A019444 -id:A019444 - cp's OEIS frontend

A382162 Number of pairs (i,j), 1 <= i < j <= n such that A019444(i) < A019444(j).

0, 1, 2, 5, 9, 12, 18, 22, 30, 39, 45, 56, 68, 76, 90, 99, 115, 132, 143, 162, 174, 195, 217, 231, 255, 280, 296, 323, 340, 369, 399, 418, 450, 483, 504, 539, 561, 598, 636, 660, 700, 725, 767, 810, 837, 882, 928, 957, 1005, 1035, 1085, 1136, 1168, 1221, 1254, 1309, 1365, 1400, 1458, 1517, 1554, 1615, 1653, 1716, 1780, 1820

Offset: 1

N. J. A. Sloane, Mar 31 2025

nonn

The count of pairs with A019444(i) > A019444(j) gives triangular numbers and is less interesting.

Cf. A019444.

A382168 Number of triples (i,j,k), 1 <= i < j < k <= n such that A019444(i) < A019444(k) < A019444(j).

Original entry on oeis.org

0, 0, 1, 1, 1, 7, 7, 17, 17, 17, 38, 38, 38, 74, 74, 119, 119, 119, 185, 185, 263, 263, 263, 368, 368, 368, 504, 504, 657, 657, 657, 847, 847, 847, 1078, 1078, 1331, 1331, 1331, 1631, 1631, 1956, 1956, 1956, 2334, 2334, 2334, 2769, 2769, 3234, 3234, 3234, 3762, 3762, 4323, 4323, 4323, 4953, 4953, 4953, 5656, 5656, 6397, 6397, 6397

Offset: 1

N. J. A. Sloane, Mar 31 2025

nonn

Counts occurrences of the pattern 132 in A019444. Note that the Catalan numbers (A000108) count permutations of 1,2,...,n that avoid the pattern 132.

Cf. A000108, A019444, A382162, A382169.

A382710 Smallest missing number after A019444(n) has been computed.

Original entry on oeis.org

2, 2, 4, 4, 4, 5, 5, 7, 7, 7, 9, 9, 9, 10, 10, 12, 12, 12, 13, 13, 15, 15, 15, 17, 17, 17, 18, 18, 20, 20, 20, 22, 22, 22, 23, 23, 25, 25, 25, 26, 26, 28, 28, 28, 30, 30, 30, 31, 31, 33, 33, 33, 34, 34, 36, 36, 36, 38, 38, 38, 39, 39, 41, 41, 41, 43, 43, 43, 44, 44, 46, 46, 46, 47, 47, 49, 49, 49, 51, 51, 51, 52, 52, 54, 54, 54, 56

Offset: 1

N. J. A. Sloane, Apr 06 2025

nonn

The RUNS transform is A282162.

Cf. A019444, A282162.

A002251 Start with the nonnegative integers; then swap L(k) and U(k) for all k >= 1, where L = A000201, U = A001950 (lower and upper Wythoff sequences).

Original entry on oeis.org

0, 2, 1, 5, 7, 3, 10, 4, 13, 15, 6, 18, 20, 8, 23, 9, 26, 28, 11, 31, 12, 34, 36, 14, 39, 41, 16, 44, 17, 47, 49, 19, 52, 54, 21, 57, 22, 60, 62, 24, 65, 25, 68, 70, 27, 73, 75, 29, 78, 30, 81, 83, 32, 86, 33, 89, 91, 35, 94, 96, 37, 99, 38, 102, 104, 40, 107, 109

Offset: 0

Michael Kleber

(n,a(n)) are Wythoff pairs: (0,0), (1,2), (3,5), (4,7), ..., where each difference occurs once.
Self-inverse when considered as a permutation or function, i.e., a(a(n)) = n. - Howard A. Landman, Sep 25 2001
If the offset is 1, the sequence can also be obtained by rearranging the natural numbers so that sum of n terms is a multiple of n, or equivalently so that the arithmetic mean of the first n terms is an integer. - Amarnath Murthy, Aug 16 2002
For n = 1, 2, 3, ..., let p(n)=least natural number not already an a(k), q(n) = n + p(n); then a(p(n)) = q(n), a(q(n)) = p(n). - Clark Kimberling
Also, indices of powers of 2 in A086482. - Amarnath Murthy, Jul 26 2003
There is a 7-state Fibonacci automaton (see a002251_1.pdf) that accepts, in parallel, the Zeckendorf representations of n and a(n). - Jeffrey Shallit, Jul 14 2023

E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 76.

