%I #20 Mar 06 2024 04:44:30

%S 1,1,1,2,1,3,1,4,2,5,1,6,1,7,4,8,1,9,1,10,5,11,1,12,2,13,7,14,1,15,1,

%T 16,8,17,3,18,1,19,10,20,1,21,1,22,11,23,1,24,2,25,13,26,1,27,6,28,14,

%U 29,1,30,1,31,16,32,7,33,1,34,17,35,1,36,1,37,19

%N Lexicographically earliest sequence of positive integers such that all equal terms appear at mutually coprime indices.

%C See A279119 for the same sequence with numbers including 0.

%C See A055396 for a similar sequence where all equal terms share a factor > 1.

%H Michael S. Branicky, <a href="/A370822/b370822.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = 1 + A279119(n). - _Rémy Sigrist_, Mar 04 2024

%e a(4)=2 because if we had a(4)=1, then i=2 and i=4, which are not coprime indices, would have the same value 1. So a(4)=2, which is a first occurrence.

%e a(9)=2 because if we had a(9)=1, i=3 and i=9, would have the same value despite not being coprime indices. a(9) can be 2 because the only other index with a 2 is a(4)=2 and 4 is coprime to 9.

%e a(15)=4 because 4 is the smallest value such that every previous index at which a 4 occurs is coprime to i=15. In this case, 4 has only occurred at i=8 and 8 is coprime to 15.

%o (Python)

%o from math import gcd, lcm

%o from itertools import combinations as C, count, islice

%o def agen(): # generator of terms

%o yield from [1, 1, 1]

%o lcms = {1: 6}

%o for n in count(4):

%o an = next(an for an in count(1) if an not in lcms or gcd(lcms[an], n) == 1)

%o yield an

%o if an not in lcms: lcms[an] = n

%o else: lcms[an] = lcm(lcms[an], n)

%o print(list(islice(agen(), 75))) # _Michael S. Branicky_, Mar 02 2024

%Y Cf. A100480, A279119, A370408.

%K nonn

%O 1,4

%A _Neal Gersh Tolunsky_, Mar 02 2024

%E a(22) and beyond from _Michael S. Branicky_, Mar 02 2024