A370822 Lexicographically earliest sequence of positive integers such that all equal terms appear at mutually coprime indices.
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, 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, 29, 1, 30, 1, 31, 16, 32, 7, 33, 1, 34, 17, 35, 1, 36, 1, 37, 19
Offset: 1
Keywords
Examples
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. 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. 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.
Links
- Michael S. Branicky, Table of n, a(n) for n = 1..10000
Programs
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Python
from math import gcd, lcm from itertools import combinations as C, count, islice def agen(): # generator of terms yield from [1, 1, 1] lcms = {1: 6} for n in count(4): an = next(an for an in count(1) if an not in lcms or gcd(lcms[an], n) == 1) yield an if an not in lcms: lcms[an] = n else: lcms[an] = lcm(lcms[an], n) print(list(islice(agen(), 75))) # Michael S. Branicky, Mar 02 2024
Formula
a(n) = 1 + A279119(n). - Rémy Sigrist, Mar 04 2024
Extensions
a(22) and beyond from Michael S. Branicky, Mar 02 2024
Comments