2020: (Start)
Numbers k such that the k-th composition in standard order has all odd parts, or numbers k such that A124758(k) is odd. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. For example, the sequence of all compositions into odd parts begins:
()              57: (1,1,3,1)          135: (5,1,1,1)
(1)             60: (1,1,1,3)          144: (3,5)
(1,1)           63: (1,1,1,1,1,1)      147: (3,3,1,1)
(3)             64: (7)                153: (3,1,3,1)
(1,1,1)         67: (5,1,1)            156: (3,1,1,3)
(3,1)           73: (3,3,1)            159: (3,1,1,1,1,1)
(1,3)           76: (3,1,3)            192: (1,7)
(1,1,1,1)       79: (3,1,1,1,1)        195: (1,5,1,1)
(5)             97: (1,5,1)            201: (1,3,3,1)
(3,1,1)        100: (1,3,3)            204: (1,3,1,3)
(1,3,1)        103: (1,3,1,1,1)        207: (1,3,1,1,1,1)
(1,1,3)        112: (1,1,5)            225: (1,1,5,1)
(1,1,1,1,1)    115: (1,1,3,1,1)        228: (1,1,3,3)
(5,1)          121: (1,1,1,3,1)        231: (1,1,3,1,1,1)
(3,3)          124: (1,1,1,1,3)        240: (1,1,1,5)
(3,1,1,1)      127: (1,1,1,1,1,1,1)    243: (1,1,1,3,1,1)
(1,5)          129: (7,1)              249: (1,1,1,1,3,1)
(1,3,1,1)      132: (5,3)              252: (1,1,1,1,1,3)
(End)
Numbers whose binary representation has the property that every run of consecutive 0's has even length. - Harry Richman, Jan 31 2024

			From _Harry Richman_, Jan 31 2024: (Start)
In the following, dots are used for zeros in the binary representation:
   n  binary(a(n))  a(n)
   0:    .......     0
   1:    ......1     1
   2:    .....11     3
   3:    ....1..     4
   4:    ....111     7
   5:    ...1..1     9
   6:    ...11..    12
   7:    ...1111    15
   8:    ..1....    16
   9:    ..1..11    19
  10:    ..11..1    25
  11:    ..111..    28
  12:    ..11111    31
  13:    .1....1    33
  14:    .1..1..    36
  15:    .1..111    39
  16:    .11....    48
  17:    .11..11    51
  18:    .111..1    57
  19:    .1111..    60
  20:    .111111    63
  21:    1......    64
  22:    1....11    67
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Clark Kimberling, A Self-Generating Set and the Golden Mean, J. Integer Sequences, 3 (2000), #00.2.8.
Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52 (2014), 195-202.
Lukasz Merta, Composition inverses of the variations of the Baum-Sweet sequence, arXiv:1803.00292 [math.NT], 2018. See l(n) p. 5.
Gus Wiseman, Statistics, classes, and transformations of standard compositions

Cf. A219608 (odd terms), A060138, A060139, A060140, A060141, A086747.
Cf. A003714 (no consecutive 1's in binary expansion).
Odd partitions are counted by A000009.
Numbers with an odd number of 1's in binary expansion are A000069.
Numbers whose binary expansion has odd length are A053738.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without odd parts are A062880.
- Sum is A070939.
- Product is A124758.
- Strict compositions are A233564.
- Heinz number is A333219.
- Number of distinct parts is A334028.

import Data.Set (singleton, deleteFindMin, insert)
a060142 n = a060142_list !! n
a060142_list = 0 : f (singleton 1) where
   f s = x : f (insert (4 * x) $ insert (2 * x + 1) s') where
       (x, s') = deleteFindMin s
-- Reinhard Zumkeller, Nov 26 2012

Take[Nest[Union[Flatten[# /. {{i_Integer -> i}, {i_Integer -> 2 i + 1}, {i_Integer -> 4 i}}]] &, {1}, 5], 32]  (* Or *)
Select[Range[124], FreeQ[Length /@ Select[Split[IntegerDigits[#, 2]], First[#] == 0 &], ?OddQ] &] (* _Birkas Gyorgy, May 29 2012 *)

is(n)=if(n<3, n<2, if(n%2,is(n\2),n%4==0 && is(n/4))) \\ Charles R Greathouse IV, Oct 21 2013

Corrected by T. D. Noe, Nov 01 2006

Definition simplified by Charles R Greathouse IV, Oct 21 2013

A007066 a(n) = 1 + ceiling((n-1)*phi^2), phi = (1+sqrt(5))/2.