T. D. Noe, 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. 18.
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
Eric Duchene, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, and Urban Larsson, Wythoff Wisdom, 43 pages, no date, unpublished.
Eric Duchene, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, and Urban Larsson, Wythoff Wisdom, unpublished, no date [Cached copy, with permission]
Alex Meadows and B. Putman, A New Twist on Wythoff's Game, arXiv preprint arXiv:1606.06819 [math.CO], 2016.
Gabriel Nivasch, More on the Sprague-Grundy function for Wythoff's game, pages 377-410 in "Games of No Chance 3", MSRI Publications Volume 56, 2009.
Jeffrey Shallit, Automaton for A002251
Jeffrey Shallit, Proving properties of some greedily-defined integer recurrences via automata theory, arXiv:2308.06544 [cs.DM], 2023.
R. Silber, Wythoff's Nim and Fibonacci Representations, Fibonacci Quarterly #14 (1977), pp. 85-88.
N. J. A. Sloane, Scatterplot of first 100 terms [The points are symmetrically placed about the diagonal, although that is hard to see here because the scales on the axes are different]
Index entries for sequences that are permutations of the natural numbers

The sequence maps between A000201 and A001950, in that a(A000201(n)) = A001950(n), a(A001950(n)) = A000201(n).
Row 0 of A018219.
Cf. A073869, A342297.

With[{n = 42}, {0}~Join~Take[Values@ #, LengthWhile[#, # == 1 &] &@ Differences@ Keys@ #] &@ Sort@ Flatten@ Map[{#1 -> #2, #2 -> #1} & @@ # &, Transpose@ {Array[Floor[# GoldenRatio] &, n], Array[Floor[# GoldenRatio^2] &, n]}]] (* Michael De Vlieger, Nov 14 2017 *)

A002251_upto(N,c=0,A=Vec(0,N))={for(n=1,N, A[n]||(#AA002251[1]=2, a(0)=0 is not included. - M. F. Hasler, Nov 27 2019, replacing earlier code from Sep 17 2014

a(n) = A019444(n+1) - 1.

Edited by Christian G. Bower, Oct 29 2002

A090909 Terms a(k) of A073869 for which a(k-1) = a(k), and a(k) and a(k+1) are distinct.

Original entry on oeis.org

2, 5, 7, 10, 13, 15, 18, 20, 23, 26, 28, 31, 34, 36, 39, 41, 44, 47, 49, 52, 54, 57, 60, 62, 65, 68, 70, 73, 75, 78, 81, 83, 86, 89, 91, 94, 96, 99, 102, 104, 107, 109, 112, 115, 117, 120, 123, 125, 128, 130, 133, 136, 138, 141, 143, 146, 149, 151, 154, 157, 159, 162

Offset: 1

Amarnath Murthy, Dec 14 2003

nonn

Is this the same as A001950? - Alec Mihailovs (alec(AT)mihailovs.com), Jul 23 2007
Identical to n + A066096(n)? - Ed Russell (times145(AT)hotmail.com), May 09 2009
From Michel Dekking, Dec 18 2024: (Start)
Proof of Mihailovs's conjecture: This follows immediately from the result in my 2023 paper in JIS that A073869 is equal to Hofstadter’s G-sequence A005206, and my recent comment in A005206 on the pairs of duplicate values in A005206.
The answer to Russell’s question is well-known, and also Detlef’s formula is well-known.
Originally, this sequence was given the name “Terms a(k) of A073869 for which a(k-1), a(k) and a(k+1) are distinct.’’ These are the triples (1,2,3),(4,5,6),(6,7,8), (9,10,11), ... occurring at  k = 3, k = 8, k = 11, k = 16,... in A005206. Note that if a duplicate pair (a(m-1), a(m)) is followed directly by another duplicate pair, then a(m-3), a(m-2) and a(m-1) are distinct, and only so. This corresponds to the block 00 occurring in the Fibonacci word obtained by projecting A005206 on the Fibonacci word (see Corollary in my recent comment in A005206). These occurrences are at the Wythoff AB numbers A003623 according to Wolfdieter Lang’s comment in A003623. Conclusion: the sequence of terms a(k) of A073869 for which a(k-1), a(k), and a(k+1) are distinct is given by the Wythoff AB-numbers. (End)

			A073869 = A005206 = 0,1,1,2,3,3,4,4,5,6,6,... The pair (1,1) occurs at k = 2.

