

A019446


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


18



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

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 n1=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 = phi1. 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) = (53)floor(phi*n)+(2*35)*n+r = 2floor(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) = 2floor(phi*n)+n+1. To incorporate also the first "singleton term" a(1)=1, we take
b(n) = V(n1) = 2floor(phi*(n1)+n1+1 = 2floor(phi*(n1))+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


LINKS

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]


FORMULA

a(1)=1; a(n) = n+1  a(a(n1)).  Benoit Cloitre, Nov 06 2002
a(n) = A005206(n1) + 1.  Reinhard Zumkeller, Feb 02 2012; corrected by Primoz Pirnat, Dec 28 2020
a(n) = A019445(n) / n.  Sean A. Irvine, Mar 17 2019


EXAMPLE

a(6)=4 since 61=[i+j*tau] for these (i,j): (5,0), (4,1), (2,2), (1,3).  Clark Kimberling, Jun 18 2002


MAPLE

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


MATHEMATICA

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


PROG

(Haskell)
a019446 n = a019446_list !! (n1)
a019446_list = 1 : zipWith () [3..] (map a019446 a019446_list)
 Reinhard Zumkeller, Feb 02 2012
(GAP) a:=[1];; for n in [2..80] do a[n]:=n+1a[a[n1]]; od; a; # Muniru A Asiru, Aug 30 2018
(Python)
from math import isqrt
def A019446(n): return (n+isqrt(5*n**2)>>1)n+1 # Chai Wah Wu, Aug 09 2022


CROSSREFS

Cf. A001622, A019444, A019445, A026351, A026352, A005206.
Sequence in context: A256502 A177151 A076935 * A097369 A257808 A249036
Adjacent sequences: A019443 A019444 A019445 * A019447 A019448 A019449


KEYWORD

nonn,easy,nice


AUTHOR

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


EXTENSIONS

Better name from David Radcliffe and John Rickard, Dec 12 2000
Edited by Dean Hickerson, Nov 09 2002


STATUS

approved