Original entry on oeis.org

1, 4, 7, 9, 12, 15, 17, 20, 22, 25, 28, 30, 33, 36, 38, 41, 43, 46, 49, 51, 54, 56, 59, 62, 64, 67, 70, 72, 75, 77, 80, 83, 85, 88, 91, 93, 96, 98, 101, 104, 106, 109, 111, 114, 117, 119, 122, 125, 127, 130, 132, 135, 138, 140, 143, 145, 148, 151, 153, 156, 159, 161, 164, 166

Offset: 1

N. J. A. Sloane, Mira Bernstein

First column of dual Wythoff array, A126714.
Positions of 0's in A189479.
Skala (2016) asks if this sequence also gives the positions of the 0's in A283310. - N. J. A. Sloane, Mar 06 2017
Upper Wythoff sequence plus 2, when shifted by 1. - Michel Dekking, Aug 26 2019
In the Fokkink-Joshi paper, this sequence is the Cloitre (0,1,2,3)-hiccup sequence, i.e., a(1) = 1; for m < n, a(n) = a(n-1)+2 if a(m) = n, else a(n) = a(n-1)+3. - Michael De Vlieger, Jul 30 2025

Clark Kimberling, "Stolarsky interspersions," Ars Combinatoria 39 (1995) 129-138.
D. R. Morrison, "A Stolarsky array of Wythoff pairs," in A Collection of Manuscripts Related to the Fibonacci Sequence. Fibonacci Assoc., Santa Clara, CA, 1980, pp. 134-136.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 9.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See pp. 3-4, 7, 10.
Clark Kimberling, Interspersions
Matthew Skala, Graph Nimors, arXiv preprint arXiv:1604.04072 [math.CO], 2016.
N. J. A. Sloane, Classic Sequences

Cf. A064437.
Apart from initial terms, same as A026356 (Cloitre (0,2,2,3)-hiccup sequence).
First column of A126714.
Complement is (essentially) A026355.
Equals 1 + A004957, also n + A004956.
First differences give A076662.
Complement of A099267. [Gerald Hillier, Dec 19 2008]
Cf. A193214 (primes). Except for the first term equal to A001950 + 2.
Cf. A026352 (Cloitre (1,1,2,3)-hiccup sequence), A064437 (Cloitre (0,1,3,2)-hiccup sequence).

a007066 n = a007066_list !! (n-1)
a007066_list = 1 : f 2 [1] where
   f x zs@(z:_) = y : f (x + 1) (y : zs) where
     y = if x `elem` zs then z + 2 else z + 3
-- Reinhard Zumkeller, Sep 26 2014, Sep 18 2011

Digits := 100: t := (1+sqrt(5))/2; A007066 := proc(n) if n <= 1 then 1 else floor(1+t*floor(t*(n-1)+1)); fi; end;

t = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0, 1}}] &, {0}, 6] (*A189479*)
Flatten[Position[t, 0]] (*A007066*)
Flatten[Position[t, 1]] (*A099267*)
With[{grs=GoldenRatio^2},Table[1+Ceiling[grs(n-1)],{n,70}]] (* Harvey P. Dale, Jun 24 2011 *)

from math import isqrt
def A007066(n): return (n+1+isqrt(5*(n-1)**2)>>1)+n if n > 1 else 1 # Chai Wah Wu, Aug 25 2022

a(n) = floor(1+phi*floor(phi*(n-1)+1)), phi=(1+sqrt(5))/2, n >= 2.
a(1)=1; for n>1, a(n)=a(n-1)+2 if n is already in the sequence, a(n)=a(n-1)+3 otherwise. - Benoit Cloitre, Mar 06 2003
a(n+1) = floor(n*phi^2) + 2, n>=1. - Michel Dekking, Aug 26 2019

A004956 a(n) = ceiling(n*phi), where phi is the golden ratio, A001622.