G. C. Greubel, Table of n, a(n) for n = 0..10000
F. M. Dekking, On Hofstadter's G-sequence, Journal of Integer Sequences 26 (2023), Article 23.9.2, 1-11.

Cf. A001622, A001950, A002251, A005206, A073869, A090908.
Cf. A004919, A004920, A004921, A004922, A004923, A004924, A004925.
Cf. A004926, A004927, A004928, A004929, A004930, A004931, A004932.
Cf. A004933, A004934, A004935, A004976, A066096.

[Floor((3+Sqrt(5))*n/2): n in [0..80]]; // G. C. Greubel, Sep 12 2023

(* First program *)
A002251= Fold[Append[#1, #2 Ceiling[#2/GoldenRatio] -Total[#1]] &, {1}, Range[2, 500]] - 1; (* Birkas Gyorgy's code of A019444, modified *)
A090909= Join[{0}, Select[Partition[A002251, 2, 1], #[[2]] > #[[1]] &][[All, 2]]] (* G. C. Greubel, Sep 12 2023 *)
(* Second program *)
Floor[GoldenRatio^2*Range[0,80]] (* G. C. Greubel, Sep 12 2023 *)

[floor(golden_ratio^2*n) for n in range(81)] # G. C. Greubel, Sep 12 2023

a(n) = floor(phi^2*n), where phi = (1+sqrt(5))/2. - Gary Detlefs, Mar 10 2011

More terms from R. J. Mathar, Sep 29 2017

Name corrected by Michel Dekking, Dec 13 2024

A019446 a(n) = ceiling(n/tau), where tau = (1+sqrt(5))/2.

Original entry on oeis.org

1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8, 9, 9, 10, 10, 11, 12, 12, 13, 13, 14, 15, 15, 16, 17, 17, 18, 18, 19, 20, 20, 21, 22, 22, 23, 23, 24, 25, 25, 26, 26, 27, 28, 28, 29, 30, 30, 31, 31, 32, 33, 33, 34, 34, 35, 36, 36, 37, 38, 38, 39, 39, 40, 41, 41, 42, 43, 43, 44, 44, 45, 46, 46

Offset: 1

R. K. Guy, Tom Halverson (halverson(AT)macalester.edu)

Average of first n terms of A019444, which is defined to be a permutation of the positive integers, p_1, p_2, ..., such that the average of each initial segment is an integer, using the greedy algorithm to define p_n.
Number of pairs (i,j) of nonnegative integers such that n-1=floor(i+j*tau). - Clark Kimberling, Jun 18 2002
The terms that occur exactly once are 1,3,6,8,..., given by A026352(n)=n+1+floor(n*tau). - Clark Kimberling, Jun 18 2002
The number n appears A001468(n) times. - Reinhard Zumkeller, Feb 02 2012
It seems that the indices of the terms that occur exactly once are listed in A276885. - Ivan N. Ianakiev, Aug 30 2018
From Michel Dekking, Oct 13 2020: (Start)
Here is a proof of the conjecture by Ivan N. Ianakiev. Let b = (b(n)) be the sequence of occurrences of the "singleton terms" in (a(n)). We have to show that b = A276885.
In the following phi := (1+sqrt(5))/2 (so phi = tau).
By its definition, the sequence (a(n)) is a generalized Beatty sequence with terms  a(n) = floor(phi*n)-n+1, since 1/phi = phi-1.  So by Lemma 8 in the paper by Allouche and Dekking, its sequence of first differences Delta = 1011010110..., given by Delta(n) = a(n+1)-a(n), is equal to y, where y = A005614 is the binary complement of the Fibonacci word. By definition, y is the fixed point of the morphism nu: 0->1, 1->10.
The crucial observation is that a term occurs exactly once in (a(n)) if and only if the word 11 of length 2 occurs in Delta (with an exception for a(1)=1). So to obtain the sequence b of occurrences of these "singleton terms", we have to study the return words of 11 in y. (The return words of 11 in y are the words occurring in y that start with 11, and having no other occurrences of 11.)
The return words of 11 are the words A:=11010, and B:=110. Since
      nu(A) = nu(11010) = 10101101, nu(B) = nu(110) = 10101,
the morphism nu induces a descendant morphism tau given by
      tau(A) = BA, tau(B) = A.
So tau is nothing else but the Fibonacci morphism on the alphabet {B,A}.
Since the words A and B have lengths 5 and 3, the first differences b(n+1)-b(n) are given by the fixed point z = 5353353533... of the Fibonacci morphism on the alphabet {5,3}.
From Lemma 8 in the paper by Allouche and Dekking we then obtain that the sequence b is a generalized Beatty sequence
      V(n) = (5-3)*floor(phi*n)+(2*3-5)*n+r = 2*floor(phi*n)+n+r, for some integer r.
Starting at the value 4, filling in n=1, we obtain that r=1, and so V(n) = 2*floor(phi*n)+n+1. To incorporate also the first "singleton term" a(1)=1, we take
      b(n) = V(n-1) = 2*floor(phi*(n-1))+n-1+1 = 2*floor(phi*(n-1))+n.
Then, indeed, b(n) = A276885(n), for n=1,2,... (see my Comment in A276885).
(End)
It seems that the indices of the records are listed in A026351. - Ivan N. Ianakiev, Mar 25 2021

