cp's OEIS Frontend

Self-inverse when considered as a permutation or function, i.e., a(a(n)) = n. - Howard A. Landman, Sep 25 2001

That each initial segment has an integer average is trivially equivalent to the sum of the first n elements always being divisible by n. - Franklin T. Adams-Watters, Jul 07 2014

Also, a lexicographically minimal sequence of distinct positive integers such that all values of a(n)-n are also distinct. - Ivan Neretin, Apr 18 2015

Comments from N. J. A. Sloane, Mar 29 2025 (Start):

Let d(n) = number of 1 <= i <= n such that a(i) < i. The d(i) sequence begins 0, 0, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6, ..., and appears to be A060144 without its initial term.

Let r(n) = 1 if a(n) < a(n+1), otherwise 0, and let f(n) = 1 if a(n) > a(n+1), otherwise 0. Then R = partial sums of r(n) and F = partial sums of f(n) count the rises and falls, respectively, in the present sequence. It appears that R and F are essentially A060143 and A060144 (again).

If a(n) is the k-th term in a monotonically strictly increasing rum of terms, set R(n) = k. It appears that the sequence R(n), n>=1, is essentially A270788.

For other sequences derived from the present one, see A382162, A382168, and A382169.

(End)

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