Original entry on oeis.org

0, 2, 4, 5, 7, 9, 10, 12, 13, 15, 17, 18, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 38, 39, 41, 43, 44, 46, 47, 49, 51, 52, 54, 56, 57, 59, 60, 62, 64, 65, 67, 68, 70, 72, 73, 75, 77, 78, 80, 81, 83, 85, 86, 88, 89

Offset: 0

N. J. A. Sloane

a(0)=0, a(1)=2; for n > 1, a(n) = a(n-1) + 2 if n is already in the sequence, a(n) = a(n-1) + 1 otherwise.
Integer solutions to the equation x = ceiling(phi*floor(x/phi)). - Benoit Cloitre, Feb 14 2004
From Benoit Cloitre, Mar 05 2007: (Start)
The following is an alternative way to obtain this sequence. NP means "term not in parentheses". Write down the natural numbers and mark the least NP, which is 1:
1* 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Take the first NP (which is 1) and parenthesize it; mark the least NP (which is 2):
(1*) 2* 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Take the 2nd NP (which is 3) and parenthesize it; mark the next NP (which is 4):
(1*) 2* (3) 4* 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Take the 4th NP (which is 6) and parenthesize it; mark the next NP (which is 5):
(1*) 2* (3) 4* 5* (6) 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Continuing in this way we obtain
(1*) 2* (3) 4* 5* (6) 7* (8) 9* 10* (11) 12* 13* (14) 15* (16) 17* (18) 19* ...
The starred entries (after the first) give the sequence. (End)
From Rick L. Shepherd, Dec 05 2009: (Start)
An equivalent statement of the sieving process described by Benoit Cloitre on Mar 05 2007:
Begin with the natural numbers N. Repeatedly perform these two steps:
i) Let k = N's least remaining term not yet used in Step ii).
ii) Remove the k-th remaining term from N.
The remaining terms of N are the (positive) terms shared by this sequence and A026351.
The terms removed from N (the complement) are A026352's terms (see also A004957).
The PARI program performs this sieving process and prints the positive terms of this sequence. (End)
In the Fokkink-Joshi paper, apart from a(0) = 0, this sequence is the Cloitre (0,2,2,1)-hiccup sequence, i.e., a(1) = 2; for m < n, a(n) = a(n-1)+2 if a(m) = n, else a(n) = a(n-1)+1. - Michael De Vlieger, Jul 30 2025

Christian Krause, Table of n, a(n) for n = 0..10000
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.
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 7.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See p. 3.

Cf. A001622, A004957, A026352 (Cloitre (1,1,2,3)-hiccup sequence).

Essentially same as A026351.

Ceiling[Range[0,100]GoldenRatio] (* Paolo Xausa, Oct 28 2023 *)
(* Second program: *)
cloitreH[j_, x_, y_, z_, w_ : 120] := Module[{c, k}, c[] := False; k = x; c[x] = True; {x}~Join~Reap[Do[If[c[n - j], k += y, k += z]; c[k] = True; Sow[k], {n, 2, w}] ][[-1, 1]] ]; {0}~Join~cloitreH[0, 2, 2, 1] (* _Michael De Vlieger, Jul 30 2025 *)

{/* paws = Print Absolute values of all elements in vector v With same Sign as sn */
paws(v,sn) = for(m=1,matsize(v)[2], if(sign(v[m])==sign(sn),\
print1(abs(v[m]),",")))}
{/* Sieve through lim numbers; make values negative to signify "removed" */
lim=100; v=vector(lim,i,i); i=0; j=0; c=1;
while(i0, k=v[i]; c=c--;
while(c0, c++)); v[j]=-v[j])); paws(v,1)\
/* Changing "1" to "-1" in paws() above prints out the terms of A026352 instead */} \\ Rick L. Shepherd, Dec 05 2009

a(n) = ceil(n*(1 + sqrt(5))/2); \\ Michel Marcus, Apr 13 2021

A013642 Numbers k such that the continued fraction for sqrt(k) has period 2.