			a(6)=4 since 6-1=[i+j*tau] for these (i,j): (5,0), (4,1), (2,2), (1,3). - _Clark Kimberling_, Jun 18 2002

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J.-P. Allouche and F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424 [math.NT], 2018.
Problem of the week, Problem 818
J. Rickard, Rearrangement of the natural numbers [Broken link]
J. Shallit, Proving properties of some greedily-defined integer recurrences via automata theory, arXiv:2308.06544 [cs.DM], August 12 2023.

Cf. A001622, A019444, A019445, A026351, A026352, A005206.

a:=[1];; for n in [2..80] do a[n]:=n+1-a[a[n-1]]; od; a; # Muniru A Asiru, Aug 30 2018

a019446 n = a019446_list !! (n-1)
a019446_list = 1 : zipWith (-) [3..] (map a019446 a019446_list)
-- Reinhard Zumkeller, Feb 02 2012

A019446:=n->ceil(2*n/(1+sqrt(5))); seq(A019446(n), 1..100); # Wesley Ivan Hurt, Jan 19 2014

Ceiling[Range[80]/GoldenRatio] (* Harvey P. Dale, Aug 02 2011 *)

from math import isqrt
def A019446(n): return (n+isqrt(5*n**2)>>1)-n+1 # Chai Wah Wu, Aug 09 2022

a(1)=1; a(n) = n+1 - a(a(n-1)). - Benoit Cloitre, Nov 06 2002
a(n) = A005206(n-1) + 1. - Reinhard Zumkeller, Feb 02 2012; corrected by Primoz Pirnat, Dec 28 2020
a(n) = A019445(n) / n. - Sean A. Irvine, Mar 17 2019

Better name from David Radcliffe and John Rickard, Dec 12 2000

Edited by Dean Hickerson, Nov 09 2002

A243700 The lexicographically earliest sequence of distinct terms with a(1) = 1 such that a(n) divides the sum of the first a(n) terms.

Original entry on oeis.org

1, 3, 2, 5, 9, 7, 8, 13, 15, 11, 14, 16, 26, 24, 41, 29, 18, 28, 20, 30, 22, 32, 25, 33, 43, 45, 31, 37, 50, 52, 54, 56, 58, 35, 87, 38, 55, 67, 40, 60, 72, 44, 63, 77, 79, 47, 70, 49, 121, 88, 53, 129, 94, 96, 98, 100, 59, 89, 105, 107, 62, 158, 113, 65, 102, 68, 103, 189

