A026272 a(n) = smallest k such that k=a(n-k-1) is the only appearance of k so far; if there is no such k, then a(n) = least positive integer that has not yet appeared.

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

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• easy
• nice

Comments

From Daniel Joyce, Apr 13 2001: (Start)

This sequence displays every positive integer exactly twice and the gap between the two occurrences of n contains exactly n other values. The first occurrence of n precedes the first occurrence of n+1.

Also related to the Wythoff array (A035513) and the Para-Fibonacci sequence (A035513) where every positive integer is displayed exactly once in the whole array. Take any integer n in A026272 and let C = number of terms from the beginning of the sequence to the second occurrence of n. Then C = (2nd term after n in the applicable sequence for n in A035513).

Also in the second occurrence of n in A026272, let N=n ( - one term) = (first term value after n in the applicable sequence for n in A035513). In this format the second occurrence of n in A026272 will produce in A035513, n itself and two of the succeeding terms of n in the Wythoff array where every positive integer can only be displayed once.

In A026272 if |a(n)-a(n+1)| > 10 then phi ~ a(n)/|a(n)-a(n+1)|. When n -> infinity it will converge to phi. (End)

Or, put a copy of n in A000027 n places further along! - Zak Seidov, May 24 2008

Another version would prefix this sequence with two leading 0's (see the Angelini reference). If we use this form and write down the indices of the two 0's, the two 1's, the two 2's, the two 3's, etc., then we get A072061. - Jacques ALARDET, Jul 26 2008

References

• Eric Angelini, "Jeux de suites", in Dossier Pour La Science, pp. 32-35, Volume 59 (Jeux math'), April/June 2008, Paris.

Links

• Zak Seidov, Table of n, a(n) for n = 1..1000
• 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.
• Jeffrey Shallit, The Hurt-Sada Array and Zeckendorf Representations, arXiv:2501.08823 [math.NT], 2025. See p. 10.

Crossrefs

Cf. A000027, A035513, A014552, A176127.

Programs

Mathematica

s=Range[1000];n=0;Do[n++;s=Insert[s,n,Position[s,n][[1]]+n+1],{500}];A026272=Take[s,1000] (* Zak Seidov, May 24 2008 *)

PARI

A026272=apply(t->t-1,A026242[3..-1]) \\ Use vecextract(A026242,"3..") in PARI versions < 2.7. - M. F. Hasler, Sep 17 2014

Python

from collections import Counter
from itertools import count, islice
def agen(): # generator of terms
    aset, alst, k, mink, counts = set(), [0], 0, 1, Counter()
    for n in count(1):
        for k in range(1, len(alst)-1):
            if k == alst[n-k-1] and counts[alst[n-k-1]] == 1: an = k; break
        else: an = mink
        yield an; aset.add(an); alst.append(an); counts.update([an])
        while mink in aset: mink += 1
print(list(islice(agen(), 66))) # Michael S. Branicky, Jun 27 2022

Formula

a(n) = A026242(n+2) - 1 = A026350(n+3) - 2 = A026354(n+4) - 3.

Extensions

Edited by Max Alekseyev, May 31 2011