Original entry on oeis.org

3, 6, 8, 11, 12, 15, 18, 20, 24, 27, 30, 35, 38, 39, 40, 42, 48, 51, 56, 63, 66, 68, 72, 80, 83, 84, 87, 90, 99, 102, 104, 105, 110, 120, 123, 132, 143, 146, 147, 148, 150, 152, 156, 168, 171, 182, 195, 198, 200, 203, 210, 224, 227, 228, 230, 231, 235, 240, 255, 258, 260, 264

Offset: 1

Clark Kimberling

nonn

This sequence is identical to the sequence of numbers of the form k = a^2 + b, where a and b are positive integers and b is a factor of 2a greater than 1, in which case the continued fraction expansion of sqrt(k) is [a; [2a/b, 2a]]. - David Terr, Jun 11 2004

Kenneth H. Rosen, Elementary Number Theory and Its Applications, Addison-Wesley, 1984, page 426 (but beware of errors!).

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)

Cf. A001066, A043549, A026604, A047219, A090848, A004957.

cf2Q[n_]:=Module[{s=Sqrt[n]},If[IntegerQ[s],1,Length[ ContinuedFraction[ s][[2]]]]==2]; Select[Range[300],cf2Q] (* Harvey P. Dale, Jun 21 2017 *)

A356327 Replace 2^k in binary expansion of n with A039834(1+k).

Original entry on oeis.org

0, 1, -1, 0, 2, 3, 1, 2, -3, -2, -4, -3, -1, 0, -2, -1, 5, 6, 4, 5, 7, 8, 6, 7, 2, 3, 1, 2, 4, 5, 3, 4, -8, -7, -9, -8, -6, -5, -7, -6, -11, -10, -12, -11, -9, -8, -10, -9, -3, -2, -4, -3, -1, 0, -2, -1, -6, -5, -7, -6, -4, -3, -5, -4, 13, 14, 12, 13, 15, 16

Offset: 0

Rémy Sigrist, Aug 03 2022

This sequence has similarities with A022290, and is related to negaFibonacci representations.

			For n = 13:
- 13 = 2^3 + 2^2 + 2^0,
- so a(13) = A039834(4) + A039834(3) + A039834(1) = -3 + 2 + 1 = 0.

Rémy Sigrist, Table of n, a(n) for n = 0..8191
Wikipedia, NegaFibonacci coding
Index entries for sequences related to Zeckendorf expansion of n

Cf. A000201, A001950, A003714, A004957, A020989, A022290, A026351, A039834, A060144, A072197, A189663, A215024, A215025, A309076, A356325, A356326.

Table[Reverse[#].Fibonacci[-Range[Length[#]]] &@ IntegerDigits[n, 2], {n, 0, 69}] (* Rémy Sigrist, Aug 05 2022 *)

a(n) = { my (v=0, k); while (n, n-=2^k=valuation(n, 2); v+=fibonacci(-1-k)); return (v) }

from sympy import fibonacci
def A356327(n): return sum(fibonacci(-a)*int(b) for a, b in enumerate(bin(n)[:1:-1],start=1)) # Chai Wah Wu, Aug 31 2022

a(n) = Sum_{k>=0} A030308(n,k)*A039834(1+k).
a(A215024(n)) = n.
a(A215025(n)) = -n.
a(A003714(n)) = A309076(n).
Empirically:
- a(n) = 0 iff n = 0 or n belongs to A072197,
- a(n) = 1 iff n belongs to A020989,
- a(2*A215024(n)) = -A000201(n) for n > 0,
- a(3*A215024(n)) = -A060143(n),
- a(floor(A215024(n)/2)) = -A060143(n),
- a(4*A215024(n)) = A001950(n) for n > 0,
- a(floor(A215024(n)/4)) = A189663(n) for n > 0,
- a(2*A215025(n)) = A026351(n),
- a(3*A215025(n)) = A019446(n) for n > 0,
- a(floor(A215025(n)/2)) = A019446(n) for n > 0,
- a(4*A215025(n)) = -A004957(n),
- a(floor(A215025(n)/4)) = -A060144(n+1) for n >= 0.

A329130 a(0)=0; for any n >= 0, if a(n) > n then a(n+1) = a(n) - n, otherwise a(n+1) = a(n) + k, where k is the total number of terms a(m) <= m with m <= n.