Offset: 1

N. J. A. Sloane, Jun 12 2014

If the sum a(1) + a(2) + ... + a(m) is not divisible by m, then m does not belong to this sequence. Sequence A019444 gives a variant of this sequence, where every positive integer is a term. - Max Alekseyev, Jun 11 2014
Positive integers that do not appear in this sequence form A243864.
Is there any index n > 3 such that a(n) <= n? - Max Alekseyev, Jun 13 2014
Comment from Max Alekseyev, Jun 19 2014 (Start)
I've added b-files for sequences A244010 and A244011. Btw, it may be worth adding two more related sequences:
(a1) Integers that appear in the current sequence, sorted (A244016).
(a2) The corresponding ratios (similar to A244011), i.e., a2(n) = A244010(a1(n)) / a1(n). (A244011) (End)
From Bill McEachen, May 21 2024: (Start)
Conjecture: For n > 1000, a(n) falls within 1% of one of the following six values. a(n) = n, 1.576385*n, 1.788185*n, 2.576385*n, 2.788185*n, or 3.576285*n, using floor at the low bound and ceiling at the high bound, inclusive.
For example, a(1153) = 1836. This is between floor(1.576385 * 1153 * 0.99) and ceiling(1.576385 * 1153 * 1.01). About 90% of values fall in the three lower slopes. (End)
Conjectured asymptotic slopes of the 6 lines in the function graph are, from low to high: 1, sqrt(3)/3+1, sqrt(3)/6+3/2, sqrt(3)/3+2, sqrt(3)/6+5/2, sqrt(3)/3+3. - Hugo Pfoertner, Dec 19 2024

			1 divides the sum of the first 1 term  (yes:  1/1=1)
3 divides the sum of the first 3 terms (yes:  6/3=2)
2 divides the sum of the first 2 terms (yes:  4/2=2)
5 divides the sum of the first 5 terms (yes: 20/5=4)
9 divides the sum of the first 9 terms (yes: 63/9=7)
7 divides the sum of the first 7 terms (yes: 35/7=5)
8 divides the sum of the first 8 terms (yes: 48/8=6)
...

Max Alekseyev, Table of n, a(n) for n = 1..100000 (first 1100 terms from Jean-Marc Falcoz)
Éric Angelini, a(n) divides the sum of the first a(n) terms of T, posting to the Sequence Fans Mailing List, Jun 11 2014 [Broken link]
Éric Angelini, a(n) divides the sum of the first a(n) terms of T, posting to the Sequence Fans Mailing List, Jun 11 2014, Cached copy.
Éric Angelini, Franklin Adams-Watters, Max Alekseyev, A. E. Povolotsky, N. J. A. Sloane, and R. G. Wilson v, a(n) divides the sum of the first a(n) terms of T, Various postings to the old Sequence Fans Mailing List, assembled by _N. J. A. Sloane_, Dec 24 2024
Hugo Pfoertner, 1.73*10^6 terms, 7z compressed b-file.

Cf. A019444, A243864 (complement, i.e. the missing terms), A244010 (partial sums), A244011 (the quotients), A244016 (sorted), A244017, A244018.

{ printA243700() = my( S=Set(), T=[], s=0, m=1, k); for(n=1,10^5, k=m; while( ((k==n || setsearch(S,n)) && Mod(s+k,n)) || if(kMax Alekseyev, Jun 13 2014

First 1100 terms were computed by Jean-Marc Falcoz.

Edited by N. J. A. Sloane, Dec 18 2024, adding comments from the lost Sequence Fans Mailing List archive. Deleted an incorrect comment.

A316156 The lexicographically earliest increasing sequence such that a(n) divides the sum of the first a(n)+1 terms.

Original entry on oeis.org

1, 2, 3, 6, 7, 8, 9, 13, 15, 17, 18, 19, 20, 31, 32, 39, 40, 43, 55, 59, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 86, 105, 106, 107, 108, 109, 110, 111, 118, 135, 136, 137, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 184, 185, 186, 187, 195

Offset: 1

Jaroslav Krizek, Aug 20 2018

nonn

Sequence b(n) of the sums of the first a(n)+1 terms of a(n) = Sum_{k=1..a(n)+1} a(k): 3, 6, 12, 36, 49, 64, 81, 169, 240, 323, 378, 437, 500, 1271, 1376, 2145, 2280, 2709, 4675, 5428, ... = A318872(1+a(n)).
Sequence c(n) of quotients when a(n) is calculated = (Sum_{k=1..a(n)+1} a(k) ) / a(n): 3, 3, 4, 6, 7, 8, 9, 13, 16, 19, 21, 23, 25, 41, 43, 55, 57, 63, 85, 92, ...
Is there a lexicographically earliest bijective sequence such that a(n) divides the sum of the first a(n)+1 terms?

			a(1) = 1 because 1 divides the sum of the first 2 (i.e., a(1) + 1) terms (a(1) + a(2)) for whatever term a(2) > a(1).
a(2) = 2 because 2 is the smallest number > a(1) and 2 divides the sum of the first 3 (i.e., a(2) + 1) terms (a(1) + a(2) + a(3)) for whatever term a(3) > a(2) such that 2 divides the sum a(1) + a(2) + a(3); the smallest number > a(2) with this property for a(3) is 3.
a(3) = 3.
a(4) = 6 because 6 is the smallest number > a(3) such that term a(3) = 3 divides the sum of the first 4 (i.e., a(3) + 1) terms.
a(5) = 7 and a(6) = 8 because a(4) < a(5) < a(6) and 6 divides sum of the first 7 (i.e., a(4) + 1) terms (a(1) + a(2) + ... + a(7)) for whatever term a(7) > a(6) such that 6 divides the sum a(1) + a(2) + ... + a(7); the smallest number with this property for a(7) is 9.
a(7) = 9.

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A019444, A244671, A316571.

Cf. A318872 (partial sums), A318873 (first differences).

povisin(v,n) = { forstep(j=n,1,-1, if(v[j] == n, return(j))); (0); }; \\ Here: povisin = position_of_n_in_strictly_increasing_v
A316156list(up_to) = { my(v316156 = vector(up_to), v318872 = vector(up_to), k, s); v316156[1] = v318872[1] = 1; for(n=2, up_to, k = 1+v316156[n-1]; if(povisin(v316156, n-1), s = v318872[n-1]; while((s+k)%(n-1), k++)); v316156[n] = k; v318872[n] = v318872[n-1] + v316156[n]); (v316156); }; \\ Antti Karttunen, Sep 16 2018

a(1) = 1; for n > 1, if n-1 is not in the sequence, a(n) = a(n-1)+1, otherwise, a(n) is the least k > a(n-1) such that A318872(n-1)+k is a multiple of n-1. - Antti Karttunen, Sep 16 2018

A340510 A permutation of the positive integers with a divisibility property (see Comments for precise definition).

Original entry on oeis.org

1, 3, 5, 2, 8, 10, 4, 13, 15, 6, 18, 7, 21, 23, 9, 26, 28, 11, 31, 12, 34, 36, 14, 39, 41, 16, 44, 17, 47, 49, 19, 52, 20, 55, 57, 22, 60, 62, 24, 65, 25, 68, 70, 27, 73, 75, 29, 78, 30, 81, 83, 32, 86, 33, 89, 91, 35, 94, 96, 37, 99, 38, 102, 104, 40, 107, 109

Offset: 1

N. J. A. Sloane, Jan 28 2021

a(1)=1; thereafter a(n) is the least positive number not yet in the sequence such that Sum_{i=1..n} a(i) == 1 mod n+1.

Alois P. Heinz, Table of n, a(n) for n = 1..10000
Muharem Avdispahić and Faruk Zejnulahi, An integer sequence with a divisibility property, Fibonacci Quarterly, Vol. 58:4 (2020), 321-333.
J. Shallit, Proving properties of some greedily-defined integer recurrences via automata theory, arXiv:2308.06544 [cs.DM], August 12 2023.

Cf. A000045, A001622, A019444.

A000045_inv := proc(n)
    local k ;
    for k from 0 do
        if combinat[fibonacci](k) = n then
            return k;
        elif combinat[fibonacci](k) > n then
            return -1;
        end if;
    end do:
end proc:
A000071_inv := proc(n)
    local k ;
    for k from 0 do
        if combinat[fibonacci](k)-1 = n then
            return k;
        elif combinat[fibonacci](k)-1 > n then
            return -1;
        end if;
    end do:
end proc:
A000201_inv := proc(n)
    local k,w ;
    for k from 1 do
        w := floor(k*(1+sqrt(5))/2) ;
        if w = n then
            return k;
        elif w > n then
            return -1;
        end if;
    end do:
end proc:
A001950_inv := proc(n)
    local k,w ;
    for k from 1 do
        w := floor(k*(3+sqrt(5))/2) ;
        if w = n then
            return k;
        elif w > n then
            return -1;
        end if;
    end do:
end proc:
A340510 := proc(n)
    local k ;
    if n = 1 then
        1;
    else
        k := A000045_inv(n) ;
        if k > 2 then
            return combinat[fibonacci](k+1) ;
        end if;
        k := A000071_inv(n) ;
        if k > 4 then
            return combinat[fibonacci](k-1)-1 ;
        end if;
        k := A000201_inv(n) ;
        if k > 0 then
            return floor(k*(3+sqrt(5))/2) ;
        end if;
        k := A001950_inv(n) ;
        return floor(k*(1+sqrt(5))/2) ;
    end if;
end proc:
seq(A340510(n),n=1..50) ; # R. J. Mathar, Jan 30 2024