Original entry on oeis.org

0, 1, 3, 1, 4, 8, 3, 8, 1, 7, 14, 4, 12, 21, 8, 18, 3, 14, 26, 8, 21, 1, 15, 30, 7, 23, 40, 14, 32, 4, 23, 43, 12, 33, 55, 21, 44, 8, 32, 57, 18, 44, 3, 30, 58, 14, 43, 73, 26, 57, 8, 40, 73, 21, 55, 1, 36, 72, 15, 52, 90

Offset: 0

James Marjamaa, Nov 05 2019

Values where a(n) = n appear to be the values of A027941.

			For n = 0,   a(0) =            0    0 >= a(0)
    n = 1,   a(1) = a(0) + 1 = 1,   1 >= a(1), I = 1
    n = 2,   a(2) = a(1) + 2 = 3,   2 <  a(2), I = 2
    n = 3,   a(3) = a(2) - 2 = 1,   3 >= a(3)
    n = 4,   a(4) = a(3) + 3 = 4,   4 >= a(4), I = 3
    n = 5,   a(5) = a(4) + 4 = 8,   5 <  a(5), I = 4
    n = 6,   a(6) = a(5) - 5 = 3,   6 >= a(6)
    n = 7,   a(7) = a(6) + 5 = 8,   7 <  a(7), I = 5
    n = 8,   a(8) = a(7) - 7 = 1,   8 >= a(8)
    n = 9,   a(9) = a(8) + 6 = 7,   9 >= a(9), I = 6
    n = 10,  a(10)= a(9) + 7 = 14,  10<  a(10), I = 7
    n = 11,  a(11)= a(10)- 10= 4,   11>= a(11)
    n = 12,  a(12)= a(11)+ 8 = 12,  12>= a(12), I = 8
    .
    .
    .

James Marjamaa, Table of n, a(n) for n = 0..10000
Rémy Sigrist, Logarithmic scatterplot of the first 1000000 terms of the ordinal transform of the ordinal transform of the sequence

Cf. A027941, A008344.

#include
void seq(int terms)
{int n = 0; int i = 0; int a = 0; int c = 0;
while (n <= terms)
{
    if (c)
       {int N = n - 1;
        a -= N;
        printf("%d\n", a);}
    else
       {a += i;
        printf("%d\n", a);
        i++;}
    if (a > n)
       {c = 1;}
    else
       {c = 0;}
    n++;
}
}
int main(void)
{
seq(1000);
}

v=k=0; for (n=0, 69, print1 (v, ", "); v=if (v>n, v-n, v+k++)) \\ Rémy Sigrist, Nov 06 2019

a(n+1) = a(n) + I, if n >= a(n) ('I' is the next consecutive integer not yet added);
       = a(n) - n, if n < a(n).
From Yan Sheng Ang, May 20 2020: (Start)
a(A004957(n)) = a(n). Hence it follows that (writing F(n) = A000045(n)):
  a(F(2*n)-1) = F(2*n) for n > 1;
  a(F(2*n)) = 1;
  a(F(2*n+1)-1) = F(2*n+1)-1 (as noted above);
  a(F(2*n+1)) = F(2*n+2);
  a(F(2*n+1)+1) = F(2*n) for n > 1.
(End)

cp's OEIS Frontend

A026352 a(n) = floor(n*tau) + n + 1 where tau is the golden ratio A001622.

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A060142 Ordered set S defined by these rules: 0 is in S and if x is in S then 2x+1 and 4x are in S.

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A007066 a(n) = 1 + ceiling((n-1)*phi^2), phi = (1+sqrt(5))/2.

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A004956 a(n) = ceiling(n*phi), where phi is the golden ratio, A001622.

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A013642 Numbers k such that the continued fraction for sqrt(k) has period 2.

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A356327 Replace 2^k in binary expansion of n with A039834(1+k).

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A329130 a(0)=0; for any n >= 0, if a(n) > n then a(n+1) = a(n) - n, otherwise a(n+1) = a(n) + k, where k is the total number of terms a(m) <= m with m <= n.

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