a[n_] := a[n] = Switch[n, 1, 1, 2, 3, 3, 5, 4, 2, _, Module[{aa, ss, dd, an}, aa = Array[a, n-1]; ss = Sort[aa]; dd = Differences[ss]; For[an = Select[Transpose[{Rest[ss], dd}], #[[2]] == 1 &][[-1, 1]]+1, True, an++, If[FreeQ[aa = Array[a, n-1], an], If[Mod[Total[aa] + an, n+1] == 1, Return[an]]]]]];
Table[Print[n, " ", a[n]]; a[n], {n, 1, 100}] (* Jean-François Alcover, Jan 31 2021 *)

Theorem 1 of Avdispahić and Zejnulahi gives an explicit formula involving Fibonacci numbers.

More terms from Alois P. Heinz, Jan 28 2021

A019445 Form a permutation of the positive integers, p_1, p_2, ..., such that the average of each initial segment is an integer, using the greedy algorithm to define p_n; sequence gives p_1 + ... + p_n.

Original entry on oeis.org

1, 4, 6, 12, 20, 24, 35, 40, 54, 70, 77, 96, 117, 126, 150, 160, 187, 216, 228, 260, 273, 308, 345, 360, 400, 442, 459, 504, 522, 570, 620, 640, 693, 748, 770, 828, 851, 912, 975, 1000, 1066, 1092, 1161, 1232, 1260, 1334, 1410, 1440, 1519, 1550

Offset: 1

R. K. Guy, Tom Halverson (halverson(AT)macalester.edu)

nonn

It appears that a(n) is divisible by n. - Michael Somos, Jan 29 2004

Somos's conjecture is proved in both Shapovalov (1996) and Venkatachala (2009). - Jeffrey Shallit, Jul 18 2023

Alois P. Heinz, Table of n, a(n) for n = 1..10000
J. Shallit, Proving properties of some greedily-defined integer recurrences via automata theory, arXiv:2308.06544 [cs.DM], August 12 2023.
A. Shapovalov, Problem M1517 (in Russian), Kvant 5 (1995), 20-21. English translation appeared in Quantum problem M185, Sept/October 1996 (beware, file is 75Mb).
The Math Forum, Problem of the Week 818.
B. J. Venkatachala, A curious bijection on natural numbers, JIS 12 (2009) 09.8.1.

Cf. A019444, A019446.

Partial sums of A019444. - Sean A. Irvine, Mar 17 2019

a(n) = n * A019446(n). - Joerg Arndt, Jul 23 2023

cp's OEIS Frontend

A382162 Number of pairs (i,j), 1 <= i < j <= n such that A019444(i) < A019444(j).

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A382168 Number of triples (i,j,k), 1 <= i < j < k <= n such that A019444(i) < A019444(k) < A019444(j).

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A382710 Smallest missing number after A019444(n) has been computed.

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A002251 Start with the nonnegative integers; then swap L(k) and U(k) for all k >= 1, where L = A000201, U = A001950 (lower and upper Wythoff sequences).

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A090909 Terms a(k) of A073869 for which a(k-1) = a(k), and a(k) and a(k+1) are distinct.

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A019446 a(n) = ceiling(n/tau), where tau = (1+sqrt(5))/2.

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A243700 The lexicographically earliest sequence of distinct terms with a(1) = 1 such that a(n) divides the sum of the first a(n) terms.

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A316156 The lexicographically earliest increasing sequence such that a(n) divides the sum of the first a(n)+1 terms